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Thermodynamics question

2007 · Shift 1 · Q9
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  5. /2007 · Shift 1 · Q9

Thermodynamics question

2007 · Shift 1 · Q9

JEE AdvancedChemistryThermodynamicsMCQ+3 / −1
The value of log 10{}_{10}10​ K for a reaction A⇌BA \rightleftharpoons BA⇌B is (Given : ΔrH298 K∘=−54.07{\Delta _r}H{^\circ _{298\,K}} = - 54.07Δr​H298K∘​=−54.07 kJ mol −1{}^{-1}−1, ΔrS298 K∘=10{\Delta _r}S{^\circ _{298\,K}} = 10Δr​S298K∘​=10 J K −1{}^{-1}−1 mol −1{}^{-1}−1 and R = 8.314 J K −1{}^{-1}−1 mol −1{}^{-1}−1; 2.303 ×\times× 8.314 ×\times× 298 = 5705)
  1. A
    5
  2. B
    10
  3. C
    95
  4. D
    100
View written solutionFree

Correct answer: B

Step-by-step derivation:

  1. Identify the goal and relevant equations. The goal is to find the value of log⁡10K\log_{10} Klog10​K for the reaction A⇌BA \rightleftharpoons BA⇌B. The key thermodynamic relationships needed are:

    • The Gibbs-Helmholtz equation, which relates standard Gibbs free energy change (ΔrG∘\Delta_r G^\circΔr​G∘), standard enthalpy change (ΔrH∘\Delta_r H^\circΔr​H∘), and standard entropy change (ΔrS∘\Delta_r S^\circΔr​S∘): ΔrG∘=ΔrH∘−TΔrS∘\Delta_r G^\circ = \Delta_r H^\circ - T \Delta_r S^\circΔr​G∘=Δr​H∘−TΔr​S∘
    • The relationship between standard Gibbs free energy change and the equilibrium constant (KKK): ΔrG∘=−RTln⁡K=−2.303RTlog⁡10K\Delta_r G^\circ = -RT \ln K = -2.303 RT \log_{10} KΔr​G∘=−RTlnK=−2.303RTlog10​K

  2. List the given values and ensure unit consistency.

    • ΔrH298K∘=−54.07\Delta_r H^\circ_{298K} = -54.07Δr​H298K∘​=−54.07 kJ mol−1^{-1}−1
    • ΔrS298K∘=10\Delta_r S^\circ_{298K} = 10Δr​S298K∘​=10 J K−1^{-1}−1 mol−1^{-1}−1
    • T=298T = 298T=298 K
    • R=8.314R = 8.314R=8.314 J K−1^{-1}−1 mol−1^{-1}−1
    • A pre-calculated value is given: 2.303×8.314×298=57052.303 \times 8.314 \times 298 = 57052.303×8.314×298=5705 J mol−1^{-1}−1.

    To maintain consistency in units, we convert ΔrH∘\Delta_r H^\circΔr​H∘ from kJ mol−1^{-1}−1 to J mol−1^{-1}−1: ΔrH∘=−54.07 kJ mol−1×1000 J1 kJ=−54070 J mol−1\Delta_r H^\circ = -54.07 \text{ kJ mol}^{-1} \times \frac{1000 \text{ J}}{1 \text{ kJ}} = -54070 \text{ J mol}^{-1}Δr​H∘=−54.07 kJ mol−1×1 kJ1000 J​=−54070 J mol−1

  3. Calculate the standard Gibbs free energy change (ΔrG∘\Delta_r G^\circΔr​G∘). Substitute the values into the Gibbs-Helmholtz equation: ΔrG∘=(−54070 J mol−1)−(298 K×10 J K−1mol−1)\Delta_r G^\circ = (-54070 \text{ J mol}^{-1}) - (298 \text{ K} \times 10 \text{ J K}^{-1} \text{mol}^{-1})Δr​G∘=(−54070 J mol−1)−(298 K×10 J K−1mol−1) ΔrG∘=−54070 J mol−1−2980 J mol−1\Delta_r G^\circ = -54070 \text{ J mol}^{-1} - 2980 \text{ J mol}^{-1}Δr​G∘=−54070 J mol−1−2980 J mol−1 ΔrG∘=−57050 J mol−1\Delta_r G^\circ = -57050 \text{ J mol}^{-1}Δr​G∘=−57050 J mol−1

  4. Calculate log⁡10K\log_{10} Klog10​K. Rearrange the equation relating ΔrG∘\Delta_r G^\circΔr​G∘ and KKK to solve for log⁡10K\log_{10} Klog10​K: log⁡10K=−ΔrG∘2.303RT\log_{10} K = -\frac{\Delta_r G^\circ}{2.303 RT}log10​K=−2.303RTΔr​G∘​ Now, substitute the value of ΔrG∘\Delta_r G^\circΔr​G∘ we just calculated and the given value for 2.303RT2.303 RT2.303RT: log⁡10K=−−57050 J mol−15705 J mol−1\log_{10} K = -\frac{-57050 \text{ J mol}^{-1}}{5705 \text{ J mol}^{-1}}log10​K=−5705 J mol−1−57050 J mol−1​ log⁡10K=570505705\log_{10} K = \frac{57050}{5705}log10​K=570557050​ log⁡10K=10\log_{10} K = 10log10​K=10

  5. Conclusion. The calculated value of log⁡10K\log_{10} Klog10​K is 10. Comparing this with the given options:

    • A: 5
    • B: 10
    • C: 95
    • D: 100 The correct option is B.
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