Question

Calculate the enthalpy change on freezing of \(1.0\ mol\) of water at \(10.0°C\) to ice at \(-10.0°C\). \(∆_{fus}H = 6.03\ kJ mol^{–1}\) at \(0°C\).
\(C_p [H_2O(l)] = 75.3\ J mol^{–1}K^{–1}\)
\(C_p [H_2O(s)] = 36.8 \ J mol^{–1}K^{–1}\)

Answer 1

Given:

• \(\Delta_\text{fus} H = 6.03\ \text{kJ mol}^{-1}\) at 0 °C (fusion, solid → liquid).
• H2O(l) = 75.3 J mol^{-1}K^{-1}
• H2O(s) = 36.8 J mol^{-1} {K}^{-1}.

Process (in steps):

1. Cool liquid water: {H2O(l, 10 °C) -> H2O(l, 0 °C)}
2. Freeze at 0 °C: {H2O(l, 0 °C) -> H2O(s, 0 °C)}.
3. Cool ice: {H2O(s, 0 °C) -> H2O(s, −10 °C)}.

Step 1: Cool liquid from 10 °C to 0 °C

\(\Delta T = 0 - 10 = -10\ \text{K}\)

\[ q_1 = n C_p \Delta T = 1 \times 75.3 \times (-10) = -753\ \text{J} = -0.753\ \text{kJ} \]

Step 2: Freezing at 0 °C

Given fusion (melting) enthalpy: \(\Delta_\text{fus} H = +6.03\ \text{kJ mol}^{-1}\).

For freezing (reverse process):

\[ q_2 = -\Delta_\text{fus} H = -6.03\ \text{kJ} \]

Step 3: Cool ice from 0 °C to −10 °C

\(\Delta T = -10\ \text{K}\)

\[ q_3 = n C_p \Delta T = 1 \times 36.8 \times (-10) = -368\ \text{J} = -0.368\ \text{kJ} \]

Total enthalpy change:

\[ \Delta H = q_1 + q_2 + q_3 = (-0.753) + (-6.03) + (-0.368)\ \text{kJ} = -7.151\ \text{kJ} \]

The enthalpy change for freezing 1.0 mol of water at 10.0 °C to ice at −10.0 °C is \[ \boxed{\Delta H \approx -7.15\ \text{kJ mol}^{-1}} \] (heat released by the system).