229338
For the equilibrium $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \quad \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ at $1 \mathrm{~atm}$ and $298 \mathrm{~K}$ :
1 standard free energy change is equal to zero $\left(\Delta G^{\circ}=0\right)$.
2 free energy change is less than zero $(\Delta \mathrm{G}<0)$.
3 standard free energy change is less than zero $\left(\Delta \mathrm{G}^{\circ}<0\right)$.
4 standard free energy change is greater than zero $\left(\Delta \mathrm{G}^{\circ}>0\right)$.
Explanation:
The process - $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ is an endothermic process, $(\Delta \mathrm{H}=+\mathrm{ve})$ and entropy increases during this change $(\Delta \mathrm{S}=+\mathrm{ve})$. Hence this process is spontaneous at all temperatures above $0^{\circ} \mathrm{C}$ $(\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H})$, so $\Delta \mathrm{G}^{\mathrm{o}}$ is negative $\left(\Delta \mathrm{G}^{\mathrm{o}}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}\right)$. Thus, free energy change $(\Delta \mathrm{G})$ will be less than zero (negative) at $1 \mathrm{~atm}$ and $298 \mathrm{~K}$.
229338
For the equilibrium $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \quad \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ at $1 \mathrm{~atm}$ and $298 \mathrm{~K}$ :
1 standard free energy change is equal to zero $\left(\Delta G^{\circ}=0\right)$.
2 free energy change is less than zero $(\Delta \mathrm{G}<0)$.
3 standard free energy change is less than zero $\left(\Delta \mathrm{G}^{\circ}<0\right)$.
4 standard free energy change is greater than zero $\left(\Delta \mathrm{G}^{\circ}>0\right)$.
Explanation:
The process - $\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ is an endothermic process, $(\Delta \mathrm{H}=+\mathrm{ve})$ and entropy increases during this change $(\Delta \mathrm{S}=+\mathrm{ve})$. Hence this process is spontaneous at all temperatures above $0^{\circ} \mathrm{C}$ $(\mathrm{T} \Delta \mathrm{S}>\Delta \mathrm{H})$, so $\Delta \mathrm{G}^{\mathrm{o}}$ is negative $\left(\Delta \mathrm{G}^{\mathrm{o}}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}\right)$. Thus, free energy change $(\Delta \mathrm{G})$ will be less than zero (negative) at $1 \mathrm{~atm}$ and $298 \mathrm{~K}$.
