Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Thermodynamics question

2024 · 31 Jan · Shift 1 · Q21
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Thermodynamics
  5. /2024 · 31 Jan · Shift 1 · Q21

Thermodynamics question

2024 · 31 Jan · Shift 1 · Q21

JEE MainChemistryThermodynamicsNumerical+4 / −1
Consider the following reaction at 298 K⋅32O2(g)⇌O3(g)⋅KP=2.47×10−29298 \mathrm{~K} \cdot \frac{3}{2} \mathrm{O}_{2(g)} \rightleftharpoons \mathrm{O}_{3(g)} \cdot \mathrm{K}_{\mathrm{P}}=2.47 \times 10^{-29}298 K⋅23​O2(g)​⇌O3(g)​⋅KP​=2.47×10−29. ΔrG⊖\Delta_r G^{\ominus}Δr​G⊖ for the reaction is ‾\underline{\hspace{2cm}}​kJ\mathrm{kJ}kJ. (Given R=8.314 JK−1 mol−1\mathrm{R}=8.314 \mathrm{~JK}^{-1} \mathrm{~mol}^{-1}R=8.314 JK−1 mol−1)
Numerical answer
View written solutionFree

Correct answer: 163

  1. Use the relation between ΔrG⊖\Delta_r G^\ominusΔr​G⊖ and KPK_PKP​:
ΔrG⊖=−RTln⁡KP\Delta_r G^\ominus = -RT \ln K_PΔr​G⊖=−RTlnKP​

Given:

  • T=298 KT = 298\,\text{K}T=298K
  • R=8.314 J K−1mol−1R = 8.314\,\text{J K}^{-1}\text{mol}^{-1}R=8.314J K−1mol−1
  • KP=2.47×10−29K_P = 2.47 \times 10^{-29}KP​=2.47×10−29
  1. Substitute the value of KPK_PKP​:
ln⁡(2.47×10−29)=ln⁡(2.47)+ln⁡(10−29)\ln(2.47 \times 10^{-29}) = \ln(2.47) + \ln(10^{-29})ln(2.47×10−29)=ln(2.47)+ln(10−29) ln⁡(2.47)≈0.904,ln⁡(10−29)=−29ln⁡10≈−29(2.303)=−66.787\ln(2.47) \approx 0.904, \qquad \ln(10^{-29}) = -29\ln 10 \approx -29(2.303) = -66.787ln(2.47)≈0.904,ln(10−29)=−29ln10≈−29(2.303)=−66.787

So,

ln⁡KP≈0.904−66.787=−65.883\ln K_P \approx 0.904 - 66.787 = -65.883lnKP​≈0.904−66.787=−65.883
  1. Now calculate ΔrG⊖\Delta_r G^\ominusΔr​G⊖:
ΔrG⊖=−(8.314)(298)(−65.883)\Delta_r G^\ominus = -(8.314)(298)(-65.883)Δr​G⊖=−(8.314)(298)(−65.883)

First,

8.314×298≈2477.68.314 \times 298 \approx 2477.68.314×298≈2477.6

Then,

ΔrG⊖≈2477.6×65.883approx163250 J mol−1\Delta_r G^\ominus \approx 2477.6 \times 65.883 approx 163250\,\text{J mol}^{-1}Δr​G⊖≈2477.6×65.883approx163250J mol−1
  1. Convert to kJ mol−1^{-1}−1:
ΔrG⊖≈163.25 kJ mol−1\Delta_r G^\ominus \approx 163.25\,\text{kJ mol}^{-1}Δr​G⊖≈163.25kJ mol−1
  1. Since the question asks for an integer,
163\boxed{163}163​

Thus, the standard Gibbs free energy change for the reaction is 163 kJ163\,\text{kJ}163kJ.

PreviousNext

More from Thermodynamics

  • 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, w, is −x J. The value of x is ​…2024 · Numerical
  • At 25∘C, the enthalpy of the following processes are given : What would be the value of X for the following reaction ? ​ (Nearest integer) H2​O(g)→H(g)+OH(g) ΔH∘=X kJ mol−1… Includes table2023 · Numerical
  • 0.3 g of ethane undergoes combustion at 27∘C in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by 0.5∘C. The heat evolved during combustion…2023 · Numerical
  • Consider the graph of Gibbs free energy G vs Extent of reaction. The number of statement/s from the following which are true with respect to points (a), (b) and (c) is ​ A. Reaction is spontaneous at (a) and (b) B.… Includes diagram2023 · Numerical
  • The value of logK for the reaction A⇌B at 298 K is ​. (Nearest integer) Given: ΔH∘=−54.07 kJ mol−1ΔS∘=10 J K−1 mol−1…2023 · Numerical
  • Consider the following data Heat of combustion of H2​(g)=−241.8 kJ mol−1 Heat of combustion of C(s)=−393.5 kJ mol−1 Heat of…2023 · Numerical
  • When a 60 W electric heater is immersed in a gas for 100 s in a constant volume container with adiabatic walls, the temperature of the gas rises by 5∘C. The heat capacity of the given gas is ​JK−1…2023 · Numerical
  • For complete combustion of ethene. C2​H4​(g)+3O2​(g)→2CO2​(g)+2H2​O(l) the amount of heat produced as measured in bomb…2023 · Numerical