Question

Calculate the enthalpy change on freezing of 1.0 mol of water at ${10.0}^{\circ }C$ to ice at $-{10.0}^{\circ }C.\Delta fusH=6.03kJmo{l}^{-1}at{0}^{\circ }C$.
$\left(C_p\right[H_{2}O\left(l\right)]=75.3Jmo{l}^{-1}{K}^{-1}$
$C_p\left[H_{2}O\right(s\left)\right]=36.8Jmo{l}^{-1}{K}^{-1}$

Answer 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 ${10}^{\circ }C$ to 1 mol of water at $0^{\circ }C$.

$\Delta H_1=n{C}_P\Delta T$

$\Delta H_1=1\times 75.3\times 10=-0.753\frac{KJ}{mole}$
(b) Energy change involved in the transformation of 1 mol of water at $0^{\circ }$ to 1 mol of ice at $0^{\circ }C$.

$\Delta H_2=\Delta H_{fusion}=-6.03\frac{KJ}{mole}$
(c) Energy change involved in the transformation of 1 mol of ice at $0^{\circ }C$ to 1 mol of ice at $-{10}^{\circ }C$.
$\Delta H_3=n{C}_P\Delta T=1\times -36.8\times 10=-0.368\frac{KJ}{mole}$

Total Heat Change= $\Delta H_1+\Delta H_2+\Delta H_3$

= $(-0.753-6.03-0.368)\frac{KJ}{mole}$ $\therefore Totalenthalpy=-7.151\frac{KJ}{mole}$

Hence, the enthalpy change involved in the transformation is $-7.151kJmo{l}^{-1}$.