The equilibrium concentrations of the species | vedclass

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(D) The reaction is $A + B \rightleftharpoons C + D$.At equilibrium,the concentrations are $[A] = 2 \, mol \, L^{-1}$,$[B] = 3 \, mol \, L^{-1}$,$[C]

Vedclass · Accepted Answer

$-1381.80$

Vedclass · Answer

(D) The reaction is $A + B ightleftharpoons C + D$.At equilibrium,the concentrations are $[A] = 2 \, mol \, L^{-1}$,$[B] = 3 \, mol \, L^{-1}$,$[C] = 10 \, mol \, L^{-1}$,and $[D] = 6 \, mol \, L^{-1}$.The equilibrium constant $K_{eq}$ is calculated as:$K_{eq} = \frac{[C][D]}{[A][B]} = \frac{10 	imes 6}{2 	imes 3} = \frac{60}{6} = 10$.The standard Gibbs free energy change $\Delta G^{\circ}$ is given by the formula:$\Delta G^{\circ} = -RT \ln K_{eq} = -2.303 \, RT \log K_{eq}$.Substituting the values $R = 2 \, cal \, mol^{-1} \, K^{-1}$,$T = 300 \, K$,and $K_{eq} = 10$:$\Delta G^{\circ} = -2.303 	imes 2 	imes 300 	imes \log(10)$.Since $\log(10) = 1$:$\Delta G^{\circ} = -2.303 	imes 600 = -1381.8 \, cal \, mol^{-1}$.

The equilibrium concentrations of the species in the reaction $A + B \rightleftharpoons C + D$ are $2, 3, 10$ and $6 \, mol \, L^{-1}$,respectively at $300 \, K$. $\Delta G^{\circ}$ for the reaction is $(R = 2 \, cal \, mol^{-1} \, K^{-1})$ (in $, cal$)

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