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MCQEasyJEE 2024Gibbs Free Energy & Equilibrium Constant

JEE Chemistry 2024 Question with Solution

Consider the following reaction at 298 K298 \, \text{K}:

32O2(g)→O3(g),Kp=2.47×10−29.\frac{3}{2} O_2(g) \rightarrow O_3(g), \quad K_p = 2.47 \times 10^{-29}.

ΔG∘\Delta G^\circ for the reaction is kJ\text{kJ}. (Given R=8.314 J K−1 mol−1R = 8.314 \, \text{J K}^{-1} \, \text{mol}^{-1})

  • A

    150150

  • B

    163163

  • C

    175175

  • D

    190190

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

  • Reaction: 32O2(g)⇌O3(g)\frac{3}{2} O_2(g) \rightleftharpoons O_3(g)
  • Kp=2.47×10−29K_p = 2.47 \times 10^{-29}
  • T=298 KT = 298 \, \text{K}
  • R=8.314×10−3 kJ mol−1 K−1R = 8.314 \times 10^{-3} \, \text{kJ mol}^{-1} \, \text{K}^{-1}

Find: ΔrG∘\Delta_r G^\circ

Use the standard relation:

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

Substitute the given values:

ΔrG∘=−(8.314×10−3)(298)ln⁡(2.47×10−29)\Delta_r G^\circ = -(8.314 \times 10^{-3})(298) \ln(2.47 \times 10^{-29})

From the solution working:

ln⁡(2.47×10−29)=−65.87\ln(2.47 \times 10^{-29}) = -65.87

Therefore,

ΔrG∘=−(8.314×10−3×298×−65.87)=163.19 kJ\Delta_r G^\circ = -(8.314 \times 10^{-3} \times 298 \times -65.87) = 163.19 \, \text{kJ}

Therefore, the standard Gibbs free energy change is 163.19 kJ163.19 \, \text{kJ}, so the correct option is B.

Direct Substitution

Given: Kp=2.47×10−29K_p = 2.47 \times 10^{-29}, T=298 KT = 298 \, \text{K}

Find: ΔG∘\Delta G^\circ

Apply directly:

ΔG∘=−RTln⁡Kp\Delta G^\circ = -RT \ln K_p =−8.314×10−3×298×ln⁡(2.47×10−29)= -8.314 \times 10^{-3} \times 298 \times \ln(2.47 \times 10^{-29}) =−8.314×10−3×298×(−65.87)= -8.314 \times 10^{-3} \times 298 \times (-65.87) =163.19 kJ= 163.19 \, \text{kJ}

Hence, the correct option is B.

Common mistakes

  • Using ΔG∘=−RTln⁡Kp\Delta G^\circ = -RT \ln K_p but missing that Kp<1K_p < 1 makes ln⁡Kp\ln K_p negative. This changes the sign of ΔG∘\Delta G^\circ. Always check the sign of the logarithm before concluding the sign of the Gibbs free energy.

  • Using R=8.314 J mol−1 K−1R = 8.314 \, \text{J mol}^{-1} \, \text{K}^{-1} and reporting the answer directly in kJ without conversion. The solution uses R=8.314×10−3 kJ mol−1 K−1R = 8.314 \times 10^{-3} \, \text{kJ mol}^{-1} \, \text{K}^{-1}. Keep units consistent throughout.

  • Replacing ln⁡\ln with log⁡\log without multiplying by 2.3032.303. Natural logarithm is required in ΔG∘=−RTln⁡Kp\Delta G^\circ = -RT \ln K_p. If common logarithm is used, convert properly.

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