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Chemical Thermodynamics — JEE Previous Year Questions

Every Chemical Thermodynamics question asked in JEE Main and JEE Advanced across the last 186 papers — 195 questions, each with its correct answer. Free to read, no account needed.

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195

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165/186

Appearance rate

89%

All 195 Chemical Thermodynamics questions

Most recent papers first.

Q1·ChemistrySingle correctJEE Advanced 2026
List-I contains various physical/chemical processes, and List-II contains combinations of changes in enthalpy (ΔH\Delta HΔH) and entropy (ΔS\Delta SΔS). Match each entry in List-I to the appropriate entry in List-II, and choose the correct option.
List-IList-II
P.Physisorption1.ΔH>0\Delta H > 0ΔH>0 and ΔS>0\Delta S > 0ΔS>0
Q.Diamond ⟶ Graphite2.ΔH<0\Delta H < 0ΔH<0 and ΔS<0\Delta S < 0ΔS<0
R.Denaturation of protein3.ΔH<0\Delta H < 0ΔH<0 and ΔS=0\Delta S = 0ΔS=0
S.Propene ⟶ Cyclopropane4.ΔH>0\Delta H > 0ΔH>0 and ΔS<0\Delta S < 0ΔS<0
5.ΔH<0\Delta H < 0ΔH<0 and ΔS>0\Delta S > 0ΔS>0
  1. (A)P → 2; Q → 3; R → 5; S → 4
  2. (B)P → 4; Q → 3; R → 5; S → 1
  3. (C)P → 2; Q → 5; R → 1; S → 4
  4. (D)P → 2; Q → 5; R → 1; S → 3

Correct answer: (C)

Step-by-step solution →
Q2·ChemistrySingle correctJEE Advanced 2026
An ideal gas (0.5 mol), initially at 2 bar pressure, is compressed at a constant temperature of 600 K in two steps: first, against a constant external pressure of PPP bar (2<P<82 < P < 82<P<8), and then against constant external pressure of 8 bar. At each step, the compression is stopped only when the pressure of the gas becomes equal to the external pressure. The total work done on the gas in these steps is WWW. Considering all possible values of PPP (2<P<82 < P < 82<P<8) and taking the gas constant as RRR (in J K−1 mol−1\mathrm{J\,K^{-1}\,mol^{-1}}JK−1mol−1), the minimum value of ∣W∣|W|∣W∣ (in J) is
  1. (A)207R207R207R
  2. (B)600R600R600R
  3. (C)630R630R630R
  4. (D)900R900R900R

Correct answer: (B)

Step-by-step solution →
Q3·ChemistryNumericalJEE Main 2026
Consider the reaction 2H2S(g)+3O2(g)→2H2O(l)+2SO2(g)2H_2S(g) + 3O_2(g) \rightarrow 2H_2O(l) + 2SO_2(g)2H2​S(g)+3O2​(g)→2H2​O(l)+2SO2​(g) The magnitude of enthalpy change for the reaction in kJ mol−1^{-1}−1 is __________. (Nearest integer) Given: ΔfH⊖(H2S)=−20.1\Delta_f H^{\ominus}(H_2S) = -20.1Δf​H⊖(H2​S)=−20.1 kJ mol−1^{-1}−1 ΔfH⊖(H2O)=−286.0\Delta_f H^{\ominus}(H_2O) = -286.0Δf​H⊖(H2​O)=−286.0 kJ mol−1^{-1}−1 ΔfH⊖(SO2)=−297.0\Delta_f H^{\ominus}(SO_2) = -297.0Δf​H⊖(SO2​)=−297.0 kJ mol−1^{-1}−1

Correct answer: 1126

Step-by-step solution →
Q4·ChemistrySingle correctJEE Main 2026
Match List - I (Isothermal process) with List - II (Expression). Given V1V_{1}V1​ and V2V_{2}V2​ are initial and final volumes respectively. Choose the correct answer from the options given below :
List - I (Isothermal process)List - II (Expression)
A.Reversible expansionI.q=0q = 0q=0
B.Free expansionII.q=nRTln⁡V2V1q = nRT \ln \frac{V_{2}}{V_{1}}q=nRTlnV1​V2​​
C.Irreversible CompressionIII.w=−pext(V1−V2)w = -p_{ext}(V_{1} - V_{2})w=−pext​(V1​−V2​)
D.Cyclic reversibleIV.qrevT=0\frac{q_{rev}}{T} = 0Tqrev​​=0
  1. (A)A-II, B-III, C-I, D-IV
  2. (B)A-II, B-I, C-IV, D-III
  3. (C)A-II, B-I, C-III, D-IV
  4. (D)A-I, B-II, C-III, D-IV

Correct answer: (C)

Step-by-step solution →
Q5·ChemistrySingle correctJEE Main 2026
Arrange the following isothermal processes in order of the magnitude of the work (p−V)(p - V)(p−V) involved between states 1 and 2. A. Expansion in single stage wAw_AwA​ B. Expansion in multi stages wBw_BwB​ C. Compression in single stage wCw_CwC​ D. Compression in multi stages wDw_DwD​ Choose the correct option.
  1. (A)∣wB∣>∣wA∣>∣wC∣>∣wD∣|w_B| > |w_A| > |w_C| > |w_D|∣wB​∣>∣wA​∣>∣wC​∣>∣wD​∣
  2. (B)∣wC∣>∣wD∣>∣wA∣>∣wB∣|w_C| > |w_D| > |w_A| > |w_B|∣wC​∣>∣wD​∣>∣wA​∣>∣wB​∣
  3. (C)∣wC∣>∣wD∣>∣wB∣>∣wA∣|w_C| > |w_D| > |w_B| > |w_A|∣wC​∣>∣wD​∣>∣wB​∣>∣wA​∣
  4. (D)∣wB∣>∣wA∣>∣wD∣>∣wC∣|w_B| > |w_A| > |w_D| > |w_C|∣wB​∣>∣wA​∣>∣wD​∣>∣wC​∣

Correct answer: (C)

Step-by-step solution →
Q6·ChemistryNumericalJEE Main 2026
Consider the reaction XXX ⇌ YYY at 300 K. If ΔH° and KKK are 28.40 kJ mol⁻¹ and 1.8 × 10⁻⁷ at the same temperature, then the magnitude of ΔS° for the reaction in J K⁻¹ mol⁻¹ is _______. (Nearest integer) (Given: R = 8.3 J K⁻¹ mol⁻¹, ln 10 = 2.3, log 3 = 0.48, log 2 = 0.30)

Correct answer: 34

Step-by-step solution →
Q7·ChemistrySingle correctJEE Main 2026
The correct order of molar heat capacities measured at 298 K and 1 bar is :
  1. (A)Copper(s) > Bromine(l) > Helium(g)
  2. (B)Bromine(l) > Copper(s) > Helium(g)
  3. (C)Helium(g) > Bromine(l) > Copper(s)
  4. (D)Helium(g) > Bromine(l) = Copper(s)

Correct answer: (B)

Step-by-step solution →
Q8·ChemistrySingle correctJEE Main 2026
Consider the following data for the reaction X₂(g) + Y₂(g) ⇌ 2XY(g) at 600 K. The ΔrG° (in kJ mol⁻¹) for the reaction is :
CompoundΔfH°(600K) (kJ mol⁻¹)S°(600K) (J mol⁻¹ K⁻¹)
XY(g)42200
X₂(g)8140
Y₂(g)80250
  1. (A)−21000
  2. (B)−10
  3. (C)−1000
  4. (D)−9.012

Correct answer: (B)

Step-by-step solution →
Q9·ChemistrySingle correctJEE Main 2026
Given below are two statements: Statement I: For an ideal gas, heat capacity at constant volume is always greater than the heat capacity at constant pressure. Statement II: In a constant volume process, no work is produced and all the heat withdrawn goes into the chaotic motion and is reflected by a temperature increase of the ideal gas. In the light of the above statements, choose the correct answer from the options given below
  1. (A)Both Statement I and Statement II are true
  2. (B)Both Statement I and Statement II are false
  3. (C)Statement I is true but Statement II is false
  4. (D)Statement I is false but Statement II is true

Correct answer: (D)

Step-by-step solution →
Q10·ChemistryNumericalJEE Main 2026
If 3.365 g of ethanol (l) is burnt completely in a bomb calorimeter at 298.15 K, the heat produced is 99.472 kJ. The ∣ΔHf∘∣|\Delta H_{f}^{\circ}|∣ΔHf∘​∣ of ethanol at 298.15 K is ______ ×102^{2}2 kJ mol−1^{-1}−1. (Nearest integer) Given: Standard enthalpy for combustion of graphite = −393.5 kJ mol−1^{-1}−1 Standard enthalpy of formation of water (l) = −285.8 kJ mol−1^{-1}−1 Molar mass in g mol−1^{-1}−1 of C, H, O are 12, 1 and 16 respectively

Correct answer: 3

Step-by-step solution →
Q11·ChemistrySingle correctJEE Main 2026
Gas 'A' undergoes change from state 'X' to state 'Y'. In this process, the heat absorbed and work done by the gas is 10 J and 18 J respectively. Now gas is brought back to state 'X' by another process during which 6 J of heat is evolved. In the reverse process of 'Y' to 'X',
  1. (A)18 J of the work is done by the gas 'A'.
  2. (B)2 J of the work is done by the gas 'A'.
  3. (C)12 J of the work is done on the gas 'A' by the surrounding.
  4. (D)14 J of the work is done on the gas 'A' by the surrounding.

Correct answer: (D)

Step-by-step solution →
Q12·ChemistryNumericalJEE Main 2026
At the transition temperature TTT, A⇌BA\rightleftharpoons BA⇌B and ΔG0=105−35log⁡T\Delta G^{0}=105-35\log TΔG0=105−35logT where AAA and BBB are two states of substance XXX. The transition temperature in °C when pressure is 1 atm is ________. (Nearest integer)

Correct answer: 727

Step-by-step solution →
Q13·ChemistrySingle correctJEE Main 2026
Consider the following data. (i) 2Al(s)+6HCl(aq)→Al2Cl6(aq)+3H2(g)+12002\mathrm{Al}(s) + 6\mathrm{HCl}(aq) \rightarrow \mathrm{Al_{2}Cl_{6}}(aq) + 3\mathrm{H_{2}}(g) + 12002Al(s)+6HCl(aq)→Al2​Cl6​(aq)+3H2​(g)+1200 kJ/mol (ii) H2(g)+Cl2(g)→2HCl(g)+164\mathrm{H_{2}}(g) + \mathrm{Cl_{2}}(g) \rightarrow 2\mathrm{HCl}(g) + 164H2​(g)+Cl2​(g)→2HCl(g)+164 kJ/mol (iii) HCl(g)+aq→HCl(aq)+83\mathrm{HCl}(g) + aq \rightarrow \mathrm{HCl}(aq) + 83HCl(g)+aq→HCl(aq)+83 kJ/mol (iv) Al2Cl6(s)+aq→Al2Cl6(aq)+663\mathrm{Al_{2}Cl_{6}}(s) + aq \rightarrow \mathrm{Al_{2}Cl_{6}}(aq) + 663Al2​Cl6​(s)+aq→Al2​Cl6​(aq)+663 kJ/mol The enthalpy of formation of anhydrous solid Al2Cl6\mathrm{Al_{2}Cl_{6}}Al2​Cl6​ is :
  1. (A)−648 kJ mol−1\mathrm{mol^{-1}}mol−1
  2. (B)−1350 kJ mol−1\mathrm{mol^{-1}}mol−1
  3. (C)−2002 kJ mol−1\mathrm{mol^{-1}}mol−1
  4. (D)−1527 kJ mol−1\mathrm{mol^{-1}}mol−1

Correct answer: (D)

Step-by-step solution →
Q14·ChemistrySingle correctJEE Main 2026
20.0 dm3^{3}3 of an ideal gas 'X' at 600 K and 0.5 MPa undergoes isothermal reversible expansion until pressure of the gas is 0.2 MPa. Which of the following option is correct ? (Given: log 2 = 0.3010 and log 5 = 0.6989)
  1. (A)w = – 9.1 kJ, ΔU = 0, ΔH = 0, q = 9.1 kJ
  2. (B)w = 9.1 J, ΔU = 9.1 J, ΔH = 0; q = 0
  3. (C)w = + 4.1 kJ, ΔU = 0, ΔH = 0; q = – 4.1 kJ
  4. (D)w = – 3.9 kJ, ΔU = 0, ΔH = 0; q = 3.9 kJ

Correct answer: (A)

Step-by-step solution →
Q15·ChemistrySingle correctJEE Main 2026
The plot of log⁡10K\log_{10}Klog10​K vs 1T\frac{1}{T}T1​ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant).
  1. (A)2.303RΔH∘\frac{2.303R}{\Delta H^\circ}ΔH∘2.303R​ , 2.303RΔS∘\frac{2.303R}{\Delta S^\circ}ΔS∘2.303R​
  2. (B)ΔS∘2.303R\frac{\Delta S^\circ}{2.303R}2.303RΔS∘​ , −ΔH∘2.303R-\frac{\Delta H^\circ}{2.303R}−2.303RΔH∘​
  3. (C)−ΔS∘R2.303-\frac{\Delta S^\circ R}{2.303}−2.303ΔS∘R​ , ΔH∘R2.303\frac{\Delta H^\circ R}{2.303}2.303ΔH∘R​
  4. (D)−ΔH∘2.303R-\frac{\Delta H^\circ}{2.303R}−2.303RΔH∘​ , ΔS∘2.303R\frac{\Delta S^\circ}{2.303R}2.303RΔS∘​

Correct answer: (B)

Step-by-step solution →
Q16·ChemistrySingle correctJEE Main 2026
The heat of atomisation of methane and ethane are 'x' kJ mol−1\text{kJ mol}^{-1}kJ mol−1 and 'y' kJ mol−1\text{kJ mol}^{-1}kJ mol−1 respectively. The longest wavelength (λ\lambdaλ) of light capable of breaking the C-C bond can be expressed in SI unit as :
  1. (A)hc1000(y−6x4)−1\frac{\text{hc}}{1000}\left(\frac{\text{y}-6\text{x}}{4}\right)^{-1}1000hc​(4y−6x​)−1
  2. (B)NAhc250(4y−6x)\frac{\text{N}_{\text{A}}\text{hc}}{250(4\text{y}-6\text{x})}250(4y−6x)NA​hc​
  3. (C)NAhc250(y−6x)\frac{\text{N}_{\text{A}}\text{hc}}{250(\text{y}-6\text{x})}250(y−6x)NA​hc​
  4. (D)NAhc(y−6x4)−1\text{N}_{\text{A}}\text{hc}\left(\text{y}-\frac{6\text{x}}{4}\right)^{-1}NA​hc(y−46x​)−1

Correct answer: (B)

Step-by-step solution →
Q17·ChemistrySingle correctJEE Main 2026
Match the List-I (Isothermal process for ideal gas system) with List-II Work done (Vf>ViV_f > V_iVf​>Vi​) Choose the correct answer from the options given below :
List-I (Isothermal process for ideal gas system)List-II Work done ($V_f > V_i$)
A.Reversible expansionI.w=0w = 0w=0
B.Free expansionII.w=−nRTln⁡VfViw = -nRT \ln \frac{V_f}{V_i}w=−nRTlnVi​Vf​​
C.Irreversible expansionIII.w=−pex(Vf−Vi)w = -p_{ex} (V_f - V_i)w=−pex​(Vf​−Vi​)
D.Irreversible compressionIV.w=−pex(Vi−Vf)w = -p_{ex} (V_i - V_f)w=−pex​(Vi​−Vf​)
  1. (A)A-IV, B-I, C-III, D-II
  2. (B)A-IV, B-II, C-III, D-I
  3. (C)A-I, B-III, C-II, D-IV
  4. (D)A-II, B-I, C-III, D-IV

Correct answer: (D)

Step-by-step solution →
Q18·ChemistrySingle correctJEE Main 2026
A cup of water at 5∘^\circ∘C (system) is placed in a microwave oven and the oven is turned on for one minute during which, the water begins to boil. Which of the following option is true ?
  1. (A)q = +ve, w = 0, Δ\DeltaΔU = -ve
  2. (B)q = +ve, w = -ve, Δ\DeltaΔU = +ve
  3. (C)q = -ve, w = -ve, Δ\DeltaΔU = -ve
  4. (D)q = +ve, w = -ve, Δ\DeltaΔU = -ve

Correct answer: (B)

Step-by-step solution →
Q19·ChemistrySingle correctJEE Main 2026
Match the LIST-I (Thermodynamic Process) with LIST-II (Magnitude in kJ) Choose the correct answer from the option given below :
List-I (Thermodynamic Process)List-II (Magnitude in kJ)
A.Work done in reversible, isothermal expansion of 2 mol of ideal gas from 2 dm3^33 to 20 dm3^33 at 300 K.I.4
B.Work done in irreversible isothermal expansion of 1 mol ideal gas from 1 m3^33 to 3 m3^33 at 300 K against A constant pressure of 3kPa.II.11.5
C.Change in internal energy for adiabatic expansion of a 1 mol ideal gas with change of temperature = 320 K and C‾V=32R\overline{C}_V = \frac{3}{2}RCV​=23​R.III.6
D.Change in enthalpy at constant pressure of 1 mole ideal gas with change of temperature = 337 K and C‾P=52R\overline{C}_P = \frac{5}{2}RCP​=25​R.IV.7
  1. (A)A-III, B-II, C-IV, D-I
  2. (B)A-II, B-III, C-I, D-IV
  3. (C)A-I, B-II, C-III, D-IV
  4. (D)A-II, B-I, C-III, D-IV

Correct answer: (B)

Step-by-step solution →
Q20·ChemistryNumericalJEE Main 2026
If the enthalpy of sublimation of Li is 155 kJ mol−1mol^{-1}mol−1, enthalpy of dissociation of F2F_{2}F2​ is 150 kJ mol−1mol^{-1}mol−1, ionization enthalpy of Li is 520 kJ mol−1mol^{-1}mol−1, electron gain enthalpy of F is −313-313−313 kJ mol−1mol^{-1}mol−1, standard enthalpy of formation of LiF is −594-594−594 kJ mol−1mol^{-1}mol−1. The magnitude of lattice enthalpy of LiF is _________ kJ mol−1mol^{-1}mol−1 (Nearest integer).

