Question

For vaporization of water at $1$ atmospheric pressure, the values of $\Delta$H and $\Delta$S are $40.63kJmo{l}^{-1}and$108.8 JK${}^{-1}$mol${}^{-1}$, respectively. The temperature when Gibb's energy change ($\Delta$G) for this transformation will be zero, is :

• A. $273.4K$
• B. $393.4K$
• C. $373.4K$
• D. $293.4K$

Answer 1

The correct option is C $373.4K$

According to Gibb's equation, $\Delta$G = $\Delta$H - T$\Delta$S when $\Delta$G = 0, $\Delta$H = T$\Delta$S
Given, $\Delta$H = 40.63 kJ mol${}^{-1}$ = 40.63 x 10${}^3$ J mol${}^{-1}$
$\Delta$S = 108.8 J K${}^{-1}$ mol${}^{-1}$
$\therefore$ T = $\frac{\Delta H}{\Delta S}=\frac{40.63\times {10}^3}{108.8}$ = $373.43K$
The temperature when Gibb's energy change (G) for this transformation will be zero, is $373.4K$
The zero value of the Gibb's free energy change indicates equilibrium state.