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ChemMedMain 2026 • 6th Apr - Shift 2

Calculate Standard Gibbs Energy Change for Dissociation Equilibrium

In a closed flask at 600 K600\text{ K}, one mole of X2Y4(g)\text{X}_2\text{Y}_4(\text{g}) attains equilibrium as given below: X2Y4(g)⇌2XY2(g)\text{X}_2\text{Y}_4(\text{g}) \rightleftharpoons 2\text{XY}_2(\text{g}) At equilibrium, 75%75\% X2Y4(g)\text{X}_2\text{Y}_4(\text{g}) was dissociated and the total pressure is 1 atm1\text{ atm}. The magnitude of ΔrG∘\Delta_r G^\circ (in kJ mol−1\text{kJ mol}^{-1}) at this temperature is ______. (Nearest Integer)

(Given : R=8.3 J mol−1K−1;ln⁡10=2.3,log⁡2=0.3,log⁡3=0.48,log⁡5=0.69,log⁡7=0.84)(\text{Given : } R = 8.3\text{ J mol}^{-1}\text{K}^{-1}; \ln 10 = 2.3, \log 2 = 0.3, \log 3 = 0.48, \log 5 = 0.69, \log 7 = 0.84)

Official Numerical Answer8

Topics & Concepts

ThermodynamicsGibbs Free EnergyChemical Equilibrium

Step-by-Step Solution

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

X2Y4(g)⇌2XY2(g)\text{X}_2\text{Y}_4(\text{g}) \rightleftharpoons 2\text{XY}_2(\text{g})

Step 1: Calculate the mole fractions and equilibrium partial pressures

Let the initial moles of X2Y4(g)\text{X}_2\text{Y}_4(\text{g}) be n0=1 molen_0 = 1\text{ mole}. Given that 75%75\% of X2Y4(g)\text{X}_2\text{Y}_4(\text{g}) dissociates at equilibrium, the degree of dissociation is: α=0.75\alpha = 0.75

At equilibrium:

  • Moles of X2Y4(g)=1−α=1−0.75=0.25 mol\text{X}_2\text{Y}_4(\text{g}) = 1 - \alpha = 1 - 0.75 = 0.25\text{ mol}
  • Moles of XY2(g)=2α=2×0.75=1.50 mol\text{XY}_2(\text{g}) = 2\alpha = 2 \times 0.75 = 1.50\text{ mol}

The total number of moles at equilibrium (ntotaln_{\text{total}}) is: ntotal=0.25+1.50=1.75 moln_{\text{total}} = 0.25 + 1.50 = 1.75\text{ mol}

Given that the total pressure P=1 atmP = 1\text{ atm}, the partial pressures of the gases are: pX2Y4=(0.251.75)×1 atm=17 atmp_{\text{X}_2\text{Y}_4} = \left(\frac{0.25}{1.75}\right) \times 1\text{ atm} = \frac{1}{7}\text{ atm}

pXY2=(1.501.75)×1 atm=67 atmp_{\text{XY}_2} = \left(\frac{1.50}{1.75}\right) \times 1\text{ atm} = \frac{6}{7}\text{ atm}

Step 2: Calculate the equilibrium constant (KpK_p)

Kp=(pXY2)2pX2Y4=(67)217=364917=367K_p = \frac{(p_{\text{XY}_2})^2}{p_{\text{X}_2\text{Y}_4}} = \frac{\left(\frac{6}{7}\right)^2}{\frac{1}{7}} = \frac{\frac{36}{49}}{\frac{1}{7}} = \frac{36}{7}

Step 3: Calculate ln⁡Kp\ln K_p

Using the conversion ln⁡x=ln⁡10×log⁡10x\ln x = \ln 10 \times \log_{10} x: log⁡10(Kp)=log⁡10(367)=log⁡10(36)−log⁡10(7)\log_{10}(K_p) = \log_{10}\left(\frac{36}{7}\right) = \log_{10}(36) - \log_{10}(7)

Since log⁡10(36)=log⁡10(22×32)=2log⁡10(2)+2log⁡10(3)\log_{10}(36) = \log_{10}(2^2 \times 3^2) = 2\log_{10}(2) + 2\log_{10}(3): log⁡10(36)=2(0.3)+2(0.48)=0.6+0.96=1.56\log_{10}(36) = 2(0.3) + 2(0.48) = 0.6 + 0.96 = 1.56

Thus: log⁡10(367)=1.56−0.84=0.72\log_{10}\left(\frac{36}{7}\right) = 1.56 - 0.84 = 0.72

Now, using ln⁡10=2.3\ln 10 = 2.3: ln⁡Kp=2.3×0.72=1.656\ln K_p = 2.3 \times 0.72 = 1.656

Step 4: Calculate the magnitude of ΔrG∘\Delta_r G^\circ

The standard Gibbs energy change is given by: ΔrG∘=−RTln⁡Kp\Delta_r G^\circ = -R T \ln K_p

Substitute the given values (R=8.3 J mol−1K−1R = 8.3\text{ J mol}^{-1}\text{K}^{-1}, T=600 KT = 600\text{ K}): ΔrG∘=−8.3×600×1.656 J mol−1\Delta_r G^\circ = -8.3 \times 600 \times 1.656\text{ J mol}^{-1} ΔrG∘=−4980×1.656 J mol−1=−8246.88 J mol−1=−8.24688 kJ mol−1\Delta_r G^\circ = -4980 \times 1.656\text{ J mol}^{-1} = -8246.88\text{ J mol}^{-1} = -8.24688\text{ kJ mol}^{-1}

The magnitude of ΔrG∘\Delta_r G^\circ is: ∣ΔrG∘∣=8.24688 kJ mol−1≈8 kJ mol−1|\Delta_r G^\circ| = 8.24688\text{ kJ mol}^{-1} \approx 8\text{ kJ mol}^{-1}

Final Answer: The magnitude of ΔrG∘\Delta_r G^\circ is 8

Calculate Standard Gibbs Energy Change for Dissociation Equilibrium | Chemistry PYQ Solution - JEE Challenger