Question

For the reaction, $C_2{H}_6\left(g\right)\rightleftharpoons C_2{H}_4\left(g\right)+H_2\left(g\right)$, $K_p=0.05$ atm, the value of $\Delta G^o$ of the reaction at ${627}^{o}C$ would be:

• A. $11.19$ $kJmo{l}^{-1}$
• B. $22.40$ $kJmo{l}^{-1}$
• C. $33.57$ $kJmo{l}^{-1}$
• D. $27.98$ $kJmo{l}^{-1}$

Answer 1

The correct option is B $22.40$ $kJmo{l}^{-1}$

The relationship between the standard free energy change $\left(\Delta G^o\right)$ and the equilibrium constant $\left(K_p\right)$ is $\Delta G^o=-RTln{K}_p$.

Here, $R$ is the ideal gas constant and $T$ is the temperature.

But, $K_p=0.05atm,T=627+273=900K,R=8.314Jmo{l}^{-1}{K}^{-1}$.

Substituting values in the above expression, we get

$\Delta G^o=-RTln{K}_p=-8.314\times 900\times ln\left(0.05\right)\approx 22400Jmo{l}^{-1}=22.4kJmo{l}^{-1}$.

Hence, the standard free energy change $\left(\Delta G^o\right)=22.4kJmo{l}^{-1}$.