Chemical Thermodynamics - Gibbs Energy Change and Equilibrium

Calculate the equilibrium constant $K$ for a reaction at $298 \text{ K}$ if the standard Gibbs energy change $\Delta_r G^{\circ}$ is $-4.606 \text{ kJ mol}^{-1}$. (Given: $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$)

Given: $\Delta_r G^{\circ} = -4.606 \text{ kJ mol}^{-1} = -4606 \text{ J mol}^{-1}$, $T = 298 \text{ K}$, $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$. Use the formula: $\Delta_r G^{\circ} = -2.303 RT \log K$. $-4606 = -2.303 \times 8.314 \times 298 \times \log K$ $\log K = \frac{4606}{2.303 \times 8.314 \times 298} \approx \frac{4606}{5705.8} \approx 0.807$ $K = 10^{0.807} \approx 6.41$

The negative value of $\Delta_r G^{\circ}$ indicates that the reaction is spontaneous under standard conditions, resulting in an equilibrium constant $K > 1$.

For the equilibrium $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$ at $500 \text{ K}$, $\Delta_r H^{\circ} = 124 \text{ kJ mol}^{-1}$ and $\Delta_r S^{\circ} = 170 \text{ J K}^{-1} \text{ mol}^{-1}$. Determine if the reaction is spontaneous at this temperature.

Convert $\Delta_r H^{\circ}$ to Joules: $124 \times 10^3 \text{ J mol}^{-1}$. Use the Gibbs-Helmholtz equation: $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$ $\Delta G^{\circ} = 124000 - (500 \times 170)$ $\Delta G^{\circ} = 124000 - 85000 = 39000 \text{ J mol}^{-1} = 39 \text{ kJ mol}^{-1}$ Since $\Delta G^{\circ} > 0$, the reaction is non-spontaneous at $500 \text{ K}$.

Even though the entropy increases ($\Delta S > 0$), the high positive enthalpy change (endothermic) makes the overall $\Delta G$ positive at this specific temperature, meaning the reaction requires more energy input to proceed.