Calculate the enthalpy change of freezing of `1.0` mol of water at `10^(@)C` to ice at `-10^(@)C, Delta_(fus)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 J mol^(-1) K^(-1)`

Options

A. `-7.151 KJ "mol^(-1)`
B. `5.81 KJ "mol"^(-1)`
C. `5.44 KJ "mol"^(-1)`
D. `-6.56 KJ "mol"^(-1)`

Answer

To calculate the enthalpy change of freezing of `1.0` mol of water at `10°C` to ice at `-10°C`, we will consider three steps involved in the process:

1. Cooling the liquid water from `10°C` to `0°C`.
2. Freezing the water at `0°C` to ice at `0°C`.
3. Cooling the ice from `0°C` to `-10°C`.

### Step 1: Calculate ΔH1 (Cooling water from 10°C to 0°C)

We use the formula:
\[ \Delta H_1 = C_p \times \Delta T \]
Where:
- \(C_p\) for liquid water = `75.3 J mol^{-1} K^{-1}`
- \(\Delta T = T_{final} - T_{initial} = 0°C - 10°C = -10 K\)

Calculating ΔH1:
\[ \Delta H_1 = 75.3 \, \text{J mol}^{-1} \, \text{K}^{-1} \times (-10 \, \text{K}) = -753 \, \text{J mol}^{-1} \]

### Step 2: Calculate ΔH2 (Freezing at 0°C)

The enthalpy change for freezing is the negative of the enthalpy of fusion:
\[ \Delta H_2 = -\Delta_{fus}H \]
Where:
- \(\Delta_{fus}H = 6.03 \, \text{kJ mol}^{-1} = 6030 \, \text{J mol}^{-1}\)

Calculating ΔH2:
\[ \Delta H_2 = -6030 \, \text{J mol}^{-1} \]

### Step 3: Calculate ΔH3 (Cooling ice from 0°C to -10°C)

We again use the formula:
\[ \Delta H_3 = C_p \times \Delta T \]
Where:
- \(C_p\) for solid water = `36.8 J mol^{-1} K^{-1}`
- \(\Delta T = -10°C - 0°C = -10 K\)

Calculating ΔH3:
\[ \Delta H_3 = 36.8 \, \text{J mol}^{-1} \, \text{K}^{-1} \times (-10 \, \text{K}) = -368 \, \text{J mol}^{-1} \]

### Total Enthalpy Change (ΔH_total)

Now, we sum up all the enthalpy changes:
\[ \Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]
Substituting the values:
\[ \Delta H_{total} = (-753 \, \text{J mol}^{-1}) + (-6030 \, \text{J mol}^{-1}) + (-368 \, \text{J mol}^{-1}) \]
Calculating:
\[ \Delta H_{total} = -753 - 6030 - 368 = -7151 \, \text{J mol}^{-1} \]

### Convert to kJ/mol
To convert to kJ/mol:
\[ \Delta H_{total} = -7151 \, \text{J mol}^{-1} \times \frac{1 \, \text{kJ}}{1000 \, \text{J}} = -7.151 \, \text{kJ mol}^{-1} \]

### Final Answer
The enthalpy change of freezing of `1.0` mol of water at `10°C` to ice at `-10°C` is:
\[ \Delta H_{total} = -7.151 \, \text{kJ mol}^{-1} \]

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