Question

When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be $\left(\Delta H_{vap}\right(forwater=4.08kJmo{l}^{-1})$

• A. $4.08kJ,10.93J{K}^{-1}$
• B. $-4.08kJ,-40.8j{k}^{-1}$
• C. $-4.08kJ,-10.93J{K}^{-1}$
• D. $4.08kJ,40.8J{K}^{-1}$

Answer 1

The correct option is C

$-4.08kJ,-10.93J{K}^{-1}$

$\Delta H=(-40.8)\times \frac{1.8}{18}=-4.08kJ;$
$\Delta S=\frac{\Delta H}{T}=\frac{-4.08\times {10}^3}{373.15}=-10.93J{K}^{-1}$