Question

Chloroform has $\Delta H_{vap}=29.2$ KJ/mol and boiling point ${61.2}^o$C. What is $\Delta S_{vap}$ of chloroform?

• A. $977.1$J $k^{-1}mo{l}^{-1}$
• B. $174.2$ J $k^{-1}$ $mo{l}^{-1}$
• C. $-87.3$ J $k^{-1}$ $mo{l}^{-1}$
• D. $87.3$ J $k^{-1}$ $mo{l}^{-1}$

Answer 1

The correct option is D $87.3$ J $k^{-1}$ $mo{l}^{-1}$

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = 0 for equilibrium reaction

$\frac{\Delta H}{T}=\Delta S$

$\frac{29.2}{273+61.2}=\Delta S$

= 0.0873 KJ $K^{-1}mo{l}^{-1}$

= 87.3 J $K^{-1}mo{l}^{-1}$