Question

If $2.0$ mole each of $A$ and $B$ were allowed to come to equilibrium at $300$K for the reaction $A+B\to C+D$, $\Delta G^o=460$ cal. The equilibrium concentration ratio of $C$ to $A$ will be:

• A. $1$
• B. $\frac{1}{2}$
• C. $0.68$
• D. $0.46$

Answer 1

Answer: (C)

$0.68$

$\Delta G^o=-RTln{K}_{eq}$

$460=-2\times 300ln{K}_{eq}$

$K_{eq}=e^{-0.766}$

$K_{eq}=0.464$

From the reaction $A+B\rightleftharpoons C+D$

Initial concentration of $A=2,B=2,C=0,D=0$

Equilibrium concentration of $A=2-x,B=2-x,C=x,D=x$

$K_{eq}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}=\frac{[C{]}^2}{[A{]}^2}$

$0.464=\frac{\left[x\right]\left[x\right]}{[2-x][2-x]}=\frac{x^2}{4-4x+x^2}$

$\therefore \frac{\left[C\right]}{\left[A\right]}=0.681$