Question

For vaporization of water at $1$ atmospheric pressure, the values of $\Delta H$ and $\Delta S$ are $40 \cdot 63\, kJ \,mol ^{-1}$ and $108 \cdot 8 \,JK ^{-1} mol ^{-1}$, respectively. The temperature when Gibbs energy change $(\Delta G)$ for this transformation will be zero, is:

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

Answer 1

Answer: (D)

$H _{2} O (\ell) \stackrel{\text { 1 atm }}{\rightleftharpoons} H _{2} O ( g )$
$\Delta H =40630\, J\, mol ^{-1}$
$\Delta S =108 \cdot 8 \,JK ^{-1} mol ^{-1}$
$\Delta G =\Delta H - T \Delta S$ when $\Delta G=0$
$\Delta H-T \Delta S=0$
$T =\frac{\Delta H }{\Delta S }=\frac{40630 \,J\, mol ^{-1}}{108 \cdot 8 J mol ^{-1}}$
$=373 \cdot 4 K$