Calcualte the enthalpy change on freezing of 1.0 mole of water at `10.0^(@)C` to ice at `-10^(@)`C. `Delta_(fs)H=6.03 kJ mol^(-1)` at `0^(@)C`.
`C_(p)[H_(2)O(l)] = 75.3 J mol^(-1) K^(-1), C_(P)[H_(2)O(s)] = 36.8 Jmol^(-1)K^(-1)`

Options

A. `6.00 kJ "mol"^(-1)`
B. `-5.81 kJ "mol"^(-1)`
C. `-7.151 kJ "mol"^(-1)`
D. `6.56 kJ "mol"^(-1)`

Answer

To calculate the enthalpy change on freezing 1.0 mole of water at \(10.0^\circ C\) to ice at \(-10^\circ C\), we need to consider three steps in the process:

1. Cooling the water from \(10.0^\circ C\) to \(0.0^\circ C\).
2. Freezing the water at \(0.0^\circ C\) to ice at \(0.0^\circ C\).
3. Cooling the ice from \(0.0^\circ C\) to \(-10.0^\circ C\).

### Step 1: Cooling water from \(10.0^\circ C\) to \(0.0^\circ C\)

The enthalpy change for this step can be calculated using the formula:
\[ \Delta H_1 = C_p \times \Delta T \]
Where:
- \(C_p\) for liquid water = \(75.3 \, \text{J mol}^{-1} \text{K}^{-1}\)
- \(\Delta T = 0 - 10 = -10 \, \text{K}\)

Substituting the values:
\[ \Delta H_1 = 75.3 \, \text{J mol}^{-1} \text{K}^{-1} \times (-10 \, \text{K}) = -753 \, \text{J} \]

### Step 2: Freezing water at \(0.0^\circ C\)

The enthalpy change for this phase transition is given by the enthalpy of fusion:
\[ \Delta H_2 = -\Delta H_{fus} = -6.03 \, \text{kJ mol}^{-1} = -6030 \, \text{J mol}^{-1} \]

### Step 3: Cooling ice from \(0.0^\circ C\) to \(-10.0^\circ C\)

The enthalpy change for this step is calculated similarly:
\[ \Delta H_3 = C_{p(s)} \times \Delta T \]
Where:
- \(C_{p(s)}\) for solid ice = \(36.8 \, \text{J mol}^{-1} \text{K}^{-1}\)
- \(\Delta T = -10 - 0 = -10 \, \text{K}\)

Substituting the values:
\[ \Delta H_3 = 36.8 \, \text{J mol}^{-1} \text{K}^{-1} \times (-10 \, \text{K}) = -368 \, \text{J} \]

### Total Enthalpy Change

Now, we sum the enthalpy changes from all three steps:
\[ \Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
Substituting the values:
\[ \Delta H_{total} = -753 \, \text{J} - 6030 \, \text{J} - 368 \, \text{J} \]
\[ \Delta H_{total} = -7151 \, \text{J} \]

To convert this to kilojoules:
\[ \Delta H_{total} = -7.151 \, \text{kJ} \]

### Final Answer
The enthalpy change on freezing 1.0 mole of water at \(10.0^\circ C\) to ice at \(-10.0^\circ C\) is \(-7.151 \, \text{kJ}\).

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