enthalpy changes for solids and liquids\nexample 3 - 6: 20.0 g of nh4no3 absorbs 5.28 kj of energy when the…

enthalpy changes for solids and liquids\nexample 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:\nnh4no3(s)→nh4+(aq)+no3−(aq)\nδhmolar = δhrxn / n\nδhrxn = qp = 5.28kj\n20g nh4no3 | 1mol / 80.05g/mol = 0.2498mol → 21.1kj/mol

enthalpy changes for solids and liquids\nexample 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:\nnh4no3(s)→nh4+(aq)+no3−(aq)\nδhmolar = δhrxn / n\nδhrxn = qp = 5.28kj\n20g nh4no3 | 1mol / 80.05g/mol = 0.2498mol → 21.1kj/mol

Answer

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})