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A reaction has an equilibrium. constant of 0.98 at 300 K and 1.2 at 400 K Select the incorrect option
[Assume $∆ H^{0}$ and $∆ S^{0}$ to be temperature independent]

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The reaction is endothermic

Enthalpy change of reaction $Δ H_{R}^{\circ} = 1200 R \times \ln \frac{1.2}{0.98}$

At standard conditions reaction will be non spontaneous at 300 Kand spontaneous at 400 K

$Δ S_{R}^{\circ} = 3 R \ln \frac{1.2}{0.98}$

answer is D.

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Detailed Solution

$\begin{array}{l} Δ H > 0 + V E \\ Δ G = - R T \ln K \end{array}$

$ΔG = + v e$

$\ln [\begin{matrix} ↓ \\ \frac{K_{2}}{K_{1}} \\ ↑ \end{matrix}] = \frac{↓}{Δ H} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$Δ H \Rightarrow \ln (\frac{1.2}{0.88}) \times R \times \frac{300 \times y_{00}}{400 - 300}$

$\frac{T_{2} - T_{1}}{T_{1} T_{2}}$

$Δ H = 1200 \times R \times \ln [\frac{1.2}{0.98}] .$

$Δ G = Δ H - T Δ S .$

$\Rightarrow Δ S = \frac{Δ H - Δ G}{T} \Rightarrow \frac{1200 R}{300} \ln (\frac{1.2}{0.98}) + R \ln (0.98)$

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