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ChemMedMain 2026 • 5th Apr - Shift 2

Calculate Standard Gibbs Free Energy Change for Reaction

Consider the following data for the reaction X2(g)+Y2(g)⇌2XY(g)X_2(g) + Y_2(g) \rightleftharpoons 2XY(g) at 600 K600\text{ K}. The ΔrG∘\Delta_r G^\circ (in kJ mol−1\text{kJ mol}^{-1}) for the reaction is :

CompoundΔfH600K∘(kJ mol−1)S600K∘(J mol−1K−1)XY(g)42200X2(g)8140Y2(g)80250\begin{array}{|c|c|c|} \hline \text{Compound} & \Delta_f H^{\circ}_{600\text{K}} (\text{kJ mol}^{-1}) & S^{\circ}_{600\text{K}} (\text{J mol}^{-1} \text{K}^{-1}) \\ \hline XY(g) & 42 & 200 \\ \hline X_2(g) & 8 & 140 \\ \hline Y_2(g) & 80 & 250 \\ \hline \end{array}

Options

A

−21000-21000

B

−10-10

Correct
C

−1000-1000

D

−9.012-9.012

Topics & Concepts

ThermodynamicsGibbs Free EnergyEnthalpyEntropy and Second Law

Step-by-Step Solution

To calculate the standard Gibbs free energy change (ΔrG∘\Delta_r G^\circ) for the given reaction:

X2(g)+Y2(g)⇌2XY(g)X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)

at T=600 KT = 600\text{ K}, we first determine the standard enthalpy change of the reaction (ΔrH∘\Delta_r H^\circ) and the standard entropy change of the reaction (ΔrS∘\Delta_r S^\circ).

Step 1: Calculate the Standard Enthalpy of Reaction (ΔrH∘\Delta_r H^\circ)

The standard enthalpy change of the reaction is given by: ΔrH∘=∑νpΔfH∘(products)−∑νrΔfH∘(reactants)\Delta_r H^\circ = \sum \nu_p \Delta_f H^\circ (\text{products}) - \sum \nu_r \Delta_f H^\circ (\text{reactants})

Substituting the given values into the equation: ΔrH∘=2⋅ΔfH∘(XY,g)−[ΔfH∘(X2,g)+ΔfH∘(Y2,g)]\Delta_r H^\circ = 2 \cdot \Delta_f H^\circ(XY, g) - \left[ \Delta_f H^\circ(X_2, g) + \Delta_f H^\circ(Y_2, g) \right] ΔrH∘=2×42 kJ mol−1−(8 kJ mol−1+80 kJ mol−1)\Delta_r H^\circ = 2 \times 42\text{ kJ mol}^{-1} - (8\text{ kJ mol}^{-1} + 80\text{ kJ mol}^{-1}) ΔrH∘=84−88=−4 kJ mol−1\Delta_r H^\circ = 84 - 88 = -4\text{ kJ mol}^{-1}

Step 2: Calculate the Standard Entropy of Reaction (ΔrS∘\Delta_r S^\circ)

The standard entropy change of the reaction is given by: ΔrS∘=∑νpS∘(products)−∑νrS∘(reactants)\Delta_r S^\circ = \sum \nu_p S^\circ (\text{products}) - \sum \nu_r S^\circ (\text{reactants})

Substituting the given values into the equation: ΔrS∘=2⋅S∘(XY,g)−[S∘(X2,g)+S∘(Y2,g)]\Delta_r S^\circ = 2 \cdot S^\circ(XY, g) - \left[ S^\circ(X_2, g) + S^\circ(Y_2, g) \right] ΔrS∘=2×200 J mol−1 K−1−(140 J mol−1 K−1+250 J mol−1 K−1)\Delta_r S^\circ = 2 \times 200\text{ J mol}^{-1}\text{ K}^{-1} - (140\text{ J mol}^{-1}\text{ K}^{-1} + 250\text{ J mol}^{-1}\text{ K}^{-1}) ΔrS∘=400−390=10 J mol−1 K−1\Delta_r S^\circ = 400 - 390 = 10\text{ J mol}^{-1}\text{ K}^{-1}

Converting ΔrS∘\Delta_r S^\circ into kJ mol−1 K−1\text{kJ mol}^{-1}\text{ K}^{-1}: ΔrS∘=101000 kJ mol−1 K−1=0.01 kJ mol−1 K−1\Delta_r S^\circ = \frac{10}{1000}\text{ kJ mol}^{-1}\text{ K}^{-1} = 0.01\text{ kJ mol}^{-1}\text{ K}^{-1}

Step 3: Calculate the Standard Gibbs Free Energy Change (ΔrG∘\Delta_r G^\circ)

Using the Gibbs free energy relation: ΔrG∘=ΔrH∘−TΔrS∘\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_r S^\circ

Substitute T=600 KT = 600\text{ K}, ΔrH∘=−4 kJ mol−1\Delta_r H^\circ = -4\text{ kJ mol}^{-1}, and ΔrS∘=0.01 kJ mol−1 K−1\Delta_r S^\circ = 0.01\text{ kJ mol}^{-1}\text{ K}^{-1}: ΔrG∘=−4 kJ mol−1−(600 K×0.01 kJ mol−1 K−1)\Delta_r G^\circ = -4\text{ kJ mol}^{-1} - (600\text{ K} \times 0.01\text{ kJ mol}^{-1}\text{ K}^{-1}) ΔrG∘=−4−6=−10 kJ mol−1\Delta_r G^\circ = -4 - 6 = -10\text{ kJ mol}^{-1}

Conclusion:

The standard Gibbs free energy change (ΔrG∘\Delta_r G^\circ) for the reaction is −10 kJ mol−1-10\text{ kJ mol}^{-1}, which corresponds to Option B.

Calculate Standard Gibbs Free Energy Change for Reaction | Chemistry PYQ Solution - JEE Challenger