Correct answer: 1031

Step-by-step solution →
Q21·ChemistrySingle correctJEE Main 2026
Which of the following graphs between pressure ‘P’ versus volume ‘V’ represent the maximum work done ?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (D)

Step-by-step solution →
Q22·ChemistrySingle correctJEE Main 2026
For the reaction, N2_22​O4_44​ ⇌ 2NO2_22​, graph is plotted as shown below. Identify correct statements. A. Standard free energy change for the reaction is –5.40 kJ mol−1^{-1}−1. B. As AG⊖^{\ominus}⊖ in graph is positive, N2_22​O4_44​ will not dissociate into NO2_22​ at all. C. Reverse reaction will go to completion. D. When 1 mole of N2_22​O4_44​ changes into equilibrium mixture, value of ΔG⊖^{\ominus}⊖ = –0.84 kJ mol−1^{-1}−1 E. When 2 mole of NO2_22​, changes into equilibrium mixture, ΔG⊖^{\ominus}⊖ for equilibrium mixture is –6.24 kJ mol−1^{-1}−1. E. When 2 mole of NO2_22​, changes into equilibrium mixture, ΔG⊖^{\ominus}⊖ for equilibrium mixture is –6.24 kJ mol−1^{-1}−1. Choose the correct answer from the options given below :
  1. (A)D and E only
  2. (B)C and E only
  3. (C)A and D only
  4. (D)B and C only

Correct answer: (A)

Step-by-step solution →
Q23·ChemistrySingle correctJEE Main 2026
Consider the following data: ΔfH⊖\Delta_{\text{f}}\text{H}^{\ominus}Δf​H⊖(methane, g) = −X kJ mol−1-\text{X kJ mol}^{-1}−X kJ mol−1 Enthalpy of sublimation of graphite = Y kJ mol−1\text{Y kJ mol}^{-1}Y kJ mol−1 Dissociation enthalpy of H2\text{H}_{2}H2​ = Z kJ mol−1\text{Z kJ mol}^{-1}Z kJ mol−1 The bond enthalpy of C −-− H bond is given by:
  1. (A)X+Y+2Z4\frac{\text{X} + \text{Y} + 2\text{Z}}{4}4X+Y+2Z​
  2. (B)X+Y+4Z2\frac{\text{X} + \text{Y} + 4\text{Z}}{2}2X+Y+4Z​
  3. (C)X+Y+Z\text{X} + \text{Y} + \text{Z}X+Y+Z
  4. (D)−X+Y+Z4\frac{-\text{X} + \text{Y} + \text{Z}}{4}4−X+Y+Z​

Correct answer: (A)

Step-by-step solution →
Q24·ChemistryNumericalJEE Main 2026
Use the following data : One mole each of A2_22​(g) and B2_22​(g) are taken in a 1L closed flask and allowed to establish the equilibrium at 500K. A2_22​(g) + B2_22​(g) ⇌ 2AB(g) The value of x (in kJ mol−1^{-1}−1) is ……….. (Nearest integer) (Given: log K=2.2 R=8.3 JK−1^{-1}−1 mol−1^{-1}−1)
SubstanceΔf\Delta_fΔf​H⊖^{\ominus}⊖(500K) / kJ mol−1^{-1}−1S⊖^{\ominus}⊖(500K) / J K−1^{-1}−1 mol−1^{-1}−1
AB(g)32222
A2_22​(g)6146
B2_22​(g)X280

Correct answer: 70

Step-by-step solution →
Q25·ChemistryNumericalJEE Advanced 2025
Considering ideal gas behavior, the expansion work done (in kJ) when 144 g of water is electrolyzed completely under constant pressure at 300 K is ______. Use: Universal gas constant (R) = 8.3 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1; Atomic mass (in amu) : H = 1, O = 16

Correct answer: 29.88

Step-by-step solution →
Q26·ChemistrySingle correctJEE Main 2025
The correct statement amongst the following is:
  1. (A)The term 'standard state' implies that the temperature is 0∘0^\circ0∘C
  2. (B)The standard state of pure gas is the pure gas at a pressure of 1 bar and temperature 273 K
  3. (C)ΔfH298⊖\Delta_f H^{\ominus}_{298}Δf​H298⊖​ is zero for O(g)
  4. (D)ΔfH500⊖\Delta_f H^{\ominus}_{500}Δf​H500⊖​ is zero for O2O_2O2​(g)

Correct answer: (D)

Step-by-step solution →
Q27·ChemistrySingle correctJEE Main 2025
The hydration energies of K+K^+K+ and Cl−Cl^-Cl− are −x-x−x and −y-y−y kJ/mol respectively. If lattice energy of KCl is −z-z−z kJ/mol, then the heat of solution of KCl is:
  1. (A)+x−y−z+x-y-z+x−y−z
  2. (B)x+y+zx+y+zx+y+z
  3. (C)z−(x+y)z-(x+y)z−(x+y)
  4. (D)−z−(x+y)-z-(x+y)−z−(x+y)

Correct answer: (C)

Step-by-step solution →
Q28·ChemistrySingle correctJEE Main 2025
Total enthalpy change for freezing of 1 mol of water at 10∘10^\circ10∘C to ice at −10∘-10^\circ−10∘C is ______. (Given: ΔfusH=x\Delta_{fus}H=xΔfus​H=x kJ/mol, Cp[H2O(l)]=yC_p[H_2O(l)]=yCp​[H2​O(l)]=y J mol−1^{-1}−1 K−1^{-1}−1, Cp[H2O(s)]=zC_p[H_2O(s)]=zCp​[H2​O(s)]=z J mol−1^{-1}−1 K−1^{-1}−1)
  1. (A)−x−10y−10z-x-10y-10z−x−10y−10z
  2. (B)−10(100x+y+z)-10(100x+y+z)−10(100x+y+z)
  3. (C)10(100x+y+z)10(100x+y+z)10(100x+y+z)
  4. (D)x−10y−10zx-10y-10zx−10y−10z

Correct answer: (B)

Step-by-step solution →
Q29·ChemistrySingle correctJEE Main 2025
Consider the given data: (a) HCl(g)+10H2O(l)→HCl.10H2OHCl(g) + 10H_2O(l) \rightarrow HCl.10H_2OHCl(g)+10H2​O(l)→HCl.10H2​O, ΔH=−69.01 kJ mol−1\Delta H = -69.01\ kJ\,mol^{-1}ΔH=−69.01 kJmol−1; (b) HCl(g)+40H2O(l)→HCl.40H2OHCl(g) + 40H_2O(l) \rightarrow HCl.40H_2OHCl(g)+40H2​O(l)→HCl.40H2​O, ΔH=−72.79 kJ mol−1\Delta H = -72.79\ kJ\,mol^{-1}ΔH=−72.79 kJmol−1. Choose the correct statement:
  1. (A)Dissolution of gas in water is an endothermic process
  2. (B)The heat of solution depends on the amount of solvent.
  3. (C)The heat of dilution for the HCl.10H2OHCl.10H_2OHCl.10H2​O to HCl.40H2OHCl.40H_2OHCl.40H2​O is 3.78 kJ mol−13.78\ kJ\,mol^{-1}3.78 kJmol−1.
  4. (D)The heat of formation of HCl solution is represented by both (a) and (b)

Correct answer: (B)

Step-by-step solution →
Q30·ChemistrySingle correctJEE Main 2025
Let us consider a reversible reaction at temperature, T. In this reaction, both ΔH\Delta HΔH and ΔS\Delta SΔS were observed to have positive values. If the equilibrium temperature is TeT_eTe​, then the reaction becomes spontaneous at:
  1. (A)T=TeT=T_eT=Te​
  2. (B)Te>TT_e>TTe​>T
  3. (C)T>TeT>T_eT>Te​
  4. (D)Te=5TT_e=5TTe​=5T

Correct answer: (C)

Step-by-step solution →
Q31·ChemistrySingle correctJEE Main 2025
One mole of an ideal gas expands isothermally and reversibly from 10 dm3^33 to 20 dm3^33 at 300 K. ΔU\Delta UΔU, qqq and work done in the process respectively are: (Given: R=8.3R=8.3R=8.3 JK−1^{-1}−1 mol−1^{-1}−1, ln⁡10=2.3\ln 10=2.3ln10=2.3, log⁡2=0.30\log 2=0.30log2=0.30, log⁡3=0.48\log 3=0.48log3=0.48)
  1. (A)0, 21.84 kJ, −1.26-1.26−1.26 kJ
  2. (B)0, −17.18-17.18−17.18 kJ, 1.718 J
  3. (C)0, 21.84 kJ, 21.84 kJ
  4. (D)0, 1.718 kJ, −1.718-1.718−1.718 kJ

Correct answer: (D)

Step-by-step solution →
Q32·ChemistryIntegerJEE Main 2025
A sample of n-octane (1.14 g) was completely burnt in excess of oxygen in a bomb calorimeter, whose heat capacity is 5 kJ K−1^{-1}−1. As a result of combustion reaction, the temperature of the calorimeter is increased by 5 K. The magnitude of the heat of combustion of octane at constant volume is __________ kJ mol−1^{-1}−1 (nearest integer).

Correct answer: 2500

Step-by-step solution →
Q33·ChemistryIntegerJEE Main 2025
A perfect gas (0.1 mol) having Cv‾=1.50 R\overline{C_v}=1.50\,RCv​​=1.50R (independent of temperature) undergoes the transformation shown in the following P-V diagram from point 1 to point 4. If each step is reversible, the total work done (w) while going from point 1 to point 4 is (−)(-)(−) __________ J (nearest integer). [Given: R = 0.082 L atm K−1^{-1}−1 mol−1^{-1}−1]

Correct answer: 304

Step-by-step solution →
Q34·ChemistryIntegerJEE Main 2025
Given: ΔHsub∘[C(graphite)]=710\Delta H^\circ_{sub}[C(\text{graphite})]=710ΔHsub∘​[C(graphite)]=710 kJ mol−1^{-1}−1, ΔC-HH∘=414\Delta_{C\text{-}H}H^\circ=414ΔC-H​H∘=414 kJ mol−1^{-1}−1, ΔH-HH∘=436\Delta_{H\text{-}H}H^\circ=436ΔH-H​H∘=436 kJ mol−1^{-1}−1, ΔC=CH∘=611\Delta_{C\text{=}C}H^\circ=611ΔC=C​H∘=611 kJ mol−1^{-1}−1. The ΔHf∘\Delta H^\circ_fΔHf∘​ for CH2=CH2CH_2{=}CH_2CH2​=CH2​ is __________ kJ mol−1^{-1}−1. (nearest integer value)

Correct answer: 25

Step-by-step solution →
Q35·ChemistrySingle correctJEE Main 2025
Given below are two statements: Statement I: When a system containing ice in equilibrium with water (liquid) is heated, heat is absorbed by the system and there is no change in the temperature of the system until whole ice gets melted. Statement II: At melting point of ice, there is absorption of heat in order to overcome intermolecular forces of attraction within the molecules of water in ice and kinetic energy of molecules is not increased at melting point. In the light of the above statements, choose the correct answer from the options given below:
  1. (A)Statement I is true but Statement II is false
  2. (B)Both Statement I and Statement II are false
  3. (C)Both Statement I and Statement II are true
  4. (D)Statement I is false but Statement II is true

Correct answer: (C)

Step-by-step solution →
Q36·ChemistrySingle correctJEE Main 2025
Which of the following graphs correctly represents the variation of thermodynamic properties of Haber's process?
  1. (A)(1)
  2. (B)(2)
  3. (C)(3)
  4. (D)(4)

Correct answer: (A)

Step-by-step solution →
Q37·ChemistrySingle correctJEE Main 2025
Arrange the following in order of magnitude of work done by the system / on the system at constant temperature: (a) ∣Wreversible∣|W_{reversible}|∣Wreversible​∣ for expansion in infinite stage. (b) ∣Wirreversible∣|W_{irreversible}|∣Wirreversible​∣ for expansion in single stage. (c) ∣Wreversible∣|W_{reversible}|∣Wreversible​∣ for compression in infinite stage. (d) ∣Wirreversible∣|W_{irreversible}|∣Wirreversible​∣ for compression in single stage. Choose the correct answer from the options given below:
  1. (A)a>b>c>da>b>c>da>b>c>d
  2. (B)d>c>a>bd>c>a>bd>c>a>b
  3. (C)c>a>d>bc>a>d>bc>a>d>b
  4. (D)a>c>b>da>c>b>da>c>b>d

Correct answer: (B)

Step-by-step solution →
Q38·ChemistrySingle correctJEE Main 2025
Two vessels A and B are connected via a stopcock. The vessel A is filled with a gas at a certain pressure. The vessel B is empty and is allowed to come to thermal equilibrium with water and no change in temperature is observed in the thermometer. After opening the stopcock the gas from vessel A expands into vessel B and no change in temperature is observed. Which of the following statement is true?
  1. (A)dw≠0dw\ne0dw=0
  2. (B)dq≠0dq\ne0dq=0
  3. (C)dU≠0dU\ne0dU=0
  4. (D)The pressure in the vessel B before opening the stopcock is zero.

Correct answer: (D)

Step-by-step solution →
Q39·ChemistrySingle correctJEE Main 2025
If C(diamond)→C(graphite)+XC(\text{diamond})\rightarrow C(\text{graphite})+XC(diamond)→C(graphite)+X kJ mol⁻¹; C(diamond)+O2(g)→CO2(g)+YC(\text{diamond})+O_2(g)\rightarrow CO_2(g)+YC(diamond)+O2​(g)→CO2​(g)+Y kJ mol⁻¹; C(graphite)+O2(g)→CO2(g)+ZC(\text{graphite})+O_2(g)\rightarrow CO_2(g)+ZC(graphite)+O2​(g)→CO2​(g)+Z kJ mol⁻¹. At constant temperature, then
  1. (A)X = Y + Z
  2. (B)X = Y − Z
  3. (C)X = Z − Y
  4. (D)X = Y = Z

Correct answer: (B)

Step-by-step solution →
Q40·ChemistryIntegerJEE Main 2025
Consider the following data: Heat of formation of CO2_22​(g) = −393.5-393.5−393.5 kJ mol−1^{-1}−1, Heat of formation of H2_22​O(ℓ\ellℓ) = −286.0-286.0−286.0 kJ mol−1^{-1}−1, Heat of combustion of benzene = −3267.0-3267.0−3267.0 kJ mol−1^{-1}−1. The heat of formation of benzene is ______ kJ mol−1^{-1}−1. (Nearest integer)

Correct answer: 48

Step-by-step solution →
Q41·ChemistryIntegerJEE Main 2025
The formation enthalpies, ΔHf∘\Delta H^\circ_fΔHf∘​ for H(g)_{(g)}(g)​ and O(g)_{(g)}(g)​ are 220.0 and 250.0 kJ mol−1^{-1}−1, respectively, at the same temperature. The average bond enthalpy of the O-H bond in water at 298.15 K is _____ kJ mol−1^{-1}−1 (nearest integer).

