For the reaction A_{(g)} rightleftharpoons B_ | vedclass

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(C) The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K_{eq}$.Given $\De

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$20$

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(C) The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^{\circ} = -RT \ln K_{eq}$.Given $\Delta G^{\circ} = -9.478 \ kJ \ mol^{-1} = -9478 \ J \ mol^{-1}$.Substituting the values: $-9478 = -8.314 	imes 495 	imes \ln K_{eq}$.$\ln K_{eq} = \frac{9478}{8.314 	imes 495} \approx 2.303$.Since $\ln 10 = 2.303$,we have $K_{eq} = 10$.For the reaction $A_{(g)} ightleftharpoons B_{(g)}$:At $t = 0$,moles of $A = 22$,moles of $B = 0$.At equilibrium,moles of $A = 22 - x$,moles of $B = x$.$K_{eq} = \frac{[B]}{[A]} = \frac{x}{22 - x} = 10$.$x = 220 - 10x \implies 11x = 220 \implies x = 20$.The amount of $B$ at equilibrium is $20 \ mmol$.

$1/T \text{ (K}^{-1})$	$\log_{10} K_p$
$0.05$	$3.5$
$0.06$	$2.5$
$0.07$	$1.5$

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