Question

For the equilibrium $H_2O(l) \rightleftharpoons H_2O(g)$ at $1$ atm and $298\, K$,

• A. standard free energy change is equal to zero ($\Delta G^\circ = 0$)
• B. free energy change is less than zero ($\Delta G < 0$)
• C. standard free energy change is less than zero ($\Delta G^{\circ} < 0$)
• D. standard free energy change is greater than zero ($\Delta G^{\circ} > 0$)

Answer 1

Answer: (B)

The process, $H _{2} O (l) \rightleftharpoons H _{2} O (g)$, is an endothermic process ( $\Delta H=+$ ve) and entropy increase during this change $(\Delta S=+$ ve). Hence, this process is spontaneous at all temperatures above $0^{\circ} C (T \Delta S>\Delta H$, so, $\Delta G$ is negative, $\Delta G=\Delta H-T \Delta S)$. Thus free energy change $(\Delta G)$ will be less than zero $(- ve )$ at $1$ atm and $298 \,K$.