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Q5.10:
Calculate the enthalpy change on freezing of 1.0 mol of water at (10.0^{\circ}\text{C}) to ice at (–10.0^{\circ}\text{C}). (\Delta_{\text{fus}}H = 6.03 \text{ kJ mol}^{-1}) at (0^{\circ}\text{C}). (C_{p} [\text{H}_{2}\text{O(l)}] = 75.3 \text{ J mol}^{-1} \text{ K}^{-1}) (C_{p} [\text{H}_{2}\text{O(s)}] = 36.8 \text{ J mol}^{-1} \text{ K}^{-1})

Solution :

Step 1: Establishing the Thermodynamic Path

The transformation of $1.0\text{ mol}$ of liquid water at $10.0^{\circ}\text{C}$ to ice at $-10.0^{\circ}\text{C}$ is a non-isothermal process. To calculate the total enthalpy change ($\Delta H_{total}$), we decompose the process into three distinct thermodynamic stages using the state function property of enthalpy [Per Hess's Law]:

  • Step A: Cooling liquid water from $10.0^{\circ}\text{C}$ ($283.15\text{ K}$) to $0.0^{\circ}\text{C}$ ($273.15\text{ K}$).
  • Step B: Phase transition (freezing) of liquid water to ice at $0.0^{\circ}\text{C}$ ($273.15\text{ K}$).
  • Step C: Cooling ice from $0.0^{\circ}\text{C}$ ($273.15\text{ K}$) to $-10.0^{\circ}\text{C}$ ($263.15\text{ K}$).

Step 2: Calculating Enthalpy Changes for Each Stage

Given parameters: $n = 1.0\text{ mol}$, $\Delta_{fus}H = 6.03\text{ kJ mol}^{-1} = 6030\text{ J mol}^{-1}$. Note that freezing is the reverse of fusion, hence $\Delta H_{freezing} = -\Delta_{fus}H = -6030\text{ J mol}^{-1}$.

Stage A ($\Delta H_A$): Cooling liquid water
$\Delta H_A = n \cdot C_{p(l)} \cdot \Delta T = 1.0\text{ mol} \cdot 75.3\text{ J K}^{-1}\text{ mol}^{-1} \cdot (273.15 - 283.15)\text{ K}$
$\Delta H_A = 75.3 \cdot (-10) = -753\text{ J}$

Stage B ($\Delta H_B$): Phase transition at $0^{\circ}\text{C}$
$\Delta H_B = n \cdot (-\Delta_{fus}H) = 1.0\text{ mol} \cdot (-6030\text{ J mol}^{-1}) = -6030\text{ J}$

Stage C ($\Delta H_C$): Cooling ice
$\Delta H_C = n \cdot C_{p(s)} \cdot \Delta T = 1.0\text{ mol} \cdot 36.8\text{ J K}^{-1}\text{ mol}^{-1} \cdot (263.15 - 273.15)\text{ K}$
$\Delta H_C = 36.8 \cdot (-10) = -368\text{ J}$

Step 3: Summation of Total Enthalpy Change

By the additive nature of enthalpy [Per First Law of Thermodynamics], the total change is:
$\Delta H_{total} = \Delta H_A + \Delta H_B + \Delta H_C$
$\Delta H_{total} = -753\text{ J} + (-6030\text{ J}) + (-368\text{ J})$
$\Delta H_{total} = -7151\text{ J}$

Liquid 10°C Liquid 0°C Ice 0°C Ice -10°C Thermodynamic Path Coordinate

Final Solution: The total enthalpy change for the process is $-7151\text{ J}$ or $-7.151\text{ kJ}$.


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