Group 2: Question 4

Worked solution

Part (a): Defining lattice energy

Lattice energy is the enthalpy change when 1 mole of a solid ionic compound is formed from its constituent ions in the gaseous state, under standard conditions:

$\text{M}^{2+}\text{(g)} + 2\text{OH}^-\text{(g)} \rightarrow \text{M(OH)}_2\text{(s)}$

It is always exothermic (negative), since oppositely charged gaseous ions attract strongly as they come together to form a solid lattice.

Part (b): The energy cycle and calculating $\Delta H_{sol}$

Dissolving a Group 2 hydroxide can be thought of as two steps: first, breaking the solid lattice apart into gaseous ions (the endothermic reverse of lattice energy), and second, hydrating those gaseous ions (exothermic):

$$\begin{aligned} \text{M(OH)}_2\text{(s)} &\xrightarrow{-\text{lattice energy}} \text{M}^{2+}\text{(g)} + 2\text{OH}^-\text{(g)} \xrightarrow{\Delta H_{hyd}(\text{M}^{2+})\ +\ 2\Delta H_{hyd}(\text{OH}^-)} \text{M}^{2+}\text{(aq)} + 2\text{OH}^-\text{(aq)} \end{aligned}$$

By Hess’s law:

$\Delta H_{sol} = -(\text{lattice energy}) + \Delta H_{hyd}(\text{M}^{2+}) + 2\Delta H_{hyd}(\text{OH}^-)$

Note the factor of 2 on the hydroxide hydration term, since 1 mole of $\text{M(OH)}_2$ releases 2 moles of $\text{OH}^-$ ions.

For $\text{Ca(OH)}_2$: $\Delta H_{sol} = -(-2540) + (-1650) + 2(-460) = 2540 - 1650 - 920$ $= 2540 - 2570 = -30\ \text{kJ mol}^{-1}$

For $\text{Ba(OH)}_2$: $\Delta H_{sol} = -(-2160) + (-1350) + 2(-460) = 2160 - 1350 - 920$ $= 2160 - 2270 = -110\ \text{kJ mol}^{-1}$

Check (independent recomputation): $1650 + 2(460) = 1650 + 920 = 2570$, so $2540 - 2570 = -30$ for $\text{Ca(OH)}_2$; $1350 + 920 = 2270$, so $2160 - 2270 = -110$ for $\text{Ba(OH)}_2$, consistent.

Part (c): Linking the calculation to the solubility trend

$\Delta H_{sol}(\text{Ca(OH)}_2) = -30\ \text{kJ mol}^{-1}$ is only slightly exothermic, whereas $\Delta H_{sol}(\text{Ba(OH)}_2) = -110\ \text{kJ mol}^{-1}$ is considerably more exothermic. Dissolving $\text{Ba(OH)}_2$ therefore releases more energy (is more energetically favourable) than dissolving $\text{Ca(OH)}_2$. Assuming the entropy change of solution is similar for both compounds, a more exothermic (more favourable) $\Delta H_{sol}$ is consistent with $\text{Ba(OH)}_2$ being the more soluble of the two, matching the observed trend of increasing hydroxide solubility down Group 2.

This arises because, going from $\text{Ca}^{2+}$ to the larger $\text{Ba}^{2+}$, the lattice energy becomes less negative by a larger amount ($2540 \rightarrow 2160$, a drop of $380\ \text{kJ mol}^{-1}$) than the cation’s hydration enthalpy does ($1650 \rightarrow 1350$, a drop of only $300\ \text{kJ mol}^{-1}$). The lattice becomes easier to break apart faster than the ions become harder to hydrate, so $\Delta H_{sol}$ becomes more negative down the group.

Final answers

• (a) Enthalpy change when 1 mole of a solid ionic compound forms from its ions in the gaseous state, standard conditions.
• (b) $\Delta H_{sol}(\text{Ca(OH)}_2) = \boxed{-30\ \text{kJ mol}^{-1}}$; $\Delta H_{sol}(\text{Ba(OH)}_2) = \boxed{-110\ \text{kJ mol}^{-1}}$
• (c) The more exothermic $\Delta H_{sol}$ for $\text{Ba(OH)}_2$ is consistent with it being more soluble than $\text{Ca(OH)}_2$.