Calculate the enthalpy change on freezing of $1.0 \ mol$ of water at $10.0^{\circ} C$ to ice at $-10.0^{\circ} C$. Given: $\Delta_{fus} H = 6.03 \ kJ \ mol^{-1}$ at $0^{\circ} C$,$C_p [H_2 O_{(l)}] = 75.3 \ J \ mol^{-1} \ K^{-1}$,$C_p [H_2 O_{(s)}] = 36.8 \ J \ mol^{-1} \ K^{-1}$.
The total enthalpy change is the sum of the following three steps:
$(a)$ Cooling $1 \ mol$ of water from $10^{\circ} C$ to $0^{\circ} C$: $\Delta H_1 = n C_p [H_2 O_{(l)}] \Delta T = 1 \ mol \times 75.3 \ J \ mol^{-1} \ K^{-1} \times (0 - 10) \ K = -753 \ J$.
$(b)$ Freezing $1 \ mol$ of water at $0^{\circ} C$ to ice at $0^{\circ} C$: $\Delta H_2 = -\Delta_{fus} H = -6.03 \ kJ \ mol^{-1} = -6030 \ J$.
$(c)$ Cooling $1 \ mol$ of ice from $0^{\circ} C$ to $-10^{\circ} C$: $\Delta H_3 = n C_p [H_2 O_{(s)}] \Delta T = 1 \ mol \times 36.8 \ J \ mol^{-1} \ K^{-1} \times (-10 - 0) \ K = -368 \ J$.
Total $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -753 \ J - 6030 \ J - 368 \ J = -7151 \ J = -7.151 \ kJ \ mol^{-1}$.
$(a)$ Cooling $1 \ mol$ of water from $10^{\circ} C$ to $0^{\circ} C$: $\Delta H_1 = n C_p [H_2 O_{(l)}] \Delta T = 1 \ mol \times 75.3 \ J \ mol^{-1} \ K^{-1} \times (0 - 10) \ K = -753 \ J$.
$(b)$ Freezing $1 \ mol$ of water at $0^{\circ} C$ to ice at $0^{\circ} C$: $\Delta H_2 = -\Delta_{fus} H = -6.03 \ kJ \ mol^{-1} = -6030 \ J$.
$(c)$ Cooling $1 \ mol$ of ice from $0^{\circ} C$ to $-10^{\circ} C$: $\Delta H_3 = n C_p [H_2 O_{(s)}] \Delta T = 1 \ mol \times 36.8 \ J \ mol^{-1} \ K^{-1} \times (-10 - 0) \ K = -368 \ J$.
Total $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = -753 \ J - 6030 \ J - 368 \ J = -7151 \ J = -7.151 \ kJ \ mol^{-1}$.
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