JEE Main 2026 — Chemistry Physical Chemistry

The standard Gibbs free energy change is given by:

$$\Delta G^{\theta} = -RT \ln K = -2.3 RT \log K$$ΔGθ=−RTlnK=−2.3RTlogK

First, calculate $\log K$logK:

$$\log K = \log(1.8 \times 10^{-7}) = \log(18 \times 10^{-8})$$logK=log(1.8×10−7)=log(18×10−8)

$$\log K = \log 18 - 8 = \log(2 \times 3^2) - 8$$logK=log18−8=log(2×32)−8

$$\log K = \log 2 + 2 \log 3 - 8$$logK=log2+2log3−8

Substituting the given values:

$$\log K = 0.30 + 2(0.48) - 8 = 0.30 + 0.96 - 8 = 1.26 - 8 = -6.74$$logK=0.30+2(0.48)−8=0.30+0.96−8=1.26−8=−6.74

Now, calculate $\Delta G^{\theta}$ΔGθ:

$$\Delta G^{\theta} = -2.3 \times 8.3 \times 300 \times (-6.74) = 38599.98 \text{ J mol}^{-1}$$ΔGθ=−2.3×8.3×300×(−6.74)=38599.98 J mol−1

Using the relation $\Delta G^{\theta} = \Delta H^{\theta} - T \Delta S^{\theta}$ΔGθ=ΔHθ−TΔSθ:

$$38599.98 = 28400 - 300 \Delta S^{\theta}$$38599.98=28400−300ΔSθ

$$300 \Delta S^{\theta} = 28400 - 38599.98 = -10199.98$$300ΔSθ=28400−38599.98=−10199.98

$$\Delta S^{\theta} = -\dfrac{10199.98}{300} \approx -34 \text{ J K}^{-1} \text{ mol}^{-1}$$ΔSθ=−30010199.98​≈−34 J K−1 mol−1

The magnitude of $\Delta S^{\theta}$ΔSθ is $34$34.

Answer: $34$34