Question

Calculate the enthalpy change for the dissolution of (NH4)2SO4(s) using the enthalpy of
formation data provided below. (remember to write the balanced equation for the dissolution
reaction including state symbols). Explain how the value obtained here might be different from the
value obtained using calorimetry.

Compound
Enthalpies of formation (kJmol-1)
(NH4)2SO4(s)
-1380.9
(NH4)* (as)
-143.5
(SO4)2-(aa)
-1009.0

Answer 1

This problem is providing us with the enthalpies of formation of all the species involved in the dissolution of ammonium sulfate. Thus, its enthalpy of dissolution is required and found to be 84.9 kJ/mol:

Enthalpy of reaction.

In chemistry, when a chemical reaction takes place, one can be able to calculate the energy released or absorbed due to the chemical change. In such a way, for the dissolution of ammonium sulfate we can write:

$(NH_4)_2SO_4(s)\rightarrow 2NH_4^+(aq)+SO_4^{2-}$

Thus, the expression to calculate the enthalpy of reaction is:

$\Delta H_{diss}=2\Delta _fH^{NH_4^+}+\Delta _fH^{SO_4^{2-}}-\Delta _fH^{(NH_4)_2SO_4}$

Hence, after plugging in the enthalpies of formation, we obtain:

$\Delta H_{diss}=2(-143.5kJ/mol)+(-1009.0kJ/mol)-(-1380.9kJ/mol)\\\\\Delta H_{diss}=84.9kJ/mol$

Which also is evidence of an endothermic reaction.

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