Question

The entropy change involved in the conversion of 1 mole of liquid water at $373K$ to vapour will be:
Given: $△H_{vap}=2.257kJ/g$

• A. $118.5J{K}^{-1}mo{l}^{-1}$
• B. $108.9J{K}^{-1}mo{l}^{-1}$
• C. $150J{K}^{-1}mo{l}^{-1}$
• D. $130.6J{K}^{-1}mo{l}^{-1}$

Answer 1

The correct option is B $108.9J{K}^{-1}mo{l}^{-1}$

The enthalpy change of vaporisation for 1 mole of water can be calculated as follows:
$△H_{vap}$$=$$2.257kJ/g\times 18g/mol=40.6kJmo{l}^{-1}$
so,
$△S_{vap}=\frac{△H_{vap}}{T}=\frac{40.6\times {10}^{3}Jmo{l}^{-1}}{373K}=108.9J{K}^{-1}mo{l}^{-1}$