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NVAEasyJEE 2023Gibbs Free Energy & Equilibrium Constant

JEE Chemistry 2023 Question with Solution

Consider the following reaction approaching equilibrium at 27∘C27^\circ \text{C} and 1 atm1 \, \text{atm} pressure:

A+B⇌C+DA + B \rightleftharpoons C + D

Kf=103K_f = 10^3, Kr=102K_r = 10^2

The standard Gibbs energy change (ΔrG∘\Delta_r G^\circ) at 27∘C27^\circ \text{C} is (−)(-) _____ kJ mol−1\text{kJ mol}^{-1} (Nearest integer).

(Given: R=8.3 J K−1 mol−1R = 8.3 \, \text{J K}^{-1} \, \text{mol}^{-1} and ln⁡(10)=2.3\ln(10) = 2.3)

Answer

Correct answer:6

Step-by-step solution

Standard Method

Given:

  • Reaction: A+B⇌C+DA + B \rightleftharpoons C + D
  • Kf=103K_f = 10^3
  • Kr=102K_r = 10^2
  • T=300 KT = 300 \, \text{K}
  • R=8.3×10−3 kJ mol−1 K−1R = 8.3 \times 10^{-3} \, \text{kJ mol}^{-1} \, \text{K}^{-1}
  • ln⁡(10)=2.3\ln(10) = 2.3

Find: ΔrG∘\Delta_r G^\circ

The standard Gibbs free energy change is related to the equilibrium constant by

ΔrG∘=−RTln⁡K\Delta_r G^\circ = -RT \ln K

For the reaction, the equilibrium constant is obtained from the forward and reverse constants:

K=KfKr=103102=10K = \frac{K_f}{K_r} = \frac{10^3}{10^2} = 10

Hence,

ln⁡K=ln⁡10=2.3\ln K = \ln 10 = 2.3

Now substitute the values:

ΔrG∘=−(8.3×10−3)×300×2.3  kJ mol−1\Delta_r G^\circ = -(8.3 \times 10^{-3}) \times 300 \times 2.3 \; \text{kJ mol}^{-1} ΔrG∘=−6  kJ mol−1\Delta_r G^\circ = -6 \; \text{kJ mol}^{-1}

Therefore, the standard Gibbs energy change is −6 kJ mol−1-6 \, \text{kJ mol}^{-1}, so the required numerical answer is 6.

Using forward and reverse equilibrium constants carefully

Given: Kf=103K_f = 10^3 and Kr=102K_r = 10^2

Find: the nearest integer value of the magnitude asked in the blank.

A common point here is that the reaction equilibrium constant is not directly KfK_f or KrK_r alone. It is calculated as

K=KfKrK = \frac{K_f}{K_r}

So,

K=103102=10K = \frac{10^3}{10^2} = 10

Now use the thermodynamic relation

ΔrG∘=−RTln⁡K\Delta_r G^\circ = -RT \ln K

With R=8.3×10−3 kJ mol−1 K−1R = 8.3 \times 10^{-3} \, \text{kJ mol}^{-1} \, \text{K}^{-1}, T=300 KT = 300 \, \text{K}, and ln⁡10=2.3\ln 10 = 2.3,

ΔrG∘=−(8.3×10−3)(300)(2.3)\Delta_r G^\circ = -(8.3 \times 10^{-3})(300)(2.3)

First,

8.3×10−3×300=2.498.3 \times 10^{-3} \times 300 = 2.49

Then,

2.49×2.3=5.7272.49 \times 2.3 = 5.727

So,

ΔrG∘≈−5.7  kJ mol−1\Delta_r G^\circ \approx -5.7 \; \text{kJ mol}^{-1}

Nearest integer:

−6  kJ mol−1-6 \; \text{kJ mol}^{-1}

The blank is written as (−)(-) _____ kJ mol−1\text{kJ mol}^{-1}, therefore the numerical entry is 6.

Common mistakes

  • Using Kf=103K_f = 10^3 directly as the equilibrium constant is incorrect. The reaction equilibrium constant must be obtained from K=KfKrK = \frac{K_f}{K_r}. Always combine forward and reverse constants before applying ΔrG∘=−RTln⁡K\Delta_r G^\circ = -RT \ln K.

  • Forgetting to convert RR into kJ mol−1 K−1\text{kJ mol}^{-1} \, \text{K}^{-1} leads to unit inconsistency. Since the answer is required in kJ mol−1\text{kJ mol}^{-1}, use R=8.3×10−3 kJ mol−1 K−1R = 8.3 \times 10^{-3} \, \text{kJ mol}^{-1} \, \text{K}^{-1} or convert at the end.

  • Entering −6-6 as the numerical answer is wrong here because the blank already includes (−)(-) before it. The question asks for the magnitude to be filled in that blank, so the required entry is 66.

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