The gas phase reaction 2A_{(g)} rightleftharp | vedclass

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Question

(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by $\Delta_{r} G^{\circ} = -RT \ln K_{p}$.Given $\Del

Vedclass · Accepted Answer

$166$

Vedclass · Answer

(B) The relationship between standard Gibbs free energy change and equilibrium constant is given by $\Delta_{r} G^{\circ} = -RT \ln K_{p}$.Given $\Delta_{r} G^{\circ} = 25200 \ J \ mol^{-1}$,$R = 8.3 \ J \ mol^{-1} \ K^{-1}$,and $T = 400 \ K$.Substituting the values: $25200 = -8.3 	imes 400 	imes \ln K_{p}$.$\ln K_{p} = -\frac{25200}{3320} \approx -7.59$.Converting to base $10$: $\log_{10} K_{p} = \frac{-7.59}{2.3} \approx -3.3$.$K_{p} = 10^{-3.3} = 5.01 	imes 10^{-4}$.For the reaction $2A_{(g)} ightleftharpoons A_{2(g)}$,$\Delta n = 1 - 2 = -1$.Using $K_{p} = K_{c}(RT)^{\Delta n}$,we have $K_{c} = K_{p}(RT)^{-\Delta n} = K_{p}(RT)^{1}$.$K_{c} = 5.01 	imes 10^{-4} 	imes (8.3 	imes 400) = 5.01 	imes 10^{-4} 	imes 3320 = 1.6632$.Expressing in terms of $10^{-2}$: $1.6632 = 166.32 	imes 10^{-2}$.Rounding to the nearest integer,we get $166$.

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