Question

For the equilibrium $H_{2}O(l)\rightleftharpoons H_{2}O(g)$ at $1$ atm and $298K$,

• 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 $298K$.