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ChemHardMain 2026 • 4th Apr - Shift 2

Standard Free Energy Change for Dissociation of Dinitrogen Pentoxide

For the following reaction at 50∘C50^\circ\text{C} and at 2 atm2\text{ atm} pressure,

2N2O5(g)⇌2N2O4(g)+O2(g)2\text{N}_2\text{O}_5(\text{g}) \rightleftharpoons 2\text{N}_2\text{O}_4(\text{g}) + \text{O}_2(\text{g})

N2O5\text{N}_2\text{O}_5 is 50%50\% dissociated.

The magnitude of standard free energy change at this temperature is xx.

x=x = _______  J mol−1\text{ J mol}^{-1} [Nearest integer].

Given : R=8.314 J mol−1 K−1R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}, log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48, ln⁡10=2.303\ln 10 = 2.303, ∘C+273=K{}^\circ\text{C} + 273 = \text{K}

Official Numerical Answer2474

Topics & Concepts

EquilibriumChemical EquilibriumGibbs Free Energy

Step-by-Step Solution

To find the magnitude of the standard free energy change (ΔG∘\Delta G^\circ) for the given dissociation reaction, we first need to determine the equilibrium constant KpK_p.

The given chemical equilibrium is: 2N2O5(g)⇌2N2O4(g)+O2(g)2\text{N}_2\text{O}_5(\text{g}) \rightleftharpoons 2\text{N}_2\text{O}_4(\text{g}) + \text{O}_2(\text{g})

Step 1: Equilibrium Composition

Let the initial number of moles of N2O5(g)\text{N}_2\text{O}_5(\text{g}) be 2 moles2\text{ moles}. Given that the degree of dissociation α=50%=0.50\alpha = 50\% = 0.50:

  • Moles of N2O5\text{N}_2\text{O}_5 at equilibrium =2(1−α)=2(1−0.50)=1 mole= 2(1 - \alpha) = 2(1 - 0.50) = 1\text{ mole}
  • Moles of N2O4\text{N}_2\text{O}_4 at equilibrium =2α=2(0.50)=1 mole= 2\alpha = 2(0.50) = 1\text{ mole}
  • Moles of O2\text{O}_2 at equilibrium =α=0.50 moles= \alpha = 0.50\text{ moles}

The total number of moles at equilibrium (ntotaln_{\text{total}}) is: ntotal=1+1+0.50=2.50 molesn_{\text{total}} = 1 + 1 + 0.50 = 2.50\text{ moles}

Step 2: Partial Pressures at Equilibrium

Given total pressure P=2 atmP = 2\text{ atm}:

  • pN2O5=(12.50)×2 atm=0.8 atmp_{\text{N}_2\text{O}_5} = \left(\frac{1}{2.50}\right) \times 2\text{ atm} = 0.8\text{ atm}
  • pN2O4=(12.50)×2 atm=0.8 atmp_{\text{N}_2\text{O}_4} = \left(\frac{1}{2.50}\right) \times 2\text{ atm} = 0.8\text{ atm}
  • pO2=(0.502.50)×2 atm=0.4 atmp_{\text{O}_2} = \left(\frac{0.50}{2.50}\right) \times 2\text{ atm} = 0.4\text{ atm}

Step 3: Calculation of Equilibrium Constant KpK_p

The expression for KpK_p is: Kp=(pN2O4)2⋅pO2(pN2O5)2K_p = \frac{(p_{\text{N}_2\text{O}_4})^2 \cdot p_{\text{O}_2}}{(p_{\text{N}_2\text{O}_5})^2}

Substitute the values of partial pressures: Kp=(0.8)2×0.4(0.8)2=0.4K_p = \frac{(0.8)^2 \times 0.4}{(0.8)^2} = 0.4

Step 4: Standard Free Energy Change (ΔG∘\Delta G^\circ)

The relation between ΔG∘\Delta G^\circ and KpK_p is: ΔG∘=−RTln⁡Kp=−2.303⋅R⋅T⋅log⁡10(Kp)\Delta G^\circ = -R T \ln K_p = -2.303 \cdot R \cdot T \cdot \log_{10}(K_p)

Given data:

  • Temperature, T=50∘C=50+273=323 KT = 50^\circ\text{C} = 50 + 273 = 323\text{ K}
  • Universal gas constant, R=8.314 J mol−1 K−1R = 8.314\text{ J mol}^{-1}\text{ K}^{-1}
  • ln⁡10=2.303\ln 10 = 2.303
  • log⁡2=0.30\log 2 = 0.30

Calculate log⁡10(Kp)\log_{10}(K_p): log⁡10(0.4)=log⁡10(410)=log⁡10(4)−log⁡10(10)=2log⁡10(2)−1\log_{10}(0.4) = \log_{10}\left(\frac{4}{10}\right) = \log_{10}(4) - \log_{10}(10) = 2\log_{10}(2) - 1 log⁡10(0.4)=2(0.30)−1=0.60−1=−0.40\log_{10}(0.4) = 2(0.30) - 1 = 0.60 - 1 = -0.40

Substitute all the values into the equation for ΔG∘\Delta G^\circ: ΔG∘=−2.303×8.314×323×(−0.40)\Delta G^\circ = -2.303 \times 8.314 \times 323 \times (-0.40) ΔG∘=2.303×8.314×323×0.40\Delta G^\circ = 2.303 \times 8.314 \times 323 \times 0.40 ΔG∘≈2473.81 J mol−1\Delta G^\circ \approx 2473.81\text{ J mol}^{-1}

Rounding to the nearest integer, the magnitude of standard free energy change is: x=2474 J mol−1x = 2474\text{ J mol}^{-1}

Standard Free Energy Change for Dissociation of Dinitrogen Pentoxide | Chemistry PYQ Solution - JEE Challenger