Calorimetry and Hess cycles

What you’ll learn

• Convert a measured temperature change into a molar enthalpy change.
• Combine experimental reactions in a Hess cycle and evaluate heat-loss errors.

Understand the idea

In solution calorimetry, $q=mc\Delta T$ estimates the energy gained or lost by the surroundings being modelled. Here $m$ is the total mass of solution whose temperature changes, not just the mass of a dissolved reagent. Use a given density when converting solution volume to mass. A temperature rise means the solution gains energy, so the reaction is exothermic and its enthalpy change is negative. Convert joules to kilojoules before reporting a molar enthalpy change.

Core practical 2 uses measured enthalpy changes to find another change indirectly. Write each equation with its states and scale or reverse both the equation and its enthalpy together. If two routes lead from the same starting substances to the same final substances, their total enthalpy changes agree. This allows a reaction that is difficult to measure directly to be found using two accessible reactions with a common endpoint.

Heat transferred to the cup or the room is absent from a simple solution-only calculation. In an exothermic experiment this generally makes the measured temperature rise too small and the calculated enthalpy less negative. Insulation, a lid and a temperature–time extrapolation can reduce particular losses; repeats alone do not eliminate them. The assumptions of negligible cup heat capacity, complete reaction and a water-like specific heat capacity should be stated when used.

Key formulae

$$q=mc\Delta T$$

$$\Delta H_{\mathrm{reaction}}=-\frac{q}{1000n}\quad(\mathrm{kJ\,mol^{-1}})$$

$$\Delta H_{A\rightarrow B}=\Delta H_{A\rightarrow C}-\Delta H_{B\rightarrow C}$$

Worked example

A calorimetry dataset gives $80.0\ \mathrm{g}$ of solution, $c=4.18\ \mathrm{J\,g^{-1}\,K^{-1}}$ and a rise of $7.50\ \mathrm{K}$. The reaction extent is $0.0400\ \mathrm{mol}$. Assume negligible heat loss and cup heat capacity. Calculate the molar enthalpy change.

1. Energy gained by the solution is $q=80.0\times4.18\times7.50=2508\ \mathrm{J}$.
2. The solution warms, so the reaction releases energy: $q_{\mathrm{reaction}}=-2.508\ \mathrm{kJ}$.
3. Divide by the reaction extent: $\Delta H=-2.508/0.0400=-62.7\ \mathrm{kJ\,mol^{-1}}$.

Answer: $-62.7\ \mathrm{kJ\,mol^{-1}}$.