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} $)

Question

If the enthalpy of sublimation of Li is $155\,\text{kJ mol}^{-1}$, enthalpy of dissociation of $ \mathrm{F_2} $ is $150\,\text{kJ mol}^{-1}$, ionization enthalpy of Li is $520\,\text{kJ mol}^{-1}$, electron gain enthalpy of F is $-313\,\text{kJ mol}^{-1}$, and standard enthalpy of formation of LiF is $-594\,\text{kJ mol}^{-1}$, then the magnitude of lattice enthalpy of LiF is ____________ kJ mol$^{-1}$ (Nearest integer).

Answer 1

The total enthalpy change is determined by the freezing process and the subsequent cooling of the substance. 1. Freezing process: The enthalpy change for freezing is $ \Delta_{\text{fus}}H = -x $, as freezing releases energy. 2. Cooling of liquid water: The enthalpy change for cooling liquid water from 10°C to 0°C is given by $ \Delta H_1 = -10y $, using the specific heat capacity of liquid water. 3. Cooling of ice: The enthalpy change for cooling ice from 0°C to -10°C is calculated as $ \Delta H_2 = -10z $, using the specific heat capacity of ice. The total enthalpy change is therefore $ \Delta H = -x - 10y - 10z $. Note that $ x $ is in kJ/mol and requires conversion to J/mol by multiplying by 100 to align units with $ y $ and $ z $ (in J/mol·K). Consequently, the total enthalpy change is correctly expressed as: \[ \Delta H = -10(100x + y + z) \]

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} $)

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The Correct Option is B

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