Q.1 g of graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atmospheric pressure according to the equation $C(graphite) + O_2(g) \rightarrow CO_2(g)$. During the reaction, temperature rises from 298 K to 299 K. If the heat capacity of the bomb calorimeter is 20.7 kJ/K, what is the enthalpy change for the above reaction at 298 K and 1 atm?

Bomb Calorimetry and Enthalpy: From Intuition to Precision

Imagine you want to know exactly how much heat a handful of cashews releases when your body burns it. You could eat them and measure your temperature rise — but that’s messy, slow, and full of biological noise. A bomb calorimeter is the chemist’s clean, controlled way to do the same thing: burn a sample completely in pure oxygen inside a sealed steel container (the “bomb”) submerged in water, and measure the temperature change of that water.

The key insight: everything stays at constant volume. The bomb is rigid — it doesn’t expand or contract. That single fact changes which thermodynamic quantity you measure directly.

What the bomb actually measures

When the sample burns, it releases heat. That heat warms the bomb, the water, and everything around it. From the temperature rise and the known heat capacity of the entire calorimeter, you calculate the heat released at constant volume, denoted $q_V$.

For any process at constant volume with no non-expansion work (like electrical work), the first law of thermodynamics says:

where $\Delta U$ is the change in internal energy of the system (the burning sample + oxygen + products). So a bomb calorimeter directly gives you $\Delta U$ for the combustion reaction.

But we usually want enthalpy, not internal energy

In real life — open beakers, industrial furnaces, your body — reactions happen at constant pressure (usually 1 atm). The heat released at constant pressure is called enthalpy change, $\Delta H$. For a combustion reaction:

For solids and liquids, $\Delta (PV)$ is tiny. But for reactions involving gases — and combustion almost always does — the volume change matters. If the number of moles of gas changes during the reaction, the system does work on (or receives work from) the surroundings.

The precise statement

Bomb calorimetry enthalpy is the enthalpy change of a reaction calculated from the internal energy change measured in a bomb calorimeter, corrected for the $P\Delta V$ work associated with any change in the number of moles of gas.

In practice:

1. Measure $\Delta U$ from the bomb calorimeter experiment.
2. Determine $\Delta n_g$ from the balanced chemical equation.
3. Compute $\Delta H = \Delta U + \Delta n_g RT$.

Why this matters for exams

You will often be given a bomb calorimeter experiment result (temperature rise, heat capacity) and asked for $\Delta H$ of combustion. The steps: …