Chemistry - Thermodynamics - II Question with Solution | TestHub

For the reaction:

$$CO(g)+H_{2}O(g)\rightleftharpoons C{O}_2(g)+H_2(g)$$

The standard Gibbs free energy change, $\Delta G^{\circ }$, is calculated as:

$$\Delta G^{\circ }=\sum \Delta G_f^{\circ }(products)-\sum \Delta G_f^{\circ }(reactants)$$

$$=[(-393.9)+0]-[(-138.3)+(-230.7)]$$

$$=-393.9+369.0$$

$$=-24.9kJmo{l}^{-1}$$

Using the relation:

$$\Delta G^{\circ }=-RT\ln K_p$$

We can calculate $\ln K_p$:

$$\ln K_p=\frac{24.9\times 10^3}{8.3\times 300}=10$$

Therefore, $K_p$ is:

$$K_p=e^{10}=2.2\times 10^4$$