Correct answer: 466

Step-by-step solution →
Q42·ChemistrySingle correctJEE Main 2025
An ideal gas undergoes a cyclic transformation starting from the point A and coming back to the same point by tracing the path A →\to→ B →\to→ C →\to→ D →\to→ A as shown in the three cases above. Choose the correct option regarding ΔU\Delta UΔU.
  1. (A)ΔU\Delta UΔU(Case-III) > ΔU\Delta UΔU(Case-II) > ΔU\Delta UΔU(Case-I)
  2. (B)ΔU\Delta UΔU(Case-I) > ΔU\Delta UΔU(Case-II) > ΔU\Delta UΔU(Case-III)
  3. (C)ΔU\Delta UΔU(Case-I) > ΔU\Delta UΔU(Case-III) > ΔU\Delta UΔU(Case-II)
  4. (D)ΔU\Delta UΔU(Case-I) = ΔU\Delta UΔU(Case-II) = ΔU\Delta UΔU(Case-III)

Correct answer: (D)

Step-by-step solution →
Q43·ChemistrySingle correctJEE Main 2025
S(g)+32O2(g)→SO3(g)+2xS(g)+\dfrac{3}{2}O_2(g)\to SO_3(g)+2xS(g)+23​O2​(g)→SO3​(g)+2x kcal; SO2(g)+12O2(g)→SO3(g)+ySO_2(g)+\dfrac{1}{2}O_2(g)\to SO_3(g)+ySO2​(g)+21​O2​(g)→SO3​(g)+y kcal. The heat of formation of SO2(g)SO_2(g)SO2​(g) is given by :
  1. (A)2xy\dfrac{2x}{y}y2x​ kcal
  2. (B)y−2xy-2xy−2x kcal
  3. (C)2x+y2x+y2x+y kcal
  4. (D)x+yx+yx+y kcal

Correct answer: (B)

Step-by-step solution →
Q44·ChemistryIntegerJEE Main 2025
Standard entropies of X₂, Y₂ and XY₅ are 70, 50 and 110 J K⁻¹ mol⁻¹ respectively. The temperature in Kelvin at which the reaction 12X2+52Y2→XY5\dfrac{1}{2}X_2+\dfrac{5}{2}Y_2\to XY_521​X2​+25​Y2​→XY5​, ΔH−=−35\Delta H^-=-35ΔH−=−35 kJ mol⁻¹, will be at equilibrium is _______ (Nearest integer)

Correct answer: 700

Step-by-step solution →
Q45·ChemistrySingle correctJEE Main 2025
Which of the following mixing of 1M base and 1M acid leads to the largest increase in temperature?
  1. (A)30 mL HCl and 30 mL NaOH
  2. (B)30 mL CH3COOHCH_3COOHCH3​COOH and 30 mL NaOH
  3. (C)50 mL HCl and 20 mL NaOH
  4. (D)45 mL CH3COOHCH_3COOHCH3​COOH and 25 mL NaOH

Correct answer: (A)

Step-by-step solution →
Q46·ChemistrySingle correctJEE Main 2025
Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction is spontaneous at boiling point of water. Choose the correct option.
  1. (A)Both ΔH and ΔS are (+ve)
  2. (B)ΔH is (−ve) but ΔS is (+ve)
  3. (C)ΔH is (+ve) but ΔS is (−ve)
  4. (D)Both ΔH and ΔS are (−ve)

Correct answer: (A)

Step-by-step solution →
Q47·ChemistryIntegerJEE Main 2025
The bond dissociation enthalpy of X2X_2X2​, ΔHbond∘\Delta H^\circ_{bond}ΔHbond∘​ calculated from the given data is __________ kJ mol−1^{-1}−1. (Nearest integer) Given: M+X−(s)→M+(g)+X−(g)M^+X^-(s)\to M^+(g)+X^-(g)M+X−(s)→M+(g)+X−(g), ΔHlattice∘=800\Delta H^\circ_{lattice}=800ΔHlattice∘​=800 kJ mol−1^{-1}−1; M(s)→M(g)M(s)\to M(g)M(s)→M(g), ΔHsub∘=100\Delta H^\circ_{sub}=100ΔHsub∘​=100 kJ mol−1^{-1}−1; M(g)→M+(g)+e−(g)M(g)\to M^+(g)+e^-(g)M(g)→M+(g)+e−(g), ΔHIE∘=500\Delta H^\circ_{IE}=500ΔHIE∘​=500 kJ mol−1^{-1}−1; X(g)+e−(g)→X−(g)X(g)+e^-(g)\to X^-(g)X(g)+e−(g)→X−(g), ΔHeg∘=−300\Delta H^\circ_{eg}=-300ΔHeg∘​=−300 kJ mol−1^{-1}−1; M(s)+12X2(g)→M+X−(s)M(s)+\tfrac12 X_2(g)\to M^+X^-(s)M(s)+21​X2​(g)→M+X−(s), ΔHf∘=−400\Delta H^\circ_f=-400ΔHf∘​=−400 kJ mol−1^{-1}−1. [M+X−M^+X^-M+X− is a pure ionic compound and X forms a diatomic molecule X2X_2X2​ in gaseous state.]

Correct answer: 200

Step-by-step solution →
Q48·ChemistrySingle correctJEE Main 2025
The effect of temperature on spontaneity of reactions are represented as: (A) ΔH=+, ΔS=−\Delta H=+,\ \Delta S=-ΔH=+, ΔS=−, any T →\to→ Non-spontaneous; (B) ΔH=+, ΔS=+\Delta H=+,\ \Delta S=+ΔH=+, ΔS=+, low T →\to→ spontaneous; (C) ΔH=−, ΔS=−\Delta H=-,\ \Delta S=-ΔH=−, ΔS=−, low T →\to→ Non-spontaneous; (D) ΔH=−, ΔS=+\Delta H=-,\ \Delta S=+ΔH=−, ΔS=+, any T →\to→ spontaneous.
  1. (A)(A), (B) and (D) only
  2. (B)(A) and (D) only
  3. (C)(B) and (C) only
  4. (D)(D) and (C) only

Correct answer: (C)

Step-by-step solution →
Q49·ChemistryIntegerJEE Main 2025
The standard enthalpy and standard entropy of decomposition of N₂O₄ to NO₂ are 55.0 kJ/mol and 175.0 J/K/mol respectively. The standard free energy change for this reaction at 25°C in J mol−1^{-1}−1 is _______ (Nearest integer)

Correct answer: 2850

Step-by-step solution →
Q50·ChemistrySingle correctJEE Main 2025
Ice at −5°C is heated to become vapor with temperature of 110°C at atmospheric pressure. The entropy change associated with this process can be obtained from :
  1. (A)∫268383Cp dT+ΔHmelting273+ΔHboiling373\int_{268}^{383}C_p\,dT+\dfrac{\Delta H_{melting}}{273}+\dfrac{\Delta H_{boiling}}{373}∫268383​Cp​dT+273ΔHmelting​​+373ΔHboiling​​
  2. (B)∫268273Cp,mTdT+ΔHm,fusionTf+∫273373Cp,mTdT+ΔHm,vaporisationTb+∫373383Cp,mTdT\int_{268}^{273}\dfrac{C_{p,m}}{T}dT+\dfrac{\Delta H_{m,fusion}}{T_f}+\int_{273}^{373}\dfrac{C_{p,m}}{T}dT+\dfrac{\Delta H_{m,vaporisation}}{T_b}+\int_{373}^{383}\dfrac{C_{p,m}}{T}dT∫268273​TCp,m​​dT+Tf​ΔHm,fusion​​+∫273373​TCp,m​​dT+Tb​ΔHm,vaporisation​​+∫373383​TCp,m​​dT
  3. (C)∫268383Cp dT+qrevT\int_{268}^{383}C_p\,dT+\dfrac{q_{rev}}{T}∫268383​Cp​dT+Tqrev​​
  4. (D)∫268273Cp,m dT+ΔHm,fusionTf+ΔHm,vaporisationTb+∫273373Cp,m dT+∫373383Cp,m dT\int_{268}^{273}C_{p,m}\,dT+\dfrac{\Delta H_{m,fusion}}{T_f}+\dfrac{\Delta H_{m,vaporisation}}{T_b}+\int_{273}^{373}C_{p,m}\,dT+\int_{373}^{383}C_{p,m}\,dT∫268273​Cp,m​dT+Tf​ΔHm,fusion​​+Tb​ΔHm,vaporisation​​+∫273373​Cp,m​dT+∫373383​Cp,m​dT

Correct answer: (B)

Step-by-step solution →
Q51·ChemistryIntegerJEE Main 2025
Consider the following cases of standard enthalpy of reaction (ΔHr∘\Delta H^\circ_rΔHr∘​ in kJ mol−1^{-1}−1): C2H6(g)+72O2(g)→2CO2(g)+3H2O(ℓ)C_2H_6(g)+\frac{7}{2}O_2(g)\to 2CO_2(g)+3H_2O(\ell)C2​H6​(g)+27​O2​(g)→2CO2​(g)+3H2​O(ℓ), ΔH1∘=−1550\Delta H^\circ_1=-1550ΔH1∘​=−1550; C(graphite)+O2(g)→CO2(g)C(graphite)+O_2(g)\to CO_2(g)C(graphite)+O2​(g)→CO2​(g), ΔH2∘=−393.5\Delta H^\circ_2=-393.5ΔH2∘​=−393.5; H2(g)+12O2(g)→H2O(ℓ)H_2(g)+\frac{1}{2}O_2(g)\to H_2O(\ell)H2​(g)+21​O2​(g)→H2​O(ℓ), ΔH3∘=−286\Delta H^\circ_3=-286ΔH3∘​=−286. The magnitude of ΔHf∘\Delta H^\circ_fΔHf∘​ of C2H6(g)C_2H_6(g)C2​H6​(g) is ______ kJ mol−1^{-1}−1. (Nearest integer)

Correct answer: 95

Step-by-step solution →
Q52·ChemistrySingle correctJEE Main 2025
Match List-I (Partial Derivative) with List-II (Thermodynamic Quantity). Choose the correct answer from the options given below:
List-I (Partial Derivative)List-II (Thermodynamic Quantity)
A.(∂G∂T)P\left(\frac{\partial G}{\partial T}\right)_P(∂T∂G​)P​I.CPC_PCP​
B.(∂H∂T)P\left(\frac{\partial H}{\partial T}\right)_P(∂T∂H​)P​II.−S-S−S
C.(∂G∂P)T\left(\frac{\partial G}{\partial P}\right)_T(∂P∂G​)T​III.CVC_VCV​
D.(∂U∂T)V\left(\frac{\partial U}{\partial T}\right)_V(∂T∂U​)V​IV.V
  1. (A)(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
  2. (B)(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  3. (C)(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
  4. (D)(A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Correct answer: (B)

Step-by-step solution →
Q53·ChemistrySingle correctJEE Main 2025
A liquid kept inside a thermally insulated closed vessel at 25°C is mechanically stirred from outside. What is the correct option for the following thermodynamic parameters?
  1. (A)ΔU > 0, q = 0, w > 0
  2. (B)ΔU = 0, q = 0, w = 0
  3. (C)ΔU > 0, q = 0, w < 0
  4. (D)ΔU < 0, q = 0, w > 0

Correct answer: (A)

Step-by-step solution →
Q54·ChemistryIntegerJEE Advanced 2024
Consider the following volume – temperature (V – T) diagram for the expansion of 5 moles of an ideal monoatomic gas. Consider only P – V work is involved, the total change in enthalpy (in Joule) for the transformation of state in the sequence X→Y→Z\mathbf{X} \to \mathbf{Y} \to \mathbf{Z}X→Y→Z is __________. [Use the given data: Molar heat capacity of the gas for the given temperature range, Cv,m_{v, m}v,m​ = 12 J K−1^{-1}−1 mol−1^{-1}−1 and gas constant, R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1]

Correct answer: 8120

Step-by-step solution →
Q55·ChemistryNumericalJEE Main 2024
When ΔHvap=30\Delta H_{vap}=30ΔHvap​=30 kJ/mol and ΔSvap=75\Delta S_{vap}=75ΔSvap​=75 J mol−1^{-1}−1 K−1^{-1}−1, then the temperature of vapour, at one atmosphere is _______ K.

Correct answer: 400

Step-by-step solution →
Q56·ChemistryNumericalJEE Main 2024
The heat of solution of anhydrous CuSO4CuSO_4CuSO4​ and CuSO4⋅5H2OCuSO_4\cdot 5H_2OCuSO4​⋅5H2​O are −70-70−70 kJ mol−1mol^{-1}mol−1 and +12+12+12 kJ mol−1mol^{-1}mol−1 respectively. The heat of hydration of CuSO4CuSO_4CuSO4​ to CuSO4⋅5H2OCuSO_4\cdot 5H_2OCuSO4​⋅5H2​O is −x-x−x kJ. The value of x is _____.

Correct answer: 82

Step-by-step solution →
Q57·ChemistryNumericalJEE Main 2024
Consider the figure provided. 111 mol of an ideal gas is kept in a cylinder, fitted with a piston, at the position A, at 18∘C18^\circ C18∘C. If the piston is moved to position B, keeping the temperature unchanged, then xxx L atm work is done in this reversible process. x=x=x= ___ L atm. (nearest integer) [Given: Absolute temperature = ∘C+273.15=\,^\circ C+273.15=∘C+273.15, R=0.08206 L atm mol−1K−1R=0.08206\,L\,atm\,mol^{-1}K^{-1}R=0.08206Latmmol−1K−1]

Correct answer: 55

Step-by-step solution →
Q58·ChemistryNumericalJEE Main 2024
ΔvapH⊖\Delta_{vap}H^{\ominus}Δvap​H⊖ for water is +40.79+40.79+40.79 kJ mol−1^{-1}−1 at 1 bar and 100°C. Change in internal energy for this vapourisation under same condition is _______ kJ mol−1^{-1}−1. (Integer answer) (Given R =8.3=8.3=8.3 JK−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 38

Step-by-step solution →
Q59·ChemistryNumericalJEE Main 2024
An ideal gas, CV=52RC_V = \tfrac{5}{2}RCV​=25​R, is expanded adiabatically against a constant pressure of 1 atm until it doubles in volume. If the initial temperature and pressure are 298 K and 5 atm respectively, then the final temperature is _______ K (nearest integer). (CVC_VCV​ is the molar heat capacity at constant volume)

Correct answer: 274

Step-by-step solution →
Q60·ChemistryNumericalJEE Main 2024
For the reaction at 298 K, 2A+B→C2A + B \rightarrow C2A+B→C. ΔH=400\Delta H = 400ΔH=400 kJ mol−1^{-1}−1 and ΔS=0.2\Delta S = 0.2ΔS=0.2 kJ mol−1^{-1}−1 K−1^{-1}−1. The reaction will become spontaneous above _______ K.

Correct answer: 2000

Step-by-step solution →
Q61·ChemistryNumericalJEE Main 2024
The heat of combustion of solid benzoic acid at constant volume is −321.30-321.30−321.30 kJ at 27∘C27^{\circ}\text{C}27∘C. The heat of combustion at constant pressure is (−321.30−xR)(-321.30 - xR)(−321.30−xR) kJ. The value of x is ___.

Correct answer: 150

Step-by-step solution →
Q62·ChemistrySingle correctJEE Main 2024
Given below are two statements: Assertion (A): Enthalpy of neutralisation of a strong monobasic acid with a strong monoacidic base is always −57-57−57 kJ mol−1^{-1}−1. Reason (R): Enthalpy of neutralisation is the amount of heat liberated when one mole of H+\text{H}^+H+ ions furnished by the acid combine with one mole of OH−\text{OH}^-OH− ions furnished by the base to form one mole of water. In the light of the above statements, choose the correct answer from the options given below.
  1. (A)(A) is true but (R) is false
  2. (B)Both (A) and (R) are true and (R) is the correct explanation of (A)
  3. (C)(A) is false but (R) is true
  4. (D)Both (A) and (R) are true but (R) is not the correct explanation of (A)

Correct answer: (B)

Step-by-step solution →
Q63·ChemistryNumericalJEE Main 2024
Combustion of 1 mole of benzene is expressed as C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)\text{C}_6\text{H}_6(l) + \tfrac{15}{2}\text{O}_2(g) \rightarrow 6\text{CO}_2(g) + 3\text{H}_2\text{O}(l)C6​H6​(l)+215​O2​(g)→6CO2​(g)+3H2​O(l). The standard enthalpy of combustion of 2 mol of benzene is −x-x−x kJ. Given: standard enthalpy of formation of C6H6(l)\text{C}_6\text{H}_6(l)C6​H6​(l) is +48.5+48.5+48.5 kJ mol−1^{-1}−1, of CO2(g)\text{CO}_2(g)CO2​(g) is −393.5-393.5−393.5 kJ mol−1^{-1}−1, and of H2O(l)\text{H}_2\text{O}(l)H2​O(l) is −286-286−286 kJ mol−1^{-1}−1. The value of xxx is __________.

Correct answer: 6535

Step-by-step solution →
Q64·ChemistryNumericalJEE Main 2024
Three moles of an ideal gas are compressed isothermally from 60 L to 20 L using constant pressure of 5 atm. Heat exchange Q for the compression is ______ Lit. atm.

Correct answer: 200

Step-by-step solution →
Q65·ChemistryNumericalJEE Main 2024
The enthalpy of formation of ethane (C2H6)(C_2H_6)(C2​H6​) from ethylene by addition of hydrogen where the bond-energies of C−HC-HC−H, C−CC-CC−C, C=CC=CC=C and H−HH-HH−H are 414 kJ414\,kJ414kJ, 347 kJ347\,kJ347kJ, 615 kJ615\,kJ615kJ and 435 kJ435\,kJ435kJ respectively is −-− ___ kJkJkJ.

Correct answer: 125

Step-by-step solution →
Q66·ChemistryNumericalJEE Main 2024
For a certain reaction at 300 K, K=10K=10K=10, then ΔG∘\Delta G^{\circ}ΔG∘ for the same reaction is −-− __________ ×10−1\times 10^{-1}×10−1 kJ mol−1^{-1}−1. (Given R=8.314R=8.314R=8.314 JK−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 57

Step-by-step solution →
Q67·ChemistrySingle correctJEE Main 2024
Choose the correct option for free expansion of an ideal gas under adiabatic condition from the following:
  1. (A)q=0, ΔT≠0, w=0q=0,\ \Delta T\ne0,\ w=0q=0, ΔT=0, w=0
  2. (B)q=0, ΔT<0, w≠0q=0,\ \Delta T<0,\ w\ne0q=0, ΔT<0, w=0
  3. (C)q≠0, ΔT=0, w=0q\ne0,\ \Delta T=0,\ w=0q=0, ΔT=0, w=0
  4. (D)q=0, ΔT=0, w=0q=0,\ \Delta T=0,\ w=0q=0, ΔT=0, w=0

Correct answer: (D)

Step-by-step solution →
Q68·ChemistryNumericalJEE Main 2024
If 5 moles of an ideal gas expands from 10 L to a volume of 100 L at 300 K under isothermal and reversible condition then work, www, is −x-x−x J. The value of xxx is ______. (Given R=8.314R=8.314R=8.314 J K−1^{-1}−1mol−1^{-1}−1)

Correct answer: 28721

Step-by-step solution →
Q69·ChemistryNumericalJEE Main 2024
An ideal gas undergoes a cyclic transformation starting from the point AAA and coming back to the same point by tracing the path A→B→C→AA\to B\to C\to AA→B→C→A as shown in the diagram. The total work done in the process is ___ J.

