If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the sig

NCERT Exemplar · Q41

Q. If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.

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Concept understanding — Bomb Calorimetry Enthalpy

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:

q_V = \Delta U

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.

Note

Constant volume means no $P\Delta V$ work is done — the system can’t push against the atmosphere. All the energy change appears as heat.

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:

\Delta H = \Delta U + \Delta (PV)

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.

\Delta H = \Delta U + \Delta n_g RT

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:

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

Watch out

A common mistake: assuming $\Delta H = \Delta U$ for all combustion reactions. This is only true when $\Delta n_g = 0$ — for example, burning carbon in oxygen:

$\text{C(s)} + \text{O}_2\text{(g)} \to \text{CO}_2\text{(g)}$ has $\Delta n_g = 0$ , so $\Delta H = \Delta U$ .

But burning methane:

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$ has $\Delta n_g = 1 - 3 = -2$ , so $\Delta H = \Delta U - 2RT$ .

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: …

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The molar enthalpy change for graphite combustion is found by scaling the heat from 1 g to 1 mole (12 g). The result is −248.4 kJ mol⁻¹, and the negative sign indicates that heat is released to the surroundings (exothermic process).

Why bomb calorimetry gives us enthalpy

When graphite burns in oxygen, the reaction is:

C(s) + O_2(g) \rightarrow CO_2(g)

The heat measured in a bomb calorimeter is actually the internal energy change ( $\Delta U$ ) at constant volume. But for combustion reactions involving gases, the difference between $\Delta H$ and $\Delta U$ is $\Delta n_g RT$ , where $\Delta n_g$ is the change in moles of gas.

Here, one mole of $O_2$ is consumed and one mole of $CO_2$ is produced — so $\Delta n_g = 0$ . That means:

\Delta H = \Delta U

No pressure–volume work is done, so the heat measured at constant volume equals the enthalpy change directly. This is a special case; it’s not always true.

Tip

Whenever the number of moles of gaseous reactants equals the number of moles of gaseous products, $\Delta H = \Delta U$ . This saves you from having to correct for $PV$ work.

Step‑by‑step calculation

1. Identify the given data

Mass of graphite burned: $m = 1\ \text{g}$
Heat released: $q = 20.7\ \text{kJ}$ (the problem says “produces”, so this is the heat given out)
Molar mass of carbon (graphite): $M = 12\ \text{g mol}^{-1}$

2. Find the heat per mole

If 1 g releases 20.7 kJ, then 12 g (1 mole) will release:

\text{Heat per mole} = 20.7\ \text{kJ g}^{-1} \times 12\ \text{g mol}^{-1} = 248.4\ \text{kJ mol}^{-1}

3. Assign the correct sign

Combustion is always exothermic — the system loses energy to the surroundings. By convention, enthalpy change for an exothermic process is negative. Therefore:

$\Delta H = -248.4\ \text{kJ mol}^{-1}$ …

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Q.If the combustion of 1g of graphite produces 20.7 kJ of heat, what will be molar enthalpy change? Give the significance of sign also.

Why bomb calorimetry gives us enthalpy

Step‑by‑step calculation

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