The variation of equilibrium constant with temperature is given below
$\begin{array}{lc} Temperature & Equilibrium Constant \\ T_{1} = 25^{\circ} C & K_{1} = 10 \\ T_{2} = 100^{\circ} C & K_{2} = 100 \end{array}$
$\begin{array}{l} The values of {ΔH}^{0}, {ΔG}^{0} at T_{1} and {ΔG}^{0} at T_{2} (in kJ {mol}^{- 1}) respectively, are close to \\ [use R = 8.314 {JK}^{- 1} {mol}^{- 1}] \end{array}$

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$28 .4, - 5 .71 and - 14 .29$

$28 .4, - 7 .14 and - 5 .71$

$0 .64, - 5 .71 and - 14 .29$

$0 .64, - 7 .14 and - 5 .71$

answer is C.

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Detailed Solution

$\log_{10} (\frac{K_{2}}{K_{1}}) = \frac{Δ H^{0}}{2.303 R} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]$

$\begin{array}{l} \Rightarrow \log (10) = \frac{Δ H^{0}}{2.303 R} [\frac{1}{298} - \frac{1}{373}] \\ \Rightarrow \frac{Δ H^{0}}{2.303 \times 8.314} = [\frac{298 \times 373}{75}] \\ \Rightarrow Δ H^{0} = 28377.1 J / mol \end{array}$

$= 28.4 \frac{k J}{m o l}$

at $T_{1}$

$\begin{array}{l} Δ G^{0} = - R T_{1} \times 2.303 \log_{10} K_{1} \\ = - 2.303 \times 8.314 \times 298 \log_{10}^{10} \\ = - 5.71 kJ / mol \\ At T_{2} \\ Δ G^{0} = - 2.303 R T_{2} \log K_{2} \\ = - 2.303 \times 8.314 \times 373 \times 2 \end{array}$