Question

$C_2{H}_6\left(g\right)\rightleftharpoons C_2{H}_4\left(g\right)+H_2\left(g\right);K_p=0.05\text{atm}$.
The value of $\Delta G^{\circ }$ of the reaction at $627{}^{\circ }\text{C}$ would be:
$(\text{Given}:\ln (0.05)\approx -3;R=8.3{\text{J K}}^{-1}{\text{mol}}^{-1})$

• A. $11.21{\text{kJ mol}}^{-1}$
• B. $22.41{\text{kJ mol}}^{-1}$
• C. $33.61{\text{kJ mol}}^{-1}$
• D. $27.98{\text{kJ mol}}^{-1}$

Answer 1

The correct option is B $22.41{\text{kJ mol}}^{-1}$

$\Delta G^{\circ }=-RT\ln K_p=-\left(8.3{\text{J K}}^{-1}{\text{mol}}^{-1}\right)\left(900\text{K}\right)\ln \left(0.05\right)=22410{\text{J mol}}^{-1}=22.41{\text{kJ mol}}^{-1}$