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4 mole $H_{2} O (g)$ , 2.5 mole $H_{2}$ , 2.5 mole $CO (g)$ and 1 mole inert gas $He$ are kept at a total pressure of 10 bar in a flask containing C (graphite) at temperature T. Calculate $∆ G$ (ink J) of the following reaction at the above condition :
$C_{graphite} + H_{2} O (g) \to H_{2} (g) + CO (g)$
Given
$∆ G_{formation}^{°}, H_{2} O (g) = - 230 kJ {mol}^{- 1}$
$∆ G_{formation}^{°}, CO (g) = - 130 kJ {mol}^{- 1}$
$and 2.303 RT = 10 kJ . \log \frac{5}{4} = 0.1$

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answer is 102.

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Detailed Solution

$C (graphite) + H_{2} O (g) \to H_{2} (g) + CO (g)$ $\overset{Δ G_{f}^{°}}{\to} 10 mole$

4 2.5 2.5

$\begin{aligned} Δ G^{\circ} = Σ {(Δ G_{f}^{\circ})}_{P} - Σ {(Δ G_{f}^{\circ})}_{R} \\ = 130 - (- 230) \\ = 230 - 130 \\ = 100 kJ {mol}^{- 1} \end{aligned}$

$Δ G = Δ G^{\circ} + 2.303 R T \log Q$

$= 100 + 2.303 R T \ln \frac{P_{H_{2}} \times P_{CO}}{P_{H_{2} O}}$

$= 100 + 2.303 R T \frac{\ln {(\frac{2.5}{10} \times 10)}^{2}}{(\frac{4}{10} \times 10)}$

$= 100 + 10 \ln \frac{6.25}{4}$

$= 100 + 20 \log \frac{5}{4}$

$\begin{aligned} = 100 + 20 (\log 5 - 2 \log 2) \\ = 100 + 20 (0.7 - 0.6) \\ = 100 + 20 \times 0.1 \\ = 100 + 2 = 102 kJ \end{aligned}$

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