Correct answer: 200

Step-by-step solution →
Q70·ChemistryNumericalJEE Main 2024
Two reactions are given below: 2Fe(s)+32O2(g)→Fe2O3(s)2Fe(s)+\dfrac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)2Fe(s)+23​O2​(g)→Fe2​O3​(s), ΔH∘=−822\Delta H^\circ=-822ΔH∘=−822 kJ/mol C(s)+12O2(g)→CO(g)C(s)+\dfrac{1}{2}O_2(g)\rightarrow CO(g)C(s)+21​O2​(g)→CO(g), ΔH∘=−110\Delta H^\circ=-110ΔH∘=−110 kJ/mol Then enthalpy change for following reaction 3C(s)+Fe2O3(s)→2Fe(s)+3CO(g)3C(s)+Fe_2O_3(s)\rightarrow 2Fe(s)+3CO(g)3C(s)+Fe2​O3​(s)→2Fe(s)+3CO(g) is ΔH=\Delta H=ΔH= ______ kJ/mol.

Correct answer: 492

Step-by-step solution →
Q71·ChemistryNumericalJEE Main 2024
Standard enthalpy of vapourisation for CCl4CCl_4CCl4​ is 30.5 kJ mol−1^{-1}−1. Heat required for vapourisation of 284 g of CCl4CCl_4CCl4​ at constant temperature is ___ kJ. (Given molar mass in g mol−1^{-1}−1: C = 12, Cl = 35.5)

Correct answer: 56

Step-by-step solution →
Q72·ChemistrySingle correctJEE Main 2024
Which of the following is not correct?
  1. (A)ΔG\Delta GΔG is negative for a spontaneous reaction
  2. (B)ΔG\Delta GΔG is positive for a spontaneous reaction
  3. (C)ΔG\Delta GΔG is zero for a reversible reaction
  4. (D)ΔG\Delta GΔG is positive for a non-spontaneous reaction

Correct answer: (B)

Step-by-step solution →
Q73·ChemistryNumericalJEE Main 2024
If three moles of an ideal gas at 300 K expand isothermally from 30 dm3^33 to 45 dm3^33 against a constant opposing pressure of 80 kPa, then the amount of heat transferred is __________ J.

Correct answer: 1200

Step-by-step solution →
Q74·ChemistryNumericalJEE Main 2024
For a certain thermochemical reaction M→NM\to NM→N at T=400T=400T=400 K, ΔH∘=77.2\Delta H^\circ=77.2ΔH∘=77.2 kJ mol−1^{-1}−1, ΔS∘=122\Delta S^\circ=122ΔS∘=122 JK−1^{-1}−1, log⁡K\log KlogK is __________ ×10−1\times10^{-1}×10−1.

Correct answer: -37

Step-by-step solution →
Q75·ChemistryNumericalJEE Advanced 2023
PARAGRAPH I The entropy versus temperature plot for phases α\alphaα and β\betaβ at 1 bar pressure is given. STS_TST​ and S0S_0S0​ are entropies of the phases at temperatures T and 0 K, respectively. The transition temperature for α\alphaα to β\betaβ phase change is 600 K and Cp,β−Cp,α=1C_{p,\beta} - C_{p,\alpha} = 1Cp,β​−Cp,α​=1 J mol−1^{-1}−1 K−1^{-1}−1. Assume (Cp,β−Cp,α)(C_{p,\beta} - C_{p,\alpha})(Cp,β​−Cp,α​) is independent of temperature in the range of 200 to 700 K. Cp,αC_{p,\alpha}Cp,α​ and Cp,βC_{p,\beta}Cp,β​ are heat capacities of α\alphaα and β\betaβ phases, respectively. The value of enthalpy change, Hβ−HαH_\beta - H_\alphaHβ​−Hα​ (in J mol−1^{-1}−1), at 300 K is ___.

Correct answer: 300

Step-by-step solution →
Q76·ChemistryNumericalJEE Advanced 2023
One mole of an ideal monoatomic gas undergoes two reversible processes (A →\rightarrow→ B and B →\rightarrow→ C) as shown in the given figure: A →\rightarrow→ B is an adiabatic process. If the total heat absorbed in the entire process (A →\rightarrow→ B and B →\rightarrow→ C) is RT2ln⁡10\mathrm{RT_2}\ln 10RT2​ln10, the value of 2log⁡V32\log \mathrm{V_3}2logV3​ is ___. [Use, molar heat capacity of the gas at constant pressure, Cp,m=52R\mathrm{C_{p,m}} = \frac{5}{2}\mathrm{R}Cp,m​=25​R ]

Correct answer: 7

Step-by-step solution →
Q77·ChemistryNumericalJEE Advanced 2023
PARAGRAPH I The entropy versus temperature plot for phases α\alphaα and β\betaβ at 1 bar pressure is given. STS_TST​ and S0S_0S0​ are entropies of the phases at temperatures T and 0 K, respectively. The transition temperature for α\alphaα to β\betaβ phase change is 600 K and Cp,β−Cp,α=1C_{p,\beta} - C_{p,\alpha} = 1Cp,β​−Cp,α​=1 J mol−1^{-1}−1 K−1^{-1}−1. Assume (Cp,β−Cp,α)(C_{p,\beta} - C_{p,\alpha})(Cp,β​−Cp,α​) is independent of temperature in the range of 200 to 700 K. Cp,αC_{p,\alpha}Cp,α​ and Cp,βC_{p,\beta}Cp,β​ are heat capacities of α\alphaα and β\betaβ phases, respectively. The value of entropy change, Sβ−SαS_\beta - S_\alphaSβ​−Sα​ (in J mol−1^{-1}−1 K−1^{-1}−1), at 300 K is ___. [Use: ln2 = 0.69 Given: Sβ−Sα=0S_\beta - S_\alpha = 0Sβ​−Sα​=0 at 0 K]

Correct answer: 0.31

Step-by-step solution →
Q78·ChemistryNumericalJEE Advanced 2023
In a one-litre flask, 6 moles of A undergoes the reaction A (g)⇌P (g)\mathrm{A\ (g)} \rightleftharpoons \mathrm{P\ (g)}A (g)⇌P (g). The progress of product formation at two temperatures (in Kelvin), T1\mathrm{T_1}T1​ and T2\mathrm{T_2}T2​, is shown in the figure: If T1=2T2\mathrm{T_1} = 2\mathrm{T_2}T1​=2T2​ and (ΔG2θ−ΔG1θ)=RT2ln⁡x,\left(\Delta \mathrm{G_2^{\theta}} - \Delta \mathrm{G_1^{\theta}}\right) = \mathrm{RT_2}\ln x,(ΔG2θ​−ΔG1θ​)=RT2​lnx, then the value of x is ...... [[ΔG1θ\Delta \mathrm{G_1^{\theta}}ΔG1θ​ and ΔG2θ\Delta \mathrm{G_2^{\theta}}ΔG2θ​ are standard Gibb's free energy change for the reaction at temperatures T1\mathrm{T_1}T1​ and T2\mathrm{T_2}T2​, respectively.]

Correct answer: 8

Step-by-step solution →
Q79·ChemistryNumericalJEE Main 2023
30.4 kJ of heat is required to melt one mole of sodium chloride and the entropy change at the melting point is 28.4 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1 at 1 atm. The melting point of sodium chloride is _______ K (Nearest Integer)

Correct answer: 1070

Step-by-step solution →
Q80·ChemistryNumericalJEE Main 2023
A2+B2→2AB, ΔH∘=−200 kJ mol−1A_{2}+B_{2}\to 2AB,\,\Delta H^{\circ}=-200\,kJ\,mol^{-1}A2​+B2​→2AB,ΔH∘=−200kJmol−1. AB, A2A_{2}A2​ and B2B_{2}B2​ are diatomic molecule. If the bond enthalpies of A2, B2A_{2},\,B_{2}A2​,B2​ and AB are in the ratio 1:0.5:11:0.5:11:0.5:1, then the bond enthalpy of A2A_{2}A2​ is _____ kJ mol−1kJ\,mol^{-1}kJmol−1 (Nearest integer).

Correct answer: 400

Step-by-step solution →
Q81·ChemistrySingle correctJEE Main 2023
Identify the correct order of standard enthalpy of formation of sodium halides.
  1. (A)NaF<NaBr<NaCl<NaINaF < NaBr < NaCl < NaINaF<NaBr<NaCl<NaI
  2. (B)NaF<NaCl<NaBr<NaINaF < NaCl < NaBr < NaINaF<NaCl<NaBr<NaI
  3. (C)NaCl<NaF<NaBr<NaINaCl < NaF < NaBr < NaINaCl<NaF<NaBr<NaI
  4. (D)NaI<NaBr<NaF<NaClNaI < NaBr < NaF < NaClNaI<NaBr<NaF<NaCl

Correct answer: (D)

Step-by-step solution →
Q82·ChemistrySingle correctJEE Main 2023
What happens when methane undergoes combustion in systems A and B respectively? (System A is an adiabatic system; System B is a diathermic container)
  1. (A)System A: Temperature rises; System B: Temperature remains same
  2. (B)System A: Temperature falls; System B: Temperature rises
  3. (C)System A: Temperature falls; System B: Temperature remains same
  4. (D)System A: Temperature remains same; System B: Temperature rises

Correct answer: (A)

Step-by-step solution →
Q83·ChemistryNumericalJEE Main 2023
The total number of intensive properties from the following is ____. Volume, Molar heat capacity, Molarity, EcellE_{cell}Ecell​, Gibbs free energy change, Molar mass, Mole.

Correct answer: 4

Step-by-step solution →
Q84·ChemistryNumericalJEE Main 2023
Solid fuel used in rocket is a mixture of Fe2O3Fe_2O_3Fe2​O3​ and Al (in ratio 1:2). The heat evolved (kJ) per gram of the mixture is __________ (Nearest integer). Given: ΔHf0(Al2O3)=−1700\Delta H_f^0(Al_2O_3) = -1700ΔHf0​(Al2​O3​)=−1700 kJ mol−1mol^{-1}mol−1, ΔHf0(Fe2O3)=−840\Delta H_f^0(Fe_2O_3) = -840ΔHf0​(Fe2​O3​)=−840 kJ mol−1mol^{-1}mol−1. Molar mass of Fe, Al and O are 56, 27 and 16 g mol−1mol^{-1}mol−1 respectively.

Correct answer: 4

Step-by-step solution →
Q85·ChemistryNumericalJEE Main 2023
The number of endothermic process/es from the following is _______ . A. I2(g)→2I(g)\text{I}_2(g)\rightarrow 2\text{I}(g)I2​(g)→2I(g) B. HCl(g)→H(g)+Cl(g)\text{HCl}(g)\rightarrow \text{H}(g)+\text{Cl}(g)HCl(g)→H(g)+Cl(g) C. H2O(l)→H2O(g)\text{H}_2\text{O}(l)\rightarrow \text{H}_2\text{O}(g)H2​O(l)→H2​O(g) D. C(s)+O2(g)→CO2(g)\text{C}(s)+\text{O}_2(g)\rightarrow \text{CO}_2(g)C(s)+O2​(g)→CO2​(g) E. Dissolution of ammonium chloride in water.

Correct answer: 4

Step-by-step solution →
Q86·ChemistrySingle correctJEE Main 2023
Given (A) 2CO(g)+O2(g)→2CO2(g)2CO(g)+O_2(g)\rightarrow2CO_2(g)2CO(g)+O2​(g)→2CO2​(g), ΔH10=−x\Delta H_1^0=-xΔH10​=−x kJ mol−1^{-1}−1 and (B) C(graphite)+O2(g)→CO2(g)C(graphite)+O_2(g)\rightarrow CO_2(g)C(graphite)+O2​(g)→CO2​(g), ΔH20=−y\Delta H_2^0=-yΔH20​=−y kJ mol−1^{-1}−1. The ΔH0\Delta H^0ΔH0 for the reaction C(graphite)+12O2(g)→CO(g)C(graphite)+\dfrac12 O_2(g)\rightarrow CO(g)C(graphite)+21​O2​(g)→CO(g) is
  1. (A)x−2y2\dfrac{x-2y}{2}2x−2y​
  2. (B)x+2y2\dfrac{x+2y}{2}2x+2y​
  3. (C)2x−y2\dfrac{2x-y}{2}22x−y​
  4. (D)2y−x2y-x2y−x

Correct answer: (A)

Step-by-step solution →
Q87·ChemistryNumericalJEE Main 2023
For complete combustion of ethene, C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)C2​H4​(g)+3O2​(g)→2CO2​(g)+2H2​O(l), the amount of heat produced as measured in a bomb calorimeter is 140614061406 kJ mol−1^{-1}−1 at 300 K. The minimum value of TΔST\Delta STΔS needed to reach equilibrium is (−)(-)(−) ____ kJ (nearest integer). Given: R=8.3R=8.3R=8.3 JK−1^{-1}−1mol−1^{-1}−1.

Correct answer: 1411

Step-by-step solution →
Q88·ChemistryNumericalJEE Main 2023
When a 60 W electric heater is immersed in a gas for 100s in a constant volume container with adiabatic walls, the temperature of the gas rises by 5 ∘C5\,^\circ C5∘C. The heat capacity of the given gas is ____ J K−1K^{-1}K−1 (Nearest integer)

Correct answer: 1200

Step-by-step solution →
Q89·ChemistryNumericalJEE Main 2023
Consider the graph of Gibbs free energy GGG vs Extent of reaction. The number of statement's from the following which are true with respect to points (a), (b)(a),\,(b)(a),(b) and (c)(c)(c) is _____. A. Reaction is spontaneous at (a)(a)(a) and (b)(b)(b) B. Reaction is at equilibrium at point (b)(b)(b) and non-spontaneous at point (c)(c)(c) C. Reaction is spontaneous at (a)(a)(a) and non-spontaneous at (c)(c)(c) D. Reaction is spontaneous at (a)(a)(a) and (c)(c)(c)

Correct answer: 2

Step-by-step solution →
Q90·ChemistryNumericalJEE Main 2023
Consider the following data: Heat of combustion of H2(g)=−241.8H_2(g) = -241.8H2​(g)=−241.8 kJ mol−1^{-1}−1; Heat of combustion of C(s)=−393.5C(s) = -393.5C(s)=−393.5 kJ mol−1^{-1}−1; Heat of combustion of C2H5OH(l)=−1234.7C_2H_5OH(l) = -1234.7C2​H5​OH(l)=−1234.7 kJ mol−1^{-1}−1. The heat of formation of C2H5OH(l)C_2H_5OH(l)C2​H5​OH(l) is (−)(-)(−) ____ kJ mol−1^{-1}−1 (Nearest integer).

Correct answer: 278

Step-by-step solution →
Q91·ChemistryNumericalJEE Main 2023
0.30.30.3 g of ethane undergoes combustion at 27°C27°C27°C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5°C0.5°C0.5°C. The heat evolved during combustion of ethane at constant pressure is _________ kJmol−1^{-1}−1. (Nearest integer) [Given: The heat capacity of the calorimeter system is 202020 kJ K−1^{-1}−1, R=8.3R=8.3R=8.3 JK−1^{-1}−1mol−1^{-1}−1. Assume ideal gas behaviour. Atomic mass of C and H are 121212 and 111 g mol−1^{-1}−1 respectively]

Correct answer: 1006

Step-by-step solution →
Q92·ChemistryNumericalJEE Main 2023
At 25∘^\circ∘C, the enthalpies of the following processes are given: H2(g)+O2(g)→2OH(g)\text{H}_2(g)+\text{O}_2(g)\rightarrow 2\text{OH}(g)H2​(g)+O2​(g)→2OH(g), ΔH∘=78\Delta H^\circ=78ΔH∘=78 kJ mol−1^{-1}−1; H2(g)+12O2(g)→H2O(g)\text{H}_2(g)+\tfrac{1}{2}\text{O}_2(g)\rightarrow \text{H}_2\text{O}(g)H2​(g)+21​O2​(g)→H2​O(g), ΔH∘=−242\Delta H^\circ=-242ΔH∘=−242 kJ mol−1^{-1}−1; H2(g)→2H(g)\text{H}_2(g)\rightarrow 2\text{H}(g)H2​(g)→2H(g), ΔH∘=436\Delta H^\circ=436ΔH∘=436 kJ mol−1^{-1}−1; 12O2(g)→O(g)\tfrac{1}{2}\text{O}_2(g)\rightarrow \text{O}(g)21​O2​(g)→O(g), ΔH∘=249\Delta H^\circ=249ΔH∘=249 kJ mol−1^{-1}−1. What would be the value of X for the reaction H2O(g)→H(g)+OH(g)\text{H}_2\text{O}(g)\rightarrow \text{H}(g)+\text{OH}(g)H2​O(g)→H(g)+OH(g), ΔH∘=X kJ mol−1\Delta H^\circ=\text{X kJ mol}^{-1}ΔH∘=X kJ mol−1? _______ (Nearest integer).

