The enthalpy change of freezing of 1 mol of water at `5^(@)C` to ice at `-5^(@)C` is
(Given `Delta_(fus)H=6kJ"mol"^(-1)` at `0^(@)C`,
`(C_(p)(H_(2)O,s)=36.8J"mol"^(-1)K^(-1))`

Options

A. `5.44kJ"mol"^(-1)`
B. `5.81kJ"mol"^(-1)`
C. `6.56kJ"mol"^(-1)`
D. `6.00kJ"mol"^(-1)`

Explanation

To find the enthalpy change of freezing of 1 mole of water at 5°C to ice at -5°C, we will break the process into three steps:

### Step 1: Cooling water from 5°C to 0°C
We need to calculate the enthalpy change when 1 mole of water is cooled from 5°C to 0°C. The formula for calculating the heat change (ΔH) is:

\[ \Delta H_1 = C_p \cdot \Delta T \]

Where:
- \(C_p\) is the heat capacity of water (liquid), which is approximately 75.3 J/mol·K.
- \(\Delta T\) is the change in temperature, which is \(0°C - 5°C = -5 K\).

Calculating ΔH1:

\[ \Delta H_1 = 75.3 \, \text{J/mol·K} \cdot (-5 \, \text{K}) = -376.5 \, \text{J/mol} \]

### Step 2: Freezing water at 0°C
Next, we need to consider the enthalpy change when water freezes at 0°C. The enthalpy change for freezing (ΔH_fusion) is given as -6 kJ/mol (since freezing is an exothermic process).

\[ \Delta H_2 = -6 \, \text{kJ/mol} = -6000 \, \text{J/mol} \]

### Step 3: Cooling ice from 0°C to -5°C
Now, we need to calculate the enthalpy change when ice is cooled from 0°C to -5°C. The heat capacity of ice (solid water) is given as 36.8 J/mol·K.

Using the same formula:

\[ \Delta H_3 = C_{p, ice} \cdot \Delta T \]

Where:
- \(C_{p, ice} = 36.8 \, \text{J/mol·K}\)
- \(\Delta T = -5°C - 0°C = -5 K\)

Calculating ΔH3:

\[ \Delta H_3 = 36.8 \, \text{J/mol·K} \cdot (-5 \, \text{K}) = -184 \, \text{J/mol} \]

### Total Enthalpy Change
Now we can find the total enthalpy change (ΔH_total) by summing up the changes from all three steps:

\[ \Delta H_{total} = \Delta H_1 + \Delta H_2 + \Delta H_3 \]

Substituting the values we calculated:

\[ \Delta H_{total} = (-376.5 \, \text{J/mol}) + (-6000 \, \text{J/mol}) + (-184 \, \text{J/mol}) \]

Calculating:

\[ \Delta H_{total} = -376.5 - 6000 - 184 = -6586.5 \, \text{J/mol} \]

Converting to kJ:

\[ \Delta H_{total} = -6.5865 \, \text{kJ/mol} \]

### Final Answer
The enthalpy change of freezing of 1 mol of water at 5°C to ice at -5°C is approximately -6.59 kJ/mol.

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