Question

One mole of ice is converted into water at $273$K. The entropies of $H_{2}O\left(s\right)$ and $H_{2}O\left(l\right)$ are $38.20$ and $60.01$ $J.mo{l}^{-1}$ $K^{-1}$ respectively. The enthalpy change for the conversion is:

• A. $3kJmo{l}^{-1}$
• B. $4kJmo{l}^{-1}$
• C. $5kJmo{l}^{-1}$
• D. $6kJmo{l}^{-1}$

Answer 1

Answer: (D)

$6kJmo{l}^{-1}$

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

At equilibrium, $\Delta G=0$

$\therefore \Delta H=T\Delta S$

$\Delta H=273\times (60.01-38.20)$

$=5954.13J$$mo{l}^{-1}=5.954$ $kJmo{l}^{-1}$

$\approx 6$ $kJ.mo{l}^{-1}$.