Question

The lattice energy of NaCl is 2786 kJ/mol, and the enthalpy of hydration of 1 mole of gaseous Na1 and 1 mole of gaseous Cl2 ions is 2783 kJ/mol. Calculate the enthalpy of solution per mole of solid NaCl.

Short answer

The enthalpy of solution per mole of solid NaCl is $\Delta H_{so\ln }=3kJ/mol$.

Step-by-step solution

Step 1:

According to Hess’s law,

$NaCl\left(s\right)\to N{a}^{+}\left(g\right)+C{l}^{-}\left(g\right)\to N{a}^{+}\left(aq\right)C{l}^{-}\left(aq\right)$

Step 2:

This means that solution enthalpy, $\Delta H_{so\ln }$is equal to

$\Delta H_{so\ln }=-\Delta H_{lattice}+\Delta H_{hyd}=786kJ/mol-783kJ/mol=3kJ/mol$