Correct answer: 499

Step-by-step solution →
Q93·ChemistryNumericalJEE Main 2023
Enthalpies of formation of CCl4(g),H2O(g),CO2(g)CCl_4(g), H_2O(g), CO_2(g)CCl4​(g),H2​O(g),CO2​(g) and HCl(g)HCl(g)HCl(g) are −105,−242,−394-105, -242, -394−105,−242,−394 and −92-92−92 kJ mol−1^{-1}−1 respectively. The magnitude of enthalpy of the reaction given below is _________ kJmol−1^{-1}−1. (nearest integer) CCl4(g)+2H2O(g)→CO2(g)+4HCl(g)CCl_4(g)+2H_2O(g)\rightarrow CO_2(g)+4HCl(g)CCl4​(g)+2H2​O(g)→CO2​(g)+4HCl(g)

Correct answer: 173

Step-by-step solution →
Q94·ChemistryNumericalJEE Main 2023
The enthalpy change for the conversion of 12Cl2(g)\frac{1}{2}Cl_2(g)21​Cl2​(g) to Cl−(aq)Cl^-(aq)Cl−(aq) is (−)(-)(−) ___ kJ mol−1kJ\,mol^{-1}kJmol−1 (Nearest integer). Given : ΔdisHCl2(g)⊖=240 kJ mol−1\Delta_{dis}H^{\ominus}_{Cl_2(g)} = 240\ kJ\,mol^{-1}Δdis​HCl2​(g)⊖​=240 kJmol−1, ΔegHCl(g)⊖=−350 kJ mol−1\Delta_{eg}H^{\ominus}_{Cl(g)} = -350\ kJ\,mol^{-1}Δeg​HCl(g)⊖​=−350 kJmol−1, ΔhydHCl−(g)⊖=−380 kJ mol−1\Delta_{hyd}H^{\ominus}_{Cl^-(g)} = -380\ kJ\,mol^{-1}Δhyd​HCl−(g)⊖​=−380 kJmol−1

Correct answer: 610

Step-by-step solution →
Q95·ChemistryNumericalJEE Main 2023
When 222 litre of ideal gas expands isothermally into vacuum to a total volume of 666 litre, the change in internal energy is _________ J. (Nearest integer)

Correct answer: 0

Step-by-step solution →
Q96·ChemistryNumericalJEE Main 2023
1 mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27 ∘27\,^{\circ}27∘C. The work done is 3 kJ mol−1^{-1}−1. The final temperature of the gas is _______ K (nearest integer). (Given CV=20C_V=20CV​=20 J mol−1^{-1}−1 K−1^{-1}−1)

Correct answer: 150

Step-by-step solution →
Q97·ChemistrySingle correctJEE Main 2023
Which of the following relations are correct? (A) ΔU=q+pΔV\Delta U=q+p\Delta VΔU=q+pΔV (B) ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta SΔG=ΔH−TΔS (C) ΔS=qrevT\Delta S=\dfrac{q_{rev}}{T}ΔS=Tqrev​​ (D) ΔH=ΔU−ΔnRT\Delta H=\Delta U-\Delta nRTΔH=ΔU−ΔnRT Choose the most appropriate answer from the options given below:
  1. (A)B and D Only
  2. (B)A and B Only
  3. (C)B and C Only
  4. (D)C and D Only

Correct answer: (C)

Step-by-step solution →
Q98·ChemistryNumericalJEE Main 2023
Consider the following reaction approaching equilibrium at 27 °C and 1 atm pressure: A + B ⇌kf=103kr=102\underset{k_r = 10^2}{\overset{k_f = 10^3}{\rightleftharpoons}}kr​=102⇌kf​=103​​ C + D. The standard Gibb's energy change (ΔrGθ\Delta_r G^\thetaΔr​Gθ) at 27 °C is (−)(-)(−) ________ kJ mol−1^{-1}−1 (Nearest integer). (Given: R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1 and ln 10 = 2.3)

Correct answer: 6

Step-by-step solution →
Q99·ChemistryNumericalJEE Main 2023
An athlete is given 100 g of glucose (C6H12O6C_6H_{12}O_6C6​H12​O6​) for energy. This is equivalent to 1800 kJ of energy. 50% of this energy is gained by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is _______ g (nearest integer). Assume that there is no other way of consuming stored energy. (Given: enthalpy of evaporation of water is 45 kJ mol−1^{-1}−1; molar mass of C, H and O are 12, 1 and 16 g mol−1^{-1}−1)

Correct answer: 360

Step-by-step solution →
Q100·ChemistryNumericalJEE Main 2023
For independent processes at 300 K: Process A (ΔH=−25\Delta H=-25ΔH=−25 kJ mol−1^{-1}−1, ΔS=−80\Delta S=-80ΔS=−80 J K−1^{-1}−1); Process B (ΔH=−22\Delta H=-22ΔH=−22 kJ mol−1^{-1}−1, ΔS=40\Delta S=40ΔS=40 J K−1^{-1}−1); Process C (ΔH=25\Delta H=25ΔH=25 kJ mol−1^{-1}−1, ΔS=−50\Delta S=-50ΔS=−50 J K−1^{-1}−1); Process D (ΔH=22\Delta H=22ΔH=22 kJ mol−1^{-1}−1, ΔS=20\Delta S=20ΔS=20 J K−1^{-1}−1). The number of non-spontaneous processes from these is _______ .

Correct answer: 2

Step-by-step solution →
Q101·ChemistryNumericalJEE Main 2023
One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is _________ J (nearest integer)

Correct answer: 6

Step-by-step solution →
Q102·ChemistryMultiple correctJEE Advanced 2022
The correct option(s) about entropy (S) is(are) [R = gas constant, F = Faraday constant, T = Temperature]
  1. (A)For the reaction, M(s)+2H+(aq)⟶H2(g)+M2+(aq)\mathrm{M(s)} + 2\mathrm{H^+(aq)} \longrightarrow \mathrm{H_2(g)} + \mathrm{M^{2+}(aq)}M(s)+2H+(aq)⟶H2​(g)+M2+(aq), if dEcelldT=RF\dfrac{\mathrm{dE_{cell}}}{\mathrm{dT}} = \dfrac{\mathrm{R}}{\mathrm{F}}dTdEcell​​=FR​, then the entropy change of the reaction is R (assume that entropy and internal energy changes are temperature independent).
  2. (B)The cell reaction, Pt(s) ∣ H2 (g, 1 bar) ∣ H+ (aq, 0.01 M) ∣∣ H+ (aq, 0.1 M) ∣ H2 (g, 1 bar) ∣ Pt(s)\mathrm{Pt(s)}\,|\,\mathrm{H_2}\,(\mathrm{g},\ 1\,\mathrm{bar})\,|\,\mathrm{H^+}\,(\mathrm{aq},\ 0.01\ \mathrm{M})\,||\,\mathrm{H^+}\,(\mathrm{aq},\ 0.1\ \mathrm{M})\,|\,\mathrm{H_2}\,(\mathrm{g},\ 1\ \mathrm{bar})\,|\,\mathrm{Pt(s)}Pt(s)∣H2​(g, 1bar)∣H+(aq, 0.01 M)∣∣H+(aq, 0.1 M)∣H2​(g, 1 bar)∣Pt(s), is in an entropy driven process.
  3. (C)For racemisation of an optically active compound, ΔS>0\Delta \mathrm{S} > 0ΔS>0.
  4. (D)ΔS>0\Delta \mathrm{S} > 0ΔS>0, for [Ni(H2O)6]2++3 en⟶[Ni(en)3]2++6H2O\left[\mathrm{Ni(H_2O)_6}\right]^{2+} + 3\ \mathrm{en} \longrightarrow \left[\mathrm{Ni(en)_3}\right]^{2+} + 6\mathrm{H_2O}[Ni(H2​O)6​]2++3 en⟶[Ni(en)3​]2++6H2​O (where en = ethylenediamine).

Correct answer: (B), (C), (D)

Step-by-step solution →
Q103·ChemistryNumericalJEE Advanced 2022
2 mol of Hg(g) is combusted in a fixed volume bomb calorimeter with excess of O2\mathrm{O_2}O2​ at 298 K and 1 atm into HgO(s). During the reaction, temperature increases from 298.0 K to 312.8 K. If heat capacity of the bomb calorimeter and enthalpy of formation of Hg(g) are 20.00 kJ K−1\mathrm{K^{-1}}K−1 and 61.32 kJ mol−1\mathrm{mol^{-1}}mol−1 at 298 K, respectively, the calculated standard molar enthalpy of formation of HgO(s) at 298 K is X kJ mol−1\mathrm{mol^{-1}}mol−1. The value of ∣X∣|X|∣X∣ is ________. [Given: Gas constant R = 8.3 J K−1\mathrm{K^{-1}}K−1 mol−1\mathrm{mol^{-1}}mol−1]

Correct answer: 90.39

Step-by-step solution →
Q104·ChemistryNumericalJEE Main 2022
When 600 mL of 0.2 M HNO3HNO_{3}HNO3​ is mixed with 400 mL of 0.1M NaOH solution in a flask, the rise in temperature of the flask is ______ × 10−210^{-2}10−2 °C. (Enthalpy of neutralisation = 57 kJ mol−1mol^{-1}mol−1 and Specific heat of water = 4.2 JK−1JK^{-1}JK−1 g−1g^{-1}g−1) (Neglect heat capacity of flask)

Correct answer: 54

Step-by-step solution →
Q105·ChemistrySingle correctJEE Main 2022
C(s) + O2_{2}2​(g) → CO2_{2}2​(g) + 400 kJ C(s) + 12\frac{1}{2}21​O2_{2}2​(g) → CO(g) + 100 kJ When coal of purity 60% is allowed to burn in presence of insufficient oxygen, 60% of carbon is converted into 'CO' and the remaining is converted into 'CO2_{2}2​'. The heat generated when 0.6 kg of coal is burnt is ______.
  1. (A)1600 kJ
  2. (B)3200 kJ
  3. (C)4400 kJ
  4. (D)6600 kJ

Correct answer: (D)

Step-by-step solution →
Q106·ChemistryNumericalJEE Main 2022
Among the following the number of state variable is _____. Internal energy (U) Volume (V) Heat (q) Enthalpy (H)

Correct answer: 3

Step-by-step solution →
Q107·ChemistrySingle correctJEE Main 2022
Which of the following relation is not correct ?
  1. (A)ΔH = ΔU − PΔV
  2. (B)ΔU = q + W
  3. (C)ΔSsys_{sys}sys​ + ΔSsurr_{surr}surr​ ≥ 0
  4. (D)ΔG = ΔH − TΔS

Correct answer: (A)

Step-by-step solution →
Q108·ChemistryNumericalJEE Main 2022
A gas (Molar mass = 280 g mol−1^{-1}−1) was burnt in excess O2_{2}2​ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45 K. If the heat capacity of calorimeter is 2.5 kJ K−1^{-1}−1 and enthalpy of combustion of gas is 9 kJ mol−1^{-1}−1 then amount of gas burnt is________g. (Nearest Integer)

Correct answer: 35

Step-by-step solution →
Q109·ChemistryNumericalJEE Main 2022
The molar heat capacity for an ideal gas at constant pressure is 20.785 J K−1mol−1K^{-1}mol^{-1}K−1mol−1. The change in internal energy is 5000 J upon heating it from 300K to 500K. The number of moles of the gas at constant volume is ___ [Nearest integer] (Given: R = 8.314 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1)

Correct answer: 2

Step-by-step solution →
Q110·ChemistryNumericalJEE Main 2022
2.4 g coal is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure. The temperature of the calorimeter rises from 298 K to 300 K. The enthalpy change during the combustion of coal is − x kJ mol−1^{-1}−1. The value of x is___________. (Nearest Integer) (Given : Heat capacity of bomb calorimeter 20.0 kJ K−1^{-1}−1 . Assume coal to be pure carbon)

Correct answer: 200

Step-by-step solution →
Q111·ChemistryNumericalJEE Main 2022
For the reaction H2F2(g)H_{2}F_{2}(g)H2​F2​(g) → H2(g)H_{2}(g)H2​(g) + F2(g)F_{2}(g)F2​(g) ΔU = –59.6 kJ mol−1mol^{-1}mol−1 at 27°C. The enthalpy change for the above reaction is (–) ___ kJ mol−1mol^{-1}mol−1 [nearest integer] Given : R = 8.314 JK−1JK^{-1}JK−1 mol−1mol^{-1}mol−1.

Correct answer: 57

Step-by-step solution →
Q112·ChemistryNumericalJEE Main 2022
The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are: –2220.0 kJ mol−1mol^{-1}mol−1, –393.5 kJ mol−1mol^{-1}mol−1 and –285.8 kJ mol−1mol^{-1}mol−1 respectively. The magnitude enthalpy of formation of propane (C3H8C_{3}H_{8}C3​H8​) is………kJ mol−1mol^{-1}mol−1. (Nearest integer)

Correct answer: 104

Step-by-step solution →
Q113·ChemistryNumericalJEE Main 2022
17.0 g of NH3NH_3NH3​ completely vapourises at − 33.42°C and 1 bar pressure and the enthalpy change in the process is 23.4 kJ mol−1mol^{-1}mol−1. The enthalpy change for the vapourisation of 85 g of NH3NH_3NH3​ under the same conditions is _______ kJ.

Correct answer: 117

Step-by-step solution →
Q114·ChemistryNumericalJEE Main 2022
2.2 g of nitrous oxide (N2_22​O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1 mL to 167.75 mL. The change in internal energy of the process, Δ\DeltaΔU is '−-−x' J. The value of 'x' is __. [nearest integer] (Given: atomic mass of N = 14 g mol−1^{-1}−1 and of O = 16 g mol−1^{-1}−1. Molar heat capacity of N2_22​O is 100 JK−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 195

Step-by-step solution →
Q115·ChemistryNumericalJEE Main 2022
For combustion of one mole of magnesium in an open container at 300 K and 1 bar pressure, ΔC_CC​H⊖^\ominus⊖ = –601.70 kJ mol−1^{-1}−1, the magnitude of change in internal energy for the reaction is ______ kJ. (Nearest integer) (Given : R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1)

Correct answer: 600

Step-by-step solution →
Q116·ChemistryNumericalJEE Main 2022
4.0 L of an ideal gas is allowed to expand isothermally into vacuum until the total volume is 20 L. The amount of heat absorbed in this expansion is __________ L atm.

Correct answer: 0

Step-by-step solution →
Q117·ChemistrySingle correctJEE Main 2022
Match List-I with List-II Choose the correct answer from the options given below:
List-IList-II
A.Spontaneous processI.ΔH<0\Delta H < 0ΔH<0
B.Process with ΔP=0\Delta P = 0ΔP=0, ΔT=0\Delta T = 0ΔT=0II.ΔGT,P<0\Delta G_{T,P} < 0ΔGT,P​<0
C.ΔHreaction\Delta H_{reaction}ΔHreaction​III.Isothermal and isobaric process
D.Exothermic processIV.[Bond energies of molecules in reactants] - [Bond energies of product molecules
  1. (A)(A) – (III), (B) – (II), (C) – (IV), (D) – (I)
  2. (B)(A) – (II), (B) – (III), (C) – (IV), (D) – (I)
  3. (C)(A) – (II), (B) – (III), (C) – (I), (D) – (IV)
  4. (D)(A) – (II), (B) – (I), (C) – (III), (D) – (IV)

Correct answer: (B)

Step-by-step solution →
Q118·ChemistryNumericalJEE Main 2022
When 5 moles of He gas expand isothermally and reversibly at 300 K from 10 litre to 20 litre, the magnitude of the maximum work obtained is ____ J. [nearest integer] (Given: R = 8.3 J K−1^{-1}−1mol−1^{-1}−1 and log 2 = 0.3010)

Correct answer: 8630

Step-by-step solution →
Q119·ChemistryNumericalJEE Main 2022
For complete combustion of methanol CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)CH_{3}OH(l) + \frac{3}{2}O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(l)CH3​OH(l)+23​O2​(g)→CO2​(g)+2H2​O(l) the amount of heat produced as measured by bomb calorimeter is 726 kJ mol−1mol^{-1}mol−1 at 27°C. The enthalpy of combustion for the reaction is –x kJ mol−1mol^{-1}mol−1, where x is ______. (Nearest integer) (Given : R = 8.3 JK−1JK^{-1}JK−1 mol−1mol^{-1}mol−1)

Correct answer: 727

Step-by-step solution →
Q120·ChemistryNumericalJEE Main 2022
A fish swimming in water body when taken out from the water body is covered with a film of water of weight 36 g. When it is subjected to cooking at 100°C, then the internal energy for vaporization in kJ mol−1mol^{-1}mol−1 is _________. [nearest integer] [Assume steam to be an ideal gas. Given AvapH⊖A_{vap}H^{\ominus}Avap​H⊖ for water at 373 K and 1 bar is 41.1 kJ mol−1mol^{-1}mol−1 ; R = 8.31 JK−1mol−1JK^{-1}mol^{-1}JK−1mol−1]

Correct answer: 38

Step-by-step solution →
Q121·ChemistryNumericalJEE Main 2022
The standard entropy change for the reaction 4Fe(s)+3O2(g)→2Fe2O3(s)4Fe(s) + 3O_2(g) \rightarrow 2Fe_2O_3(s)4Fe(s)+3O2​(g)→2Fe2​O3​(s) is −550 JK−1JK^{-1}JK−1 at 298 K. [Given : The standard enthalpy change for the reaction is −165 kJ mol−1mol^{-1}mol−1]. The temperature in K at which the reaction attains equilibrium is _________. (Nearest Integer)

