Question

enthalpy changes for solids and liquids
example 3 - 6: 20.0 g of nh4no3 absorbs 5.28 kj of energy when the compound dissolves in a constant - pressure calorimeter. calculate the molar enthalpy change for the dissolving process:
nh4no3(s)→nh4+(aq)+no3−(aq)
δhmolar = δhrxn / n
δhrxn = qp = 5.28kj
20g nh4no3 | 1mol / 80.05g/mol = 0.2498mol → 21.1kj/mol

Explanation:

Step1: Calculate moles of NH₄NO₃

The molar mass of NH₄NO₃ is \(M=(14 + 4\times1+14 + 3\times16)\text{ g/mol}=80.05\text{ g/mol}\). The number of moles \(n\) of NH₄NO₃ with a mass \(m = 20.0\text{ g}\) is \(n=\frac{m}{M}=\frac{20.0\text{ g}}{80.05\text{ g/mol}}\approx0.2498\text{ mol}\).

Step2: Calculate molar - enthalpy change

The heat absorbed by the reaction \(q_p=\Delta H_{rxn}=5.28\text{ kJ}\). The molar enthalpy change \(\Delta H_{molar}=\frac{\Delta H_{rxn}}{n}\). Substituting the values, we get \(\Delta H_{molar}=\frac{5.28\text{ kJ}}{0.2498\text{ mol}}\approx21.1\text{ kJ/mol}\).

Answer:

\(21.1\text{ kJ/mol}\)