Total enthalpy change for freezing of 1 mol water at 10°C to ice at -10°C is ______
(Given: $ \Delta_{\text{fus}}H = x \, \text{kJ/mol} $, $ C_p[\text{H}_2\text{O}(l)] = y \, \text{J mol}^{-1} \text{K}^{-1} $, and $ C_p[\text{H}_2\text{O}(s)] = z \, \text{J mol}^{-1} \text{K}^{-1} $)
Show Hint
- \( -x - 10y - 10z \)
- \( -10(100x + y + z) \)
- \( 10(100x + y + z) \)
- \( x - 10y - 10z \)
The Correct Option is B
Approach Solution - 1
To find the total enthalpy change, we need to consider both the freezing process and the cooling of the substance:
1. Freezing process: The enthalpy change for freezing is \( \Delta_{\text{fus}}H = -x \), since freezing is an exothermic process.
2. Cooling of liquid water: The enthalpy change for cooling the water from 10°C to 0°C is calculated using the specific heat capacity of liquid water: \[ \Delta H_1 = -10y \]
3. Cooling of ice: The enthalpy change for cooling the ice from 0°C to -10°C is calculated using the specific heat capacity of ice: \[ \Delta H_2 = -10z \]
Thus, the total enthalpy change can be expressed as: \[ \Delta H = -x - 10y - 10z \] The term \( x \) is given in kJ/mol and should be multiplied by 100 to convert it to J/mol to match the units of \( y \) and \( z \) (which are in J/mol·K).
Thus the correct expression for the total enthalpy change is: \[ \Delta H = -10(100x + y + z) \]
Approach Solution -2
The reaction steps are given as: \[ \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(l)} \quad \text{(1) } 10^\circ C \quad \text{(1)} \] \[ \text{H}_2\text{O(l)} \rightarrow \text{H}_2\text{O(s)} \quad \text{(2) } 0^\circ C \quad \text{(2)} \] \[ \text{H}_2\text{O(s)} \rightarrow \text{H}_2\text{O(s)} \quad \text{(3) } -10^\circ C \quad \text{(3)} \] The steps are associated with the following changes: \[ (1) \rightarrow -y \times 10 \] \[ (2) \rightarrow -x \times 1000 \] \[ (3) \rightarrow -z \times 10 \] Thus, the net enthalpy change is: \[ \Delta H_{\text{net}} = -10(y + 1000x + z) \]
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