Correct answer: 300

Step-by-step solution →
Q122·ChemistrySingle correctJEE Main 2022
At 25∘^{\circ}∘C and 1 atm pressure, the enthalpy of combustion of benzene (1) and acetylene (g) are -3268 kJ mol−1^{-1}−1 and -1300 kJ mol−1^{-1}−1, respectively. The change in enthalpy for the reaction 3 C2_22​H2_22​(g) →\rightarrow→ C6_66​H6_66​(l), is
  1. (A)+ 324 kJ mol−1^{-1}−1
  2. (B)+632 kJ mol−1^{-1}−1
  3. (C)- 632 kJ mol−1^{-1}−1
  4. (D)- 732 kJ mol−1^{-1}−1

Correct answer: (C)

Step-by-step solution →
Q123·ChemistryNumericalJEE Main 2022
2O3_33​(g) ⇌\rightleftharpoons⇌ 3O2_22​(g) At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (–) __J mol−1^{-1}−1 (Nearest integer) [Given: ln 1.35 = 0.3 and R = 8.3 J K−1^{-1}−1 mol−1^{-1}−1]

Correct answer: 747

Step-by-step solution →
Q124·ChemistrySingle correctJEE Main 2022
At 25∘25^\circ25∘C and 1 atm pressure, the enthalpies of combustion are as given below: The enthalpy of formation of ethane is
SubstanceH2_22​C(graphite)C2_22​H6_66​(g)
ΔCH⊖kJmol−1\dfrac{\Delta_\mathrm{C}H^{\ominus}}{\mathrm{kJmol}^{-1}}kJmol−1ΔC​H⊖​−286.0-286.0−286.0−394.0-394.0−394.0−1560.0-1560.0−1560.0
  1. (A)+54.0 kJ mol−1^{-1}−1
  2. (B)−-−68.0 kJ mol−1^{-1}−1
  3. (C)−-−86.0 kJ mol−1^{-1}−1
  4. (D)+97.0 kJ mol−1^{-1}−1

Correct answer: (C)

Step-by-step solution →
Q125·ChemistryNumericalJEE Advanced 2021
For the reaction X(s)⇌Y(s)+Z(g)\mathbf{X}(s) \rightleftharpoons \mathbf{Y}(s) + \mathbf{Z}(g)X(s)⇌Y(s)+Z(g), the plot of ln⁡pzp⊖\ln\frac{p_z}{p^{\ominus}}lnp⊖pz​​ versus 104T\frac{10^4}{T}T104​ is given below (in solid line), where pzp_zpz​ is the pressure (in bar) of the gas Z\mathbf{Z}Z at temperature TTT and P⊖=1P^{\ominus} = 1P⊖=1 bar. (Given, d(ln⁡K)d(1T)=−ΔH⊖R\frac{d(\ln K)}{d\left(\frac{1}{T}\right)} = -\frac{\Delta H^{\ominus}}{R}d(T1​)d(lnK)​=−RΔH⊖​, where the equilibrium constant, K =pzp⊖= \frac{p_z}{p^{\ominus}}=p⊖pz​​ and the gas constant, R = 8.314 J K−1^{-1}−1 mol−1^{-1}−1) The value of ΔS⊖\Delta \mathrm{S}^{\ominus}ΔS⊖ (in J K−1^{-1}−1 mol−1^{-1}−1) for the given reaction, at 1000 K is______ .

Correct answer: 141.33 or 141.34

Step-by-step solution →
Q126·ChemistryNumericalJEE Advanced 2021
For the reaction X(s)⇌Y(s)+Z(g)\mathbf{X}(s) \rightleftharpoons \mathbf{Y}(s) + \mathbf{Z}(g)X(s)⇌Y(s)+Z(g), the plot of ln⁡pzp⊖\ln\frac{p_z}{p^{\ominus}}lnp⊖pz​​ versus 104T\frac{10^4}{T}T104​ is given below (in solid line), where pzp_zpz​ is the pressure (in bar) of the gas Z\mathbf{Z}Z at temperature TTT and P⊖=1P^{\ominus} = 1P⊖=1 bar. (Given, d(ln⁡K)d(1T)=−ΔH⊖R\frac{d(\ln K)}{d\left(\frac{1}{T}\right)} = -\frac{\Delta H^{\ominus}}{R}d(T1​)d(lnK)​=−RΔH⊖​, where the equilibrium constant, K =pzp⊖= \frac{p_z}{p^{\ominus}}=p⊖pz​​ and the gas constant, R = 8.314 J K−1^{-1}−1 mol−1^{-1}−1) The value of standard enthalpy, ΔH⊖\Delta \mathrm{H}^{\ominus}ΔH⊖ (in kJ mol−1^{-1}−1) for the reaction is_____ .

Correct answer: 166.28

Step-by-step solution →
Q127·ChemistryMultiple correctJEE Advanced 2021
An ideal gas undergoes a reversible isothermal expansion from state I\mathbf{I}I to state II\mathbf{II}II followed by a reversible adiabatic expansion from state II\mathbf{II}II to state III\mathbf{III}III. The correct plot(s) representing the changes from state I\mathbf{I}I to state III\mathbf{III}III is(are) (ppp : pressure, VVV : volume, TTT : temperature, HHH : enthalpy, SSS : entropy)
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A), (B), (D)

Step-by-step solution →
Q128·ChemistryIntegerJEE Advanced 2021
One mole of an ideal gas at 900 K, undergoes two reversible processes, I\mathbf{I}I followed by II\mathbf{II}II, as shown below. If the work done by the gas in the two processes are same, the value of ln⁡V3V2\ln \frac{V_{3}}{V_{2}}lnV2​V3​​ is ___. (UUU: internal energy, SSS: entropy, ppp: pressure, VVV: volume, RRR: gas constant) (Given: molar heat capacity at constant volume, CV,mC_{V,\mathrm{m}}CV,m​ of the gas is 52R\frac{5}{2}R25​R)

Correct answer: 10

Step-by-step solution →
Q129·ChemistryNumericalJEE Main 2021
For the reaction 2NO2_{2}2​(g) ⇌ N2_{2}2​O4_{4}4​(g), when ∆S = −176.0 JK−1^{-1}−1 and ∆H = −57.8 kJ mol−1^{-1}−1, the magnitude of ∆G at 298 K for the reaction is ______ kJ mol−1^{-1}−1. (Nearest integer)

Correct answer: 5

Step-by-step solution →
Q130·ChemistrySingle correctJEE Main 2021
The incorrect expression among the following is:
  1. (A)ΔGSystemΔSTotal=−T\dfrac{\Delta \mathrm{G_{System}}}{\Delta \mathrm{S_{Total}}} = -\mathrm{T}ΔSTotal​ΔGSystem​​=−T (at constant P)
  2. (B)ln⁡K=ΔHo−TΔSoRT\ln \mathrm{K} = \dfrac{\Delta \mathrm{H^{o}} - \mathrm{T}\Delta \mathrm{S^{o}}}{\mathrm{RT}}lnK=RTΔHo−TΔSo​
  3. (C)K=e−ΔGo/RT\mathrm{K} = e^{-\Delta \mathrm{G^{o}}/\mathrm{RT}}K=e−ΔGo/RT
  4. (D)For isothermal process wreversible=− nRTln⁡VfVi\mathrm{w_{reversible}} = -\ \mathrm{nRT} \ln \dfrac{\mathrm{V}_f}{\mathrm{V}_i}wreversible​=− nRTlnVi​Vf​​

Correct answer: (B)

Step-by-step solution →
Q131·ChemistryNumericalJEE Main 2021
According to the following figure, the magnitude of the enthalpy change of the reaction A + B →\rightarrow→ M + N in kJ mol−1^{-1}−1 is equal to _____. (Integer answer)

Correct answer: 45

Step-by-step solution →
Q132·ChemistryNumericalJEE Main 2021
Data given for the following reaction is as follows: FeO(s)_{(s)}(s)​ + C(graphite)_{(graphite)}(graphite)​ → Fe(s)_{(s)}(s)​ + CO(g)_{(g)}(g)​ Substance | ΔH° (kJ mol−1^{-1}−1) | ΔS° (J mol−1^{-1}−1K−1^{-1}−1) FeO(s)_{(s)}(s)​ | −266.3 | 57.49 C(graphite)_{(graphite)}(graphite)​ | 0 | 5.74 Fe(s)_{(s)}(s)​ | 0 | 27.28 CO(g)_{(g)}(g)​ | −110.5 | 197.6 The minimum temperature in K at which the reaction becomes spontaneous is _____ . (Integer answer)

Correct answer: 964

Step-by-step solution →
Q133·ChemistryNumericalJEE Main 2021
200 mL of 0.2 M HCl is mixed with 300 mL of 0.1 M NaOH. The molar heat of neutralization of this reaction is −57.1-57.1−57.1 kJ. The increase in temperature in ∘C^\circ C∘C of the system on mixing is x×10−2x \times 10^{-2}x×10−2. The value of x is ________ . (Nearest integer) [Given : Specific heat of water = 4.18 J g−1g^{-1}g−1 K−1K^{-1}K−1 Density of water = 1.00 g cm−3cm^{-3}cm−3] (Assume no volume change on mixing)

Correct answer: 82

Step-by-step solution →
Q134·ChemistryNumericalJEE Main 2021
The Born-Haber cycle for KCl is evaluated with the following data : ΔfH⊖\Delta_f H^\ominusΔf​H⊖ for KCl=−436.7 kJ mol−1KCl = -436.7\ kJ\ mol^{-1}KCl=−436.7 kJ mol−1; ΔsubH⊖\Delta_{sub} H^\ominusΔsub​H⊖ for K=89.2 kJ mol−1K = 89.2\ kJ\ mol^{-1}K=89.2 kJ mol−1; ΔionizationH⊖\Delta_{ionization} H^\ominusΔionization​H⊖ for K=419.0 kJ mol−1K = 419.0\ kJ\ mol^{-1}K=419.0 kJ mol−1; Δelectron gainH⊖\Delta_{electron\ gain} H^\ominusΔelectron gain​H⊖ for Cl(g)=−348.6 kJ mol−1Cl_{(g)} = -348.6\ kJ\ mol^{-1}Cl(g)​=−348.6 kJ mol−1; ΔbondH⊖\Delta_{bond} H^\ominusΔbond​H⊖ for Cl2=243.0 kJ mol−1Cl_2 = 243.0\ kJ\ mol^{-1}Cl2​=243.0 kJ mol−1 The magnitude of lattice enthalpy of KCl in kJ mol−1kJ\ mol^{-1}kJ mol−1 is ________ (Nearest integer)

Correct answer: 718

Step-by-step solution →
Q135·ChemistryNumericalJEE Main 2021
For water ΔvapH=41\Delta_{vap} H = 41Δvap​H=41 kJ mol−1^{-1}−1 at 373 K and 1 bar pressure. Assuming that water vapour is an ideal gas that occupies a much larger volume than liquid water, the internal energy change during evaporation of water is ______ kJ mol−1^{-1}−1 [Use : R = 8.3 J mol−1^{-1}−1 K−1^{-1}−1]

Correct answer: 38

Step-by-step solution →
Q136·ChemistryNumericalJEE Main 2021
When 400 mL of 0.2 M H2_22​SO4_44​ solution is mixed with 600 mL of of 0.1 M NaOH solution the increase in temperature of the final solution is ____________×10−2^{-2}−2 K. (Round off to the Nearest Integer). [Use: H+^++(aq) + OH−^-− (aq) → H2_22​O : Δr\Delta_rΔr​H = −57.1kJmol−1^{-1}−1 Specific heat of H2_22​O = 4.18JK−1^{-1}−1 g−1^{-1}−1 Density of H2_22​O = 1.0 gcm−3^{-3}−3 Assume no change in volume of solution on mixing]

Correct answer: 82

Step-by-step solution →
Q137·ChemistryNumericalJEE Main 2021
For water at 100°C and 1 bar ΔvapH−ΔvapU\Delta_{vap}H - \Delta_{vap}UΔvap​H−Δvap​U = ________ ×102\times 10^2×102 J mol−1^{-1}−1 (Round off to the Nearest integer) [ Use: R = 8.31 J mol−1^{-1}−1 K−1^{-1}−1 ] [ Assume volume of H2H_2H2​O(l) is much smaller than volume of H2H_2H2​O(g). Assume H2H_2H2​O(g) can be treated as an ideal gas]

Correct answer: 31

Step-by-step solution →
Q138·ChemistryNumericalJEE Main 2021
At 298 K, the enthalpy of fusion of a solid (X) is 2.8kJ mol−1^{-1}−1 and the enthalpy of vaporization of the liquid (X) is 98.2 kJ mol−1^{-1}−1. The enthalpy of sublimation of the substance (X) in kJ mol−1^{-1}−1 is ___________. (in nearest integer)

Correct answer: 101

Step-by-step solution →
Q139·ChemistryNumericalJEE Main 2021
A system does 200 J of work and at the same time absorbs 150 J of heat. The magnitude of the change in internal energy is _________J. (Nearest integer)

Correct answer: 50

Step-by-step solution →
Q140·ChemistryNumericalJEE Main 2021
If the standard molar enthalpy change for combustion of graphite powder is −2.48×102-2.48 \times 10^{2}−2.48×102 kJ mol−1^{-1}−1, the amount of heat generated on combustion of 1 g of graphite powder is __________ kJ. (Nearest integer)

Correct answer: 21

Step-by-step solution →
Q141·ChemistryNumericalJEE Main 2021
For a given chemical reaction A→B\mathrm{A} \rightarrow \mathrm{B}A→B at 300 K the free energy change is − 49.4-\,49.4−49.4 kJ mol−1\mathrm{mol^{-1}}mol−1 and the enthalpy of reaction is 51.4 kJ mol−1\mathrm{mol^{-1}}mol−1. The entropy change of the reaction is ________ J K−1 mol−1\mathrm{J\ K^{-1}\ mol^{-1}}J K−1 mol−1.

Correct answer: 336

Step-by-step solution →
Q142·ChemistryNumericalJEE Main 2021
For the reaction C2_{2}2​H6_{6}6​ → C2_{2}2​H4_{4}4​ + H2_{2}2​ the reaction enthalpy Δr_{r}r​H = ______ kJ mol−1^{-1}−1. (Round off to the Nearest Integer). [Given : Bond enthalpies in kJ mol−1^{-1}−1 : C–C : 347, C=C : 611; C–H : 414, H–H : 436]

Correct answer: 128

Step-by-step solution →
Q143·ChemistryNumericalJEE Main 2021
The gas phase reaction 2A(g) ⇌ A2_{2}2​(g) at 400 K has ΔG° = + 25.2 kJ mol−1^{-1}−1. The equilibrium constant KC_{C}C​ for this reaction is ______ × 10−2^{-2}−2. (Round off to the Nearest integer) [Use : R = 8.3 J mol−1^{-1}−1K−1^{-1}−1 , ln 10 = 2.3 log10_{10}10​ 2 = 0.30, 1 atm = 1 bar] [antilog (−0.3) = 0.501]

Correct answer: 2

Step-by-step solution →
Q144·ChemistrySingle correctJEE Main 2021
During which of the following processes, does entropy decrease ? (A) Freezing of water to ice at 0°C (B) Freezing of water to ice at –10°C (C) N2_{2}2​(g) + 3H2_{2}2​(g) →\rightarrow→ 2NH3_{3}3​(g) (D) Adsorption of CO(g) and lead surface (E) Dissolution of NaCl in water
  1. (A)(A), (B), (C) and (D) only
  2. (B)(B) and (C) only
  3. (C)(A) and (E) only
  4. (D)(A), (C) and (E) only

Correct answer: (A)

Step-by-step solution →
Q145·ChemistryNumericalJEE Main 2021
The standard enthalpies of formation of Al2O3\mathrm{Al_2O_3}Al2​O3​ and CaO are −1675-1675−1675 kJ mol−1\mathrm{mol^{-1}}mol−1 and −635-635−635 kJ mol−1\mathrm{mol^{-1}}mol−1 respectively. For the reaction 3CaO+2Al→3Ca+Al2O3\mathrm{3CaO + 2Al \rightarrow 3Ca + Al_2O_3}3CaO+2Al→3Ca+Al2​O3​ the standard reaction enthalpy ΔrH0=\Delta_\mathrm{r} H^0 =Δr​H0= ________ kJ. (Round off to the Nearest Integer).

Correct answer: 230

Step-by-step solution →
Q146·ChemistryNumericalJEE Main 2021
At 25°C, 50 g of iron reacts with HCl to form FeCl2_{2}2​. The evolved hydrogen gas expands against a constant pressure of 1 bar. The work done by the gas during this expansion is _______ J. (Round off to the Nearest Integer) [Given : R = 8.314 J mol−1^{-1}−1 K−1^{-1}−1. Assume, hydrogen is an ideal gas] [Atomic mass off Fe is 55.85 u]

Correct answer: 2218

Step-by-step solution →
Q147·ChemistryNumericalJEE Main 2021
For the reaction A(g) ⇌ B(g) at 495 K, ΔrG∘\Delta_{r}G^{\circ}Δr​G∘ = −9.478 kJ mol−1^{-1}−1. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B is the equilibrium mixture is _______ millimoles. (Round off to the Nearest Integer). [R = 8.314 J mol−1^{-1}−1 K−1^{-1}−1; ℓn 10 = 2.303]

Correct answer: 20

Step-by-step solution →
Q148·ChemistryNumericalJEE Main 2021
For a chemical reaction A + B ⇌ C + D (Δr_rr​H⊖^{\ominus}⊖ = 80 kJ mol−1^{-1}−1) the entropy change Δr_rr​S⊖^{\ominus}⊖ depends on the temperature T (in K) as Δr_rr​S⊖^{\ominus}⊖ = 2T (J K−1^{-1}−1 mol−1^{-1}−1). Minimum temperature at which it will become spontaneous is _________K.

