Consider the reaction A rightleftharpoons B a | vedclass

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(D) The equilibrium constant $K_{eq}$ for the reaction $A \rightleftharpoons B$ is given by $K_{eq} = \frac{P_B}{P_A}$.From the graph,at $1000 \ K$,$P

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(D) The equilibrium constant $K_{eq}$ for the reaction $A ightleftharpoons B$ is given by $K_{eq} = \frac{P_B}{P_A}$.From the graph,at $1000 \ K$,$P_B = 10 \ bar$ and $P_A = 1 \ bar$,so $K_{1000} = \frac{10}{1} = 10$.At $2000 \ K$,$P_B = 100 \ bar$ and $P_A = 1 \ bar$,so $K_{2000} = \frac{100}{1} = 100$.The standard Gibbs energy change is given by $\Delta G^0 = -RT \ln(K_{eq})$.Therefore,the ratio is:$\frac{\Delta G_{1000}^0}{\Delta G_{2000}^0} = \frac{-R(1000) \ln(10)}{-R(2000) \ln(100)}$$= \frac{1000 	imes \ln(10)}{2000 	imes \ln(10^2)}$$= \frac{1000 	imes \ln(10)}{2000 	imes 2 \ln(10)}$$= \frac{1000}{4000} = 0.25$.

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