Question

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

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

Answer 1

Answer: (B)

373.4 K

Answer 2

${\Delta G = \Delta H - T \Delta S \Delta G =0 }$
Given $\Delta H = T \Delta S,$
$T = \frac{40.63 \times 10^{3}}{108.8 } = 373.4 K$