Correct answer: 200

Step-by-step solution →
Q149·ChemistryNumericalJEE Main 2021
The average S−F bond energy in kJ mol−1mol^{-1}mol−1 of SF6SF_6SF6​ is ____________. (Rounded off to the nearest integer) [Given : The values of standard enthalpy of formation of SF6SF_6SF6​(g), S(g) and F(g) are - 1100, 275 and 80 kJ mol−1mol^{-1}mol−1 respectively.]

Correct answer: 309

Step-by-step solution →
Q150·ChemistryNumericalJEE Main 2021
The ionization enthalpy of Na+^{+}+ formation from Na(g)_{(g)}(g)​ is 495.8 kJ mol−1^{-1}−1, while the electron gain enthalpy of Br is −325.0 kJ mol−1^{-1}−1. Given the lattice enthalpy of NaBr is −728.4 kJ mol−1^{-1}−1. The energy for the formation of NaBr ionic solid is (−)_______ × 10−1^{-1}−1 kJ mol−1^{-1}−1.

Correct answer: 5576

Step-by-step solution →
Q151·ChemistryNumericalJEE Main 2021
The reaction of cyanamide, NH2_{2}2​CN(s)_{(s)}(s)​ with oxygen was run in a bomb calorimeter and ΔU was found to be −742.24 kJ mol−1^{-1}−1. The magnitude of ΔH298_{298}298​ for the reaction NH2_{2}2​CN(s)_{(s)}(s)​ + 32\frac{3}{2}23​O2_{2}2​(g) → N2(g)_{2(g)}2(g)​ + O2(g)_{2(g)}2(g)​ + H2_{2}2​O(l)_{(l)}(l)​ is _______ kJ. (Rounded off to the nearest integer) [Assume ideal gases and R = 8.314 J mol−1^{-1}−1 K−1^{-1}−1]

Correct answer: 741

Step-by-step solution →
Q152·ChemistryNumericalJEE Main 2021
Five moles of an ideal gas at 293 K is expanded isothermally from an initial pressure of 2.1 MPa to 1.3 MPa against at constant external 4.3 MPa. The heat transferred in this process is ____kJ mol−1^{-1}−1. (Rounded-off of the nearest integer) [Use R = 8.314 J mol−1^{-1}−1 K−1^{-1}−1]

Correct answer: 15

Step-by-step solution →
Q153·ChemistryNumericalJEE Main 2021
For the reaction A(g)→B(g)\mathrm{A_{(g)}} \rightarrow \mathrm{B_{(g)}}A(g)​→B(g)​, the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of ΔrG\mathrm{\Delta_r G}Δr​G for the reaction at 300 K and 1 atm in J mol−1\mathrm{J\,mol^{-1}}Jmol−1 is −xR-xR−xR, where x is ______. (Rounded off to the nearest integer) [R=8.31 J mol−1K−1[\mathrm{R} = 8.31\,\mathrm{J\,mol^{-1}K^{-1}}[R=8.31Jmol−1K−1 and ln⁡10=2.3]\ln 10 = 2.3]ln10=2.3]

Correct answer: 1380

Step-by-step solution →
Q154·ChemistryNumericalJEE Main 2021
Assuming ideal behaviour, the magnitude of log K for the following reaction at 25ºC is x × 10−110^{-1}10−1 . The value of x is ____________.(Integer answer) 3HC≡CH(g)⇌C6H6(ℓ)3HC \equiv CH_{(g)} \rightleftharpoons C_{6}H_{6(\ell)}3HC≡CH(g)​⇌C6​H6(ℓ)​ [Given : ΔfG°(HC≡CH)\Delta_{f}G°(HC \equiv CH)Δf​G°(HC≡CH) = – 2.04 × 10510^{5}105] mol−1mol^{-1}mol−1; ΔfG°(C6H6)\Delta_{f}G°(C_{6}H_{6})Δf​G°(C6​H6​) = – 1.24 × 10510^{5}105 J mol−1mol^{-1}mol−1 ; R = 8.314 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1 ]

Correct answer: 855

Step-by-step solution →
Q155·ChemistryNumericalJEE Advanced 2020
Tin is obtained from cassiterite by reduction with coke. Use the data given below to determine the minimum temperature (in K) at which the reduction of cassiterite by coke would take place. At 298 K : ΔfH°(SnO2(s))\Delta_{f}H°(SnO_{2}(s))Δf​H°(SnO2​(s)) = –581.0 kJ mol−1mol^{-1}mol−1, ΔfH°(CO2(g))\Delta_{f}H°(CO_{2}(g))Δf​H°(CO2​(g)) = –394.0 kJ mol−1mol^{-1}mol−1 S°(SnO2(s))S°(SnO_{2}(s))S°(SnO2​(s)) = 56.0 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1, S°(Sn(s))S°(Sn(s))S°(Sn(s)) = 52.0 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1, S°(C(s))S°(C(s))S°(C(s)) = 6.0 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1, S°(CO2(g))S°(CO_{2}(g))S°(CO2​(g)) = 210.0 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1. Assume that the enthalpies and the entropies are temperature independent.

Correct answer: 935.00

Step-by-step solution →
Q156·ChemistryMultiple correctJEE Advanced 2020
In thermodynamics the P-V work done is given by w=−∫dV Pextw = -\int dV\, P_{ext}w=−∫dVPext​ . For a system undergoing a particular process, the work done is , w=−∫dV(RTV−b−aV2)w = -\int dV \left( \frac{RT}{V-b} - \frac{a}{V^2} \right)w=−∫dV(V−bRT​−V2a​). This equation is applicable to a
  1. (A)System that satisfies the van der Waals equation of state.
  2. (B)Process that is reversible and isothermal.
  3. (C)Process that is reversible and adiabatic.
  4. (D)Process that is irreversible and at constant pressure.

Correct answer: (A), (B), (C)

Step-by-step solution →
Q157·ChemistrySingle correctJEE Main 2020
The variation of equilibrium constant with temperature is given below: The value of ΔHo\Delta H^oΔHo, ΔGo\Delta G^oΔGo at T1T_1T1​ and ΔGo\Delta G^oΔGo at T2T_2T2​ (in kJ mol−1^{-1}−1) respectively, are close to [Use R=8.314JK−1mol−1R = 8.314 J K^{-1} mol^{-1}R=8.314JK−1mol−1]
TemperatureEquilibrium Constant
T1=25oCT_1 = 25^oCT1​=25oCK1=10K_1 = 10K1​=10
T2=100oCT_2 = 100^oCT2​=100oCK2=100K_2 = 100K2​=100
  1. (A)28.7, − 7.14 and − 5.71
  2. (B)0.64 − 7.14 and − 5.71
  3. (C)28.4, − 5.71 and − 14.29
  4. (D)0.64, − 5.71 and −14.29

Correct answer: (C)

Step-by-step solution →
Q158·ChemistrySingle correctJEE Main 2020
Lattice enthalpy and enthalpy of solution of NaCl are 788 kJ mol−1^{-1}−1 and 4 kJ mol−1^{-1}−1, respectively. The hydration enthalpy of NaCl is:
  1. (A)784 kJ mol−1^{-1}−1
  2. (B)-784 kJ mol−1^{-1}−1
  3. (C)780 kJ mol−1^{-1}−1
  4. (D)-780 kJ mol−1^{-1}−1

Correct answer: (B)

Step-by-step solution →
Q159·ChemistrySingle correctJEE Main 2020
Five moles of an ideal gas at 1 bar and 298 K is expanded into vacuum to double the volume. The work done is :
  1. (A)CV(T2−T1)C_V(T_2 - T_1)CV​(T2​−T1​)
  2. (B)−RTln⁡V2/V1-RT \ln V_2/V_1−RTlnV2​/V1​
  3. (C)zero
  4. (D)−RT V2/V1-RT\, V_2/V_1−RTV2​/V1​

Correct answer: (C)

Step-by-step solution →
Q160·ChemistrySingle correctJEE Main 2020
For one mole of an ideal gas, which of these statements must be true? (a) U and H each depends only on temperature (b) Compressibility factor z is not equal to 1 (c) CP,m−CV,m=RC_{P,m} - C_{V,m} = RCP,m​−CV,m​=R (d) dU=CVdTdU = C_V dTdU=CV​dT for any process
  1. (A)(a), (c) and (d)
  2. (B)(b), (c) and (d)
  3. (C)(a) and (c)
  4. (D)(c) and (d)

Correct answer: (A)

Step-by-step solution →
Q161·ChemistrySingle correctJEE Main 2020
If enthalpy of atomisation for Br2_{2}2​(A) is x kJ/mol and bond enthalpy for Br2_{2}2​ is y kJ/mol, the relation between them:
  1. (A)is x = y
  2. (B)is x > y
  3. (C)does not exist
  4. (D)is x < y

Correct answer: (B)

Step-by-step solution →
Q162·ChemistryNumericalJEE Main 2020
The magnitude of work done by a gas that undergoes a reversible expansion along the path ABC shown in the figure is ________

Correct answer: 48.00

Step-by-step solution →
Q163·ChemistryNumericalJEE Main 2020
At constant volume, 4 mol of an ideal gas when heated from 300 K to 500 K changes its internal energy by 5000 J. The molar heat capacity at constant volume is ____

Correct answer: 6.25

Step-by-step solution →
Q164·ChemistryNumericalJEE Main 2020
For the reaction A(l) ⟶\longrightarrow⟶ 2B(g) ΔU\Delta UΔU = 2.1 Kcal, ΔS\Delta SΔS = 20 cal K−1^{-1}−1 at 300 K Hence ΔG\Delta GΔG in Kcal is

Correct answer: -2.70

Step-by-step solution →
Q165·ChemistryMultiple correctJEE Advanced 2019
Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to the standard enthalpy of formation.
  1. (A)2C(g) + 3H2_{2}2​(g) ⟶\longrightarrow⟶ C2_{2}2​H6_{6}6​(g)
  2. (B)32\frac{3}{2}23​O2_{2}2​(g) ⟶\longrightarrow⟶ O3_{3}3​(g)
  3. (C)2H2_{2}2​(g) + O2_{2}2​(g) ⟶\longrightarrow⟶ 2H2_{2}2​O(ℓ\ellℓ)
  4. (D)18\frac{1}{8}81​S8_{8}8​(s) + O2_{2}2​(g) ⟶\longrightarrow⟶ SO2_{2}2​(g)

Correct answer: (B), (D)

Step-by-step solution →
Q166·ChemistrySingle correctJEE Main 2019
The INCORRECT match in the following is:
  1. (A)ΔG∘<0,K>1\Delta G^{\circ} < 0, K > 1ΔG∘<0,K>1
  2. (B)ΔG∘<0,K<1\Delta G^{\circ} < 0, K < 1ΔG∘<0,K<1
  3. (C)ΔG∘=0,K=1\Delta G^{\circ} = 0, K = 1ΔG∘=0,K=1
  4. (D)ΔG∘>0,K<1\Delta G^{\circ} > 0, K < 1ΔG∘>0,K<1

Correct answer: (B)

Step-by-step solution →
Q167·ChemistrySingle correctJEE Main 2019
An ideal gas is allowed to expand from 1 L to 10 L against a constant external pressure of 1bar. The work done in kJ is:
  1. (A)−9.0-9.0−9.0
  2. (B)−0.9-0.9−0.9
  3. (C)−2.0-2.0−2.0
  4. (D)+10.0+10.0+10.0

Correct answer: (B)

Step-by-step solution →
Q168·ChemistrySingle correctJEE Main 2019
Enthalpy of sublimation of iodine is 24 cal g−1^{-1}−1 at 200°C. If specific heat of I2_22​(s) and I2_22​ (vap) are 0.055 and 0.031 cal g−1^{-1}−1K−1^{-1}−1 respectively, then enthalpy of sublimation of iodine at 250°C in cal g−1^{-1}−1 is :
  1. (A)2.85
  2. (B)11.4
  3. (C)5.7
  4. (D)22.8

Correct answer: (D)

Step-by-step solution →
Q169·ChemistrySingle correctJEE Main 2019
The difference between ΔH and ΔU (ΔH - ΔU), when the combustion of one mole of heptane(I) is carried out a temperature T is equal to
  1. (A)-4 RT
  2. (B)-3 RT
  3. (C)3 RT
  4. (D)4 RT

Correct answer: (A)

Step-by-step solution →
Q170·ChemistrySingle correctJEE Main 2019
A process will be spontaneous at all temperatures if:
  1. (A)ΔH < 0 and ΔS < 0
  2. (B)ΔH < 0 and ΔS > 0
  3. (C)ΔH > 0 and ΔS > 0
  4. (D)ΔH > 0 and ΔS < 0

Correct answer: (B)

Step-by-step solution →
Q171·ChemistrySingle correctJEE Main 2019
During compression of a spring the work done is 10kJ and 2kJ escaped to the surroundings as heat. The change in internal energy, ΔU\Delta UΔU (in kJ) is:
  1. (A)8
  2. (B)12
  3. (C)-12
  4. (D)-8

Correct answer: (A)

Step-by-step solution →
Q172·ChemistrySingle correctJEE Main 2019
Among the following the set of parameters that represents path functions, is: (a) q + w (b) q (c) w (d) H – TS
  1. (A)(b) and (c)
  2. (B)(b), (c) and (d)
  3. (C)(a), (b) and (c)
  4. (D)(a) and (d)

Correct answer: (A)

Step-by-step solution →
Q173·ChemistrySingle correctJEE Main 2019
Which one of the following equations does not correctly represent the first law of thermodynamics for the given processes involving an ideal gas? (Assume non–expansion work is zero)
  1. (A)Adiabatic process: ΔU=−w\Delta U = -wΔU=−w
  2. (B)Isochoric process: ΔU=q\Delta U = qΔU=q
  3. (C)Cyclic process: q=−wq = -wq=−w
  4. (D)Isothermal process: q=−wq = -wq=−w

Correct answer: (A)

Step-by-step solution →
Q174·ChemistrySingle correctJEE Main 2019
For silver CP(JK−1mol−1)=23+0.01TC_P (JK^{-1} mol^{-1}) = 23 + 0.01TCP​(JK−1mol−1)=23+0.01T. If the temperature (T) of 3 moles of silver is raised from 300 K to 1000 K at 1 atm pressure, the value of ΔH\Delta HΔH will be close to:
  1. (A)13 kJ
  2. (B)62 kJ
  3. (C)16 kJ
  4. (D)21 kJ

Correct answer: (B)

Step-by-step solution →
Q175·ChemistrySingle correctJEE Main 2019
5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If CV=28 JK−1mol−1C_V=28\ JK^{-1}mol^{-1}CV​=28 JK−1mol−1, calculate ΔU\Delta UΔU and ΔpV\Delta pVΔpV for this process. (R = 8.0 J K−1^{-1}−1 mol−1^{-1}−1)
  1. (A)ΔU=2.8kJ;Δ(pV)=0.8kJ\Delta U=2.8kJ;\Delta(pV)=0.8kJΔU=2.8kJ;Δ(pV)=0.8kJ
  2. (B)ΔU=14kJ;Δ(pV)=4kJ\Delta U=14kJ;\Delta(pV)=4kJΔU=14kJ;Δ(pV)=4kJ
  3. (C)ΔU=14kJ;Δ(pV)=18kJ\Delta U=14kJ;\Delta(pV)=18kJΔU=14kJ;Δ(pV)=18kJ
  4. (D)ΔU=14kJ;Δ(pV)=0.8J\Delta U=14kJ;\Delta(pV)=0.8JΔU=14kJ;Δ(pV)=0.8J

Correct answer: (B)

Step-by-step solution →
Q176·ChemistrySingle correctJEE Main 2019
The combination of plots which do not represent isothermal expansion of an ideal gas is:
  1. (A)(b) and (d)
  2. (B)(a) and (c)
  3. (C)(b) and (c)
  4. (D)(a) and (d)

Correct answer: (A)

Step-by-step solution →
Q177·ChemistrySingle correctJEE Main 2019
For a disatomic ideal gas in a closed system, which of the following plots does not correctly describe the relation between various thermodynamic quantities?
  1. (A)(A)
  2. (B)(B)
  3. (C)(C)
  4. (D)(D)

Correct answer: (A)

Step-by-step solution →
Q178·ChemistrySingle correctJEE Main 2019
Given : (i) C(graphite)+O2(g)→CO2(g)C(graphite) + O_2(g) \rightarrow CO_2(g)C(graphite)+O2​(g)→CO2​(g); ΔrH°=x\Delta rH° = xΔrH°=x kJ mol−1^{-1}−1 (ii) C(graphite)+12O2(g)→CO(g)C(graphite) + \frac{1}{2}O_2(g) \rightarrow CO(g)C(graphite)+21​O2​(g)→CO(g); ΔrH°=y\Delta rH° = yΔrH°=y kJ mol−1^{-1}−1 (iii) CO(g)+12O2(g)→CO2(g)CO(g) + \frac{1}{2}O_2(g) \rightarrow CO_2(g)CO(g)+21​O2​(g)→CO2​(g); ΔrH°=z\Delta rH° = zΔrH°=z kJ mol−1^{-1}−1 Based on the above thermochemical equations, find out which one of the following algebraic relationships is correct?
  1. (A)x=y+zx = y + zx=y+z
  2. (B)z=x+yz = x + yz=x+y
  3. (C)y=2z−xy = 2z - xy=2z−x
  4. (D)x=y−zx = y - zx=y−z

