The equilibrium concentrations of the species | vedclass

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Question

(B) For the reaction $A + B ightleftharpoons C + D$,the equilibrium constant $K_c$ is calculated as:$K_c = \frac{[C][D]}{[A][B]} = \frac{10 	imes 1

Vedclass · Accepted Answer

$-1381.8$

Vedclass · Answer

(B) For the reaction $A + B ightleftharpoons C + D$,the equilibrium constant $K_c$ is calculated as:$K_c = \frac{[C][D]}{[A][B]} = \frac{10 	imes 15}{3 	imes 5} = \frac{150}{15} = 10$At equilibrium,the Gibbs free energy change $\Delta G$ is given by the relation $\Delta G = -RT \ln K_c$ or $\Delta G = -2.303 \ RT \log_{10} K_c$.Given $R = 2 \ cal \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,and $K_c = 10$:$\Delta G = -2.303 	imes 2 	imes 300 	imes \log_{10}(10)$$\Delta G = -2.303 	imes 600 	imes 1 = -1381.8 \ cal$.

The equilibrium concentrations of the species in the reaction $A + B \rightleftharpoons C + D$ are $3, 5, 10$ and $15 \ mol \ L^{-1}$ respectively at $300 \ K$. The $\Delta G$ for the reaction is (in $cal$)

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