• question_answer 63) For the reaction, $2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)$at 300 K, the value of $\Delta {{G}^{o}}$is$~-690.9\text{ }R.$ The equilibrium constant value for the reaction at that temperature is (R is gas constant) A)  $10\,\text{at}{{\text{m}}^{-1}}$ B)  $10\,\text{atm}$ C)  10 D)  1

Solution :

For the reaction, $2S{{O}_{2}}(g)+{{O}_{2}}(g)2S{{O}_{3}}(g)$ Given that, $\Delta {{G}^{o}}=-690.9R,T=300\,K$ From the formula, $\Delta {{G}^{o}}=-RT\,\ln \,K$ $-690.9\,R=-R\times 300\times \ln \,K$ or $\ln \,k=\frac{690.9}{300}=2.303$ or $=2.303\,\log \,K=2.303$ or $K={{10}^{1}}=10$ unit of $K=(atm){{\Delta }^{n}}={{(atm)}^{2-3}}={{(atm)}^{-1}}$ $\therefore$ $K=10\,at{{m}^{-1}}$

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