Chemical Energetics: Question 2

Worked solution

Part (a): Statement of Hess’s law

Hess’s law states that the total enthalpy change for a chemical reaction is the same regardless of the route taken from reactants to products, provided the initial and final conditions are identical.

Part (b): Energy cycle and calculation of $\Delta H^{\ominus}$

The cycle links the direct route with an indirect route via the elements calcium, carbon (graphite) and oxygen in their standard states.

Direct route (the reaction itself): $\text{CaO(s)} + \text{CO}_2\text{(g)} \xrightarrow{\ \Delta H^{\ominus}\ } \text{CaCO}_3\text{(s)}$

Indirect route, both branches starting from the same elements, $\text{Ca(s)} + \text{C(graphite)} + \tfrac{3}{2}\text{O}_2\text{(g)}$:

1. The elements form $\text{CaO(s)}$ and $\text{CO}_2\text{(g)}$ separately, releasing $\Delta H_f^{\ominus}[\text{CaO}] + \Delta H_f^{\ominus}[\text{CO}_2]$.
2. The elements form $\text{CaCO}_3\text{(s)}$ directly, releasing $\Delta H_f^{\ominus}[\text{CaCO}_3]$.

By Hess’s law, since both branches start at the same elements and finish at the same product ($\text{CaCO}_3\text{(s)}$), the two routes must give the same total enthalpy change: $\Delta H_f^{\ominus}[\text{CaO}] + \Delta H_f^{\ominus}[\text{CO}_2] + \Delta H^{\ominus} = \Delta H_f^{\ominus}[\text{CaCO}_3]$

Rearranging for the unknown, $\Delta H^{\ominus}$: $\Delta H^{\ominus} = \Delta H_f^{\ominus}[\text{CaCO}_3] - \left(\Delta H_f^{\ominus}[\text{CaO}] + \Delta H_f^{\ominus}[\text{CO}_2]\right)$

Substituting the values from the table: $\Delta H^{\ominus} = -1207 - \left[(-635) + (-394)\right] = -1207 - (-1029)$

$\Delta H^{\ominus} = -1207 + 1029 = -178\ \text{kJ mol}^{-1}$

Check (independent recomputation): $(-635)+(-394) = -1029$; then $-1207-(-1029) = -1207+1029$. Adding $1029$ to $-1207$: $1207-1029=178$, and since $1029<1207$ the result stays negative, giving $-178\ \text{kJ mol}^{-1}$, consistent.

Part (c): Exothermic or endothermic?

The reaction is exothermic, because $\Delta H^{\ominus} = -178\ \text{kJ mol}^{-1}$ is negative, meaning energy is released to the surroundings as the products form.

Part (d): Heat released from a given mass of CaO

Amount, in mol, of $\text{CaO}$ in $2.805\ \text{g}$: $n = \frac{2.805}{56.1} = 0.0500\ \text{mol}$

Since $1$ mol of CaO reacting releases $178\ \text{kJ}$ (the magnitude of $\Delta H^{\ominus}$ found in (b)), $0.0500$ mol releases: $0.0500 \times 178 = 8.90\ \text{kJ}$

Check (independent recomputation): $2.805 \div 56.1 = 0.0500$ exactly; $0.0500\times178 = 8.90\ \text{kJ}$, consistent.

Final answers

• (a) Hess’s law: the enthalpy change for a reaction is independent of the route taken, given the same start and end states.
• (b) $\Delta H^{\ominus} = \boxed{-178\ \text{kJ mol}^{-1}}$
• (c) Exothermic ($\Delta H^{\ominus} < 0$).
• (d) Heat released $= \boxed{8.90\ \text{kJ}}$