Correct answer: (A)

Step-by-step solution →
Q179·ChemistrySingle correctJEE Main 2019
The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by ΔrG0=A−BT\Delta_{r}G^{0} = A - BTΔr​G0=A−BT Where A and B are non-zero constant. Which of the following is TRUE about this reaction?
  1. (A)Endothermic if A > 0
  2. (B)Exothermic if A > 0 and B < 0
  3. (C)Endothermic if A < 0 and B > 0
  4. (D)Exothermic if B < 0

Correct answer: (A)

Step-by-step solution →
Q180·ChemistrySingle correctJEE Main 2019
The reaction: MgO(s)+C(s)→Mg(s)+CO(g)MgO(s) + C(s) \rightarrow Mg(s) + CO(g)MgO(s)+C(s)→Mg(s)+CO(g), for which ΔrH0=+491.1\Delta_{r}H^{0} = +491.1Δr​H0=+491.1 kJ mol−1^{-1}−1 and ΔrS0=198.0\Delta_{r}S^{0} = 198.0Δr​S0=198.0 JK−1^{-1}−1 mol−1^{-1}−1 is not feasible at 298 K. Temperature above which reaction will be feasible is:
  1. (A)2040.5 K
  2. (B)1890.0K
  3. (C)2480. K
  4. (D)2380.K

Correct answer: (C)

Step-by-step solution →
Q181·ChemistrySingle correctJEE Main 2019
For the chemical reaction X⇌YX \rightleftharpoons YX⇌Y, the standard reaction Gibbs energy depends on temperature T (in K) as ΔrG0(in kJ mol−1)=120−38T\Delta_rG^0(\text{in kJ mol}^{-1}) = 120 - \dfrac{3}{8}TΔr​G0(in kJ mol−1)=120−83​T The major component of the reaction mixture at T is:
  1. (A)Y if T = 300 K
  2. (B)Y if T = 280 K
  3. (C)X if T = 350 K
  4. (D)X if T = 315 K

Correct answer: (D)

Step-by-step solution →
Q182·ChemistrySingle correctJEE Main 2019
Two blocks of the same metal having same mass and at temperature T1T_1T1​ and T2T_2T2​ respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, ΔS\Delta SΔS, for this process is:
  1. (A)Cpln⁡[(T1+T2)24T1T2]C_p \ln\left[\dfrac{(T_1+T_2)^2}{4T_1T_2}\right]Cp​ln[4T1​T2​(T1​+T2​)2​]
  2. (B)2Cpln⁡[(T1+T2)1/2T1T2]2C_p \ln\left[\dfrac{(T_1+T_2)^{1/2}}{T_1T_2}\right]2Cp​ln[T1​T2​(T1​+T2​)1/2​]
  3. (C)2Cpln⁡[(T1+T2)4T1T2]2C_p \ln\left[\dfrac{(T_1+T_2)}{4T_1T_2}\right]2Cp​ln[4T1​T2​(T1​+T2​)​]
  4. (D)2Cpln⁡[(T1+T2)2T1T2]2C_p \ln\left[\dfrac{(T_1+T_2)}{2T_1T_2}\right]2Cp​ln[2T1​T2​(T1​+T2​)​]

Correct answer: (A)

Step-by-step solution →
Q183·ChemistrySingle correctJEE Main 2019
A process that has ΔH=200\Delta H = 200ΔH=200 J mol−1^{-1}−1 and ΔS=40\Delta S = 40ΔS=40 JK−1^{-1}−1 mol−1^{-1}−1. Out of the values given below, choose the minimum temperature above which the process will be spontaneous:
  1. (A)20 K
  2. (B)12 K
  3. (C)5 K
  4. (D)4 K

Correct answer: (C)

Step-by-step solution →
Q184·ChemistrySingle correctJEE Main 2019
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is: (Specific heat of water liquid and water vapours are 4.2 kJ K−1^{-1}−1 kg−1^{-1}−1 and 2.0 kJ K−1^{-1}−1 kg−1^{-1}−1, heat of liquid fusion and vapourisation of water are 334 kJ kg−1^{-1}−1 and 2491 kJ kg−1^{-1}−1, respectively) (log 273 = 2.436, log 373 = 2.572, log 383 = 2.583)
  1. (A)7.90 kJ K−1^{-1}−1 kg−1^{-1}−1
  2. (B)2.64 kJ K−1^{-1}−1 kg−1^{-1}−1
  3. (C)8.49 kJ K−1^{-1}−1 kg−1^{-1}−1
  4. (D)9.26 kJ K−1^{-1}−1 kg−1^{-1}−1

Correct answer: (D)

Step-by-step solution →
Q185·ChemistryNumericalJEE Advanced 2018
The surface of copper gets tarnished by the formation of copper oxide. N2N_{2}N2​ gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2N_{2}N2​ gas contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per the reaction given below: 2Cu(s)+H2O(g)→Cu2O(s)+H2(g)2Cu(s) + H_{2}O(g) \rightarrow Cu_{2}O(s) + H_{2}(g)2Cu(s)+H2​O(g)→Cu2​O(s)+H2​(g) pH2p_{H_{2}}pH2​​ is the minimum partial pressure of H2H_{2}H2​ (in bar) needed to prevent the oxidation at 1250 K. The value of ln⁡(pH2)\ln\left(p_{H_{2}}\right)ln(pH2​​) is ____. (Given: total pressure = 1 bar, RRR (universal gas constant) = 8 J K−1K^{-1}K−1 mol−1mol^{-1}mol−1, ln⁡(10)\ln(10)ln(10) = 2.3. Cu(s) and Cu2O(s)Cu_{2}O(s)Cu2​O(s) are mutually immiscible. At 1250 K: 2Cu(s)+12O2(g)→Cu2O(s)2Cu(s) + \frac{1}{2} O_{2}(g) \rightarrow Cu_{2}O(s)2Cu(s)+21​O2​(g)→Cu2​O(s); ΔG⊖=−78,000\Delta G^{\ominus} = -78,000ΔG⊖=−78,000 J mol−1mol^{-1}mol−1 H2(g)+12O2(g)→H2O(g)H_{2}(g) + \frac{1}{2} O_{2}(g) \rightarrow H_{2}O(g)H2​(g)+21​O2​(g)→H2​O(g); ΔG⊖=−1,78,000\Delta G^{\ominus} = -1,78,000ΔG⊖=−1,78,000 J mol−1mol^{-1}mol−1; GGG is the Gibbs energy)

Correct answer: -14.6

Step-by-step solution →
Q186·ChemistryMultiple correctJEE Advanced 2018
For a reaction, A⇌PA \rightleftharpoons PA⇌P, the plots of [A] and [P] with time at temperatures T1T_{1}T1​ and T2T_{2}T2​ are given below. If T2>T1T_{2} > T_{1}T2​>T1​, the correct statement(s) is (are) (Assume ΔH⊖\Delta H^{\ominus}ΔH⊖ and ΔS⊖\Delta S^{\ominus}ΔS⊖ are independent of temperature and ratio of ln⁡K\ln KlnK at T1T_{1}T1​ to ln⁡K\ln KlnK at T2T_{2}T2​ is greater than T2/T1T_{2}/T_{1}T2​/T1​. Here HHH, SSS, GGG and KKK are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)
  1. (A)ΔH⊖<0,ΔS⊖<0\Delta H^{\ominus} < 0, \Delta S^{\ominus} < 0ΔH⊖<0,ΔS⊖<0
  2. (B)ΔG⊖<0,ΔH⊖>0\Delta G^{\ominus} < 0, \Delta H^{\ominus} > 0ΔG⊖<0,ΔH⊖>0
  3. (C)ΔG⊖<0,ΔS⊖<0\Delta G^{\ominus} < 0, \Delta S^{\ominus} < 0ΔG⊖<0,ΔS⊖<0
  4. (D)ΔG⊖<0,ΔS⊖>0\Delta G^{\ominus} < 0, \Delta S^{\ominus} > 0ΔG⊖<0,ΔS⊖>0

Correct answer: (A), (C)

Step-by-step solution →
Q187·ChemistryMultiple correctJEE Advanced 2018
A reversible cyclic process for an ideal gas is shown below. Here, PPP, VVV, and TTT are pressure, volume and temperature, respectively. The thermodynamic parameters qqq, www, HHH and UUU are heat, work, enthalpy and internal energy, respectively. The correct option(s) is (are)
  1. (A)qAC=ΔUBCq_{AC} = \Delta U_{BC}qAC​=ΔUBC​ and wAB=P2(V2−V1)w_{AB} = P_{2}(V_{2}-V_{1})wAB​=P2​(V2​−V1​)
  2. (B)wBC=P2(V2−V1)w_{BC} = P_{2}(V_{2}-V_{1})wBC​=P2​(V2​−V1​) and qBC=ΔHACq_{BC} = \Delta H_{AC}qBC​=ΔHAC​
  3. (C)ΔHCA<ΔUCA\Delta H_{CA} < \Delta U_{CA}ΔHCA​<ΔUCA​ and qAC=ΔUBCq_{AC} = \Delta U_{BC}qAC​=ΔUBC​
  4. (D)qBC=ΔHACq_{BC} = \Delta H_{AC}qBC​=ΔHAC​ and ΔHCA>ΔUCA\Delta H_{CA} > \Delta U_{CA}ΔHCA​>ΔUCA​

Correct answer: (B), (C)

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Q188·ChemistrySingle correctJEE Advanced 2017
The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K are ΔfG0[C(graphite)]=0 kJ mol−1\Delta_{f}G^{0}\left[\mathrm{C(graphite)}\right] = 0\ \mathrm{kJ\,mol^{-1}}Δf​G0[C(graphite)]=0 kJmol−1 ΔfG0[C(diamond)]=2.9 kJ mol−1\Delta_{f}G^{0}\left[\mathrm{C(diamond)}\right] = 2.9\ \mathrm{kJ\,mol^{-1}}Δf​G0[C(diamond)]=2.9 kJmol−1 The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2×10−6 m3 mol−12\times10^{-6}\ \mathrm{m^{3}\,mol^{-1}}2×10−6 m3mol−1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is [Useful information: 1 J = 1 kg m2^{2}2s−2^{-2}−2; 1 Pa = 1 kg m−1^{-1}−1s−2^{-2}−2; 1 bar = 10510^{5}105 Pa]
  1. (A)14501 bar
  2. (B)58001 bar
  3. (C)1450 bar
  4. (D)29001 bar

Correct answer: (A)

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Q189·ChemistryMultiple correctJEE Advanced 2017
An ideal gas is expanded from (p1,V1,T1)(p_{1}, V_{1}, T_{1})(p1​,V1​,T1​) to (p2,V2,T2)(p_{2}, V_{2}, T_{2})(p2​,V2​,T2​) under different conditions. The correct statement(s) among the following is(are)
  1. (A)The work done on the gas is maximum when it is compressed irreversibly from (p2,V2)(p_{2}, V_{2})(p2​,V2​) to (p1,V1)(p_{1}, V_{1})(p1​,V1​) against constant pressure p1p_{1}p1​
  2. (B)The work done by the gas is less when it is expanded reversibly from V1V_{1}V1​ to V2V_{2}V2​ under adiabatic conditions as compared to that when expanded reversibly from V1V_{1}V1​ to V2V_{2}V2​ under isothermal conditions
  3. (C)The change in internal energy of the gas is (i) zero, if it is expanded reversibly with T1=T2T_{1} = T_{2}T1​=T2​ , and (ii) positive, if it is expanded reversibly under adiabatic conditions with T1≠T2T_{1} \neq T_{2}T1​=T2​
  4. (D)If the expansion is carried out freely, it is simultaneously both isothermal as well as adiabatic

Correct answer: (A), (B), (D)

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Q190·ChemistrySingle correctJEE Advanced 2016
One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding (ΔSsurr\Delta S_{surr}ΔSsurr​) in JK−1JK^{-1}JK−1 is (1 L atm = 101.3 J)
  1. (A)5.763
  2. (B)1.013
  3. (C)−1.013-1.013−1.013
  4. (D)−5.763-5.763−5.763

Correct answer: (C)

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Q191·ChemistryMatrix matchJEE Advanced 2015
Match the thermodynamic processes given under Column I with the expression given under Column II:
Column IColumn II
A.Freezing of water at 273 K and 1 atmP.q=0q = 0q=0
B.Expansion of 1 mol of an ideal gas into a vacuum under isolated conditionsQ.w=0w = 0w=0
C.Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated containerR.ΔSsys<0\Delta S_{\text{sys}} < 0ΔSsys​<0
D.Reversible heating of H2(g)\mathrm{H_{2}(g)}H2​(g) at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atmS.ΔU=0\Delta U = 0ΔU=0
T.ΔG=0\Delta G = 0ΔG=0

Correct answer: A-(R,T); B-(P,Q,S); C-(P,Q,S); D-(P,Q,S,T)

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Q192·ChemistryMultiple correctJEE Advanced 2015
When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7 ∘5.7\,^{\circ}5.7∘C was measured for the beaker and its contents (Expt. 1\mathbf{Expt.\ 1}Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0-57.0−57.0 kJ mol−1^{-1}−1), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2\mathbf{Expt.\ 2}Expt. 2), 100 mL of 2.0 M acetic acid (Ka=2.0×10−5K_{a} = 2.0 \times 10^{-5}Ka​=2.0×10−5) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1\mathbf{Expt.\ 1}Expt. 1) where a temperature rise of 5.6 ∘5.6\,^{\circ}5.6∘C was measured. (Consider heat capacity of all solutions as 4.2 J g−1^{-1}−1 K−1^{-1}−1 and density of all solutions as 1.0 g mL−1^{-1}−1) Enthalpy of dissociation (in kJ mol−1^{-1}−1) of acetic acid obtained from the Expt. 2\mathbf{Expt.\ 2}Expt. 2 is
  1. (A)1.0
  2. (B)10.0
  3. (C)24.5
  4. (D)51.4

Correct answer: (A)

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Q193·ChemistrySingle correctJEE Advanced 2014
For the process H2O(ℓ)⟶H2O(g)\mathrm{H_2O}(\ell) \longrightarrow \mathrm{H_2O}(\mathrm{g})H2​O(ℓ)⟶H2​O(g) at T=100∘CT = 100^\circ\mathrm{C}T=100∘C and 1 atmosphere pressure, the correct choice is
  1. (A)ΔSsystem>0\Delta S_{\mathrm{system}} > 0ΔSsystem​>0 and ΔSsurrounding>0\Delta S_{\mathrm{surrounding}} > 0ΔSsurrounding​>0
  2. (B)ΔSsystem>0\Delta S_{\mathrm{system}} > 0ΔSsystem​>0 and ΔSsurrounding<0\Delta S_{\mathrm{surrounding}} < 0ΔSsurrounding​<0
  3. (C)ΔSsystem<0\Delta S_{\mathrm{system}} < 0ΔSsystem​<0 and ΔSsurrounding>0\Delta S_{\mathrm{surrounding}} > 0ΔSsurrounding​>0
  4. (D)ΔSsystem<0\Delta S_{\mathrm{system}} < 0ΔSsystem​<0 and ΔSsurrounding<0\Delta S_{\mathrm{surrounding}} < 0ΔSsurrounding​<0

Correct answer: (B)

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Q194·ChemistryMultiple correctJEE Advanced 2014
An ideal gas in a thermally insulated vessel at internal pressure =P1= P_1=P1​, volume =V1= V_1=V1​ and absolute temperature =T1= T_1=T1​ expands irreversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are P2P_2P2​, V2V_2V2​ and T2T_2T2​, respectively. For this expansion,
  1. (A)q=0q = 0q=0
  2. (B)T2=T1T_2 = T_1T2​=T1​
  3. (C)P2V2=P1V1P_2V_2 = P_1V_1P2​V2​=P1​V1​
  4. (D)P2V2γ=P1V1γP_2V_2^{\gamma} = P_1V_1^{\gamma}P2​V2γ​=P1​V1γ​

Correct answer: (A), (B), (C)

Step-by-step solution →
Q195·ChemistrySingle correctJEE Advanced 2013
The standard enthalpies of formation of CO2CO_2CO2​(g), H2OH_2OH2​O(l) and glucose(s) at 25∘^{\circ}∘C are –400 kJ/mol, –300 kJ/mol and –1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25∘^{\circ}∘C is
  1. (A)+2900 kJ
  2. (B)–2900 kJ
  3. (C)–16.11 kJ
  4. (D)+16.11 kJ

Correct answer: (C)

Step-by-step solution →

Chemical Thermodynamics — frequently asked

How many questions from Chemical Thermodynamics appear in JEE?

Chemical Thermodynamics has appeared in 165 of the last 186 JEE Main and JEE Advanced papers — about 89% of them — contributing 195 questions in total across those papers.

Is Chemical Thermodynamics an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 89% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Chemical Thermodynamics questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Chemistry chapters

  • Coordination Compounds 318
  • p-Block Elements 276
  • Redox Reactions and Electrochemistry 265
  • Chemical Bonding and Molecular Structure 207
  • d- and f-Block Elements 202
  • Aldehydes and Ketones 199
  • Equilibrium 199
  • Solutions 198

All 36 Chemistry chapters →

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