Question

How do you calculate the molar enthalpy of condensation (\[\Delta {H_{cond}}\]) for ammonia when \[100g\] of \[N{H_3}\] gas turn into a liquid at its boiling point?

Answer 1

Hint: The condensation is a process when a substance undergoes a phase transformation from gas phase to liquid phase. During this phase transition heat is released.

Complete step by step answer:
The heat of condensation is defined as the amount of heat released when a substance goes from gas phase to liquid phase. The molar enthalpy is the molar heat of the substance. Thus the molar enthalpy is defined as the amount of heat released when one mole of substance is converted from gas to liquid.
The molar enthalpy of condensation is just the reverse of molar enthalpy of vaporization. The molar enthalpy of vaporization is the amount of heat absorbed when one mole of substance is converted from liquid to gas phase.
The substance in this case is ammonia. The molar heat of vaporization of ammonia is given in literature as \[23.35KJ/mol\] . Thus the molar heat of condensation for one mole of ammonia is \[ - 23.35KJ/mol\].
Now we need to determine the number of moles of ammonia present in \[100g\] of ammonia. The molar mass of ammonia (\[N{H_3}\])= atomic mass of \[N\] + \[3{\text{ }} \times \] atomic mass of \[H\]
$ = 14.0067 + 3 \times 1.0079 = 17.03g/mol$.
The moles of ammonia in $100g$ = $\dfrac{{100g}}{{17.03g/mol}} = 5.87moles$.
So \[23.35KJ\] of heat is absorbed for condensation of one mole of ammonia. Hence the heat released for condensation of \[5.87moles\] of ammonia is = \[\dfrac{{ - 23.35KJ}}{{1mol}} \times 5.87moles = - 137.06KJ.\]

Hence, the molar enthalpy of condensation (\[\Delta {H_{cond}}\]) for ammonia, when \[100g\] of \[N{H_3}\] gas turn into a liquid at its boiling point is \[ - 137.06KJ/mol\].

Note: As vaporization and condensation of a substance are exactly reverse processes so the magnitude of the molar heat of vaporization is the same as the numerical value of the molar heat of condensation. They only differ in sign i.e. $\Delta {H_{vap}} = - \Delta {H_{cond}}$.