Question

Calculate the heat of crystallization for NaCl from the given data:The enthalpy of fusion of NaCl = 28.3 kJ/mol and the heat capacity of NaCl is approximately constant at 40.0 J/mol.K for temperatures close to room temperature. The melting point of NaCl is 801 °C and the density of the solid NaCl is 2.16 g/cm³.

Answer 1

To calculate the heat of crystallization for NaCl, we need to determine the heat released when NaCl transitions from its liquid state to its solid state at its melting point. We can use the following equation:H_crystallization = -H_fusionwhere H_crystallization is the heat of crystallization, and H_fusion is the enthalpy of fusion.We are given the enthalpy of fusion of NaCl as 28.3 kJ/mol. Therefore, the heat of crystallization is:H_crystallization = -28.3 kJ/molThe heat of crystallization for NaCl is -28.3 kJ/mol. This means that 28.3 kJ of heat is released per mole of NaCl when it crystallizes from its liquid state to its solid state at its melting point.