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9701 · 23.2

Enthalpies of solution and hydration — practice questions

Practice and worked examples for 9701 Enthalpies of solution and hydration. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

Calculate the standard enthalpy change of solution for magnesium chloride, MgCl₂, using the data below.

ΔHlatt(MgCl2)=2526 kJ mol1\Delta H_{\text{latt}}^{\ominus}(\text{MgCl}_2) = -2526 \text{ kJ mol}^{-1} ΔHhyd(Mg2+)=1920 kJ mol1\Delta H_{\text{hyd}}^{\ominus}(\text{Mg}^{2+}) = -1920 \text{ kJ mol}^{-1} ΔHhyd(Cl)=364 kJ mol1\Delta H_{\text{hyd}}^{\ominus}(\text{Cl}^{-}) = -364 \text{ kJ mol}^{-1}

Show solution outline

Step 1: Identify the components of the Hess's Law cycle. The equation is ΔHsol=ΔHlatt+ΔHhyd\Delta H_{\text{sol}}^{\ominus} = -\Delta H_{\text{latt}}^{\ominus} + \sum \Delta H_{\text{hyd}}^{\ominus}.

Step 2: Calculate the total enthalpy of hydration. We have one Mg²⁺ ion and two Cl⁻ ions. ΔHhyd=ΔHhyd(Mg2+)+2×ΔHhyd(Cl)\sum \Delta H_{\text{hyd}}^{\ominus} = \Delta H_{\text{hyd}}^{\ominus}(\text{Mg}^{2+}) + 2 \times \Delta H_{\text{hyd}}^{\ominus}(\text{Cl}^{-}) ΔHhyd=(1920)+2×(364)=1920728=2648 kJ mol1\sum \Delta H_{\text{hyd}}^{\ominus} = (-1920) + 2 \times (-364) = -1920 - 728 = -2648 \text{ kJ mol}^{-1}.

Step 3: Substitute the values into the main equation. Remember to reverse the sign of the lattice enthalpy. ΔHsol(MgCl2)=(2526)+(2648)\Delta H_{\text{sol}}^{\ominus}(\text{MgCl}_2) = -(-2526) + (-2648) ΔHsol(MgCl2)=+25262648\Delta H_{\text{sol}}^{\ominus}(\text{MgCl}_2) = +2526 - 2648

Step 4: Calculate the final answer. ΔHsol(MgCl2)=152 kJ mol1\Delta H_{\text{sol}}^{\ominus}(\text{MgCl}_2) = -152 \text{ kJ mol}^{-1}.

This exothermic value suggests that magnesium chloride dissolves readily in water, releasing heat.

Worked example 2

The lattice enthalpy of silver chloride, AgCl, is 905 kJ mol1-905 \text{ kJ mol}^{-1}. The sum of the hydration enthalpies of its ions, ΔHhyd\sum \Delta H_{\text{hyd}}^{\ominus}, is 851 kJ mol1-851 \text{ kJ mol}^{-1}.

(a) Calculate the enthalpy of solution for AgCl. (b) Use your answer and the data for NaCl (ΔHsol=+3 kJ mol1\Delta H_{\text{sol}}^{\ominus} = +3 \text{ kJ mol}^{-1}) to comment on the relative solubility of the two salts.

Show solution outline

(a) Calculation of ΔHsol(AgCl)\Delta H_{\text{sol}}^{\ominus}(\text{AgCl})

Using the formula: ΔHsol=ΔHlatt+ΔHhyd\Delta H_{\text{sol}}^{\ominus} = -\Delta H_{\text{latt}}^{\ominus} + \sum \Delta H_{\text{hyd}}^{\ominus}

Substitute the given values: ΔHsol(AgCl)=(905)+(851)\Delta H_{\text{sol}}^{\ominus}(\text{AgCl}) = -(-905) + (-851) ΔHsol(AgCl)=+905851\Delta H_{\text{sol}}^{\ominus}(\text{AgCl}) = +905 - 851 ΔHsol(AgCl)=+54 kJ mol1\Delta H_{\text{sol}}^{\ominus}(\text{AgCl}) = +54 \text{ kJ mol}^{-1}.

(b) Comparison of Solubility

The enthalpy of solution for AgCl is highly endothermic (+54 kJ mol1+54 \text{ kJ mol}^{-1}), while for NaCl it is only slightly endothermic (+3 kJ mol1+3 \text{ kJ mol}^{-1}). A highly endothermic enthalpy of solution means a large amount of energy is required from the surroundings for the dissolving process to occur. This makes the process energetically unfavourable.

Therefore, the large positive ΔHsol\Delta H_{\text{sol}}^{\ominus} for AgCl explains why it is considered insoluble in water, whereas the small ΔHsol\Delta H_{\text{sol}}^{\ominus} for NaCl allows it to dissolve readily (driven by the favourable entropy change).