Calculate the enthalpy change on freezing of $1.0 \mathrm{~mol}$ of water at $10.0^{\circ} \mathrm{C}$ to ice at $-10.0^{\circ} \mathrm{C} . \Delta_{\text {fus }} H=6.03 \mathrm{~kJ} \mathrm{~mol}^{-1}$ at $0^{\circ} \mathrm{C}$.
$\mathrm{C}_{p}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\right]=75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
$C_{\rho}\left[\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]=36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$
Total enthalpy change involved in the transformation is the sum of the following changes:
(a) Energy change involved in the transformation of 1 mol of water at 10°C to 1 mol of water at 0°C.
(b) Energy change involved in the transformation of 1 mol of water at 0° to 1 mol of ice at 0°C.
(c) Energy change involved in the transformation of 1 mol of ice at 0°C to 1 mol of ice at –10°C.
Total $\Delta \mathrm{H}=C_{p}\left[\mathrm{H}_{2} \mathrm{OCl}\right] \Delta T+\Delta H_{\text {freczing }}+C_{p}\left[\mathrm{H}_{2} \mathrm{O}_{(s)}\right] \Delta T$
$=\left(75.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(0-10) \mathrm{K}+\left(-6.03 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\right)+\left(36.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)(-10-0) \mathrm{K}$
$=-753 \mathrm{~J} \mathrm{~mol}^{-1}-6030 \mathrm{~J} \mathrm{~mol}^{-1}-368 \mathrm{~J} \mathrm{~mol}^{-1}$
$=-7151 \mathrm{~J} \mathrm{~mol}^{-1}$
$=-7.151 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Hence, the enthalpy change involved in the transformation is $-7.151 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
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