For the reaction, 2 SO_{2(g)} + O_{2(g)} righ | vedclass

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(A) For the reaction, $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$.Given that, $\Delta G^{\circ} = -690.9 R$ and $T = 300 \ K$.The relation

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$10 \ atm^{-1}$

Vedclass · Answer

(A) For the reaction, $2 SO_{2(g)} + O_{2(g)} ightleftharpoons 2 SO_{3(g)}$.Given that, $\Delta G^{\circ} = -690.9 R$ and $T = 300 \ K$.The relationship between standard Gibbs free energy change and equilibrium constant is $\Delta G^{\circ} = -RT \ln K$.Substituting the values: $-690.9 R = -R 	imes 300 	imes \ln K$.Dividing both sides by $-R$: $690.9 = 300 	imes \ln K$.$\ln K = \frac{690.9}{300} = 2.303$.Since $\ln K = 2.303 \log K$, we have $2.303 \log K = 2.303$.Therefore, $\log K = 1$, which implies $K = 10^1 = 10$.The unit of $K$ is $(atm)^{\Delta n}$, where $\Delta n = 2 - (2 + 1) = -1$.Thus, $K = 10 \ atm^{-1}$.

For the reaction, $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ at $300 \ K$, the value of $\Delta G^{\circ}$ is $-690.9 R$. The equilibrium constant value for the reaction at that temperature is ($R$ is gas constant).

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