Q.$\Delta U^\ominus$ of combustion of methane is $-X$ kJ mol$^{-1}$. The value of $\Delta H^\ominus$ is

Concept: Bomb calorimetry measures $\Delta U$ (constant volume). To get $\Delta H$ (constant pressure), use $\Delta H = \Delta U + \Delta n_g RT$.

Reasoning:

1. Combustion of methane: $\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$.
2. Count gaseous moles: reactants = $1 + 2 = 3$ mol gas; products = $1$ mol gas ($\text{CO}_2$). So $\Delta n_g = 1 - 3 = -2$. …

The key to this problem lies in the relationship between enthalpy change ($\Delta H$) and internal energy change ($\Delta U$) for a reaction. That relationship is:

where $\Delta n_g$ is the change in the number of moles of gaseous substances (products minus reactants). The $RT$ term accounts for the work done by or on the system due to volume change at constant pressure.

For combustion in a bomb calorimeter, we directly measure $\Delta U$ (constant volume). But the standard enthalpy of combustion $\Delta H^\ominus$ is defined at constant pressure. The difference between them is the $P\Delta V$ work.

Let’s apply this to methane combustion.

1. Write the balanced chemical equation for the complete combustion of methane:

Notice that water is produced as a liquid, not a gas. This is crucial because only gaseous species contribute to $\Delta n_g$.

2. Calculate $\Delta n_g$ — the change in moles of gas:

   • Moles of gaseous reactants: $1\ (\text{CH}_4) + 2\ (\text{O}_2) = 3$ moles
   • Moles of gaseous products: $1\ (\text{CO}_2)$ only (water is liquid)
   • So $\Delta n_g = 1 - 3 = -2$
   Watch out

   A common mistake is to count water vapour as a gas. At standard conditions (298 K, 1 bar), water from combustion is liquid. If you mistakenly treat it as $\text{H}_2\text{O}(g)$, you'd get $\Delta n_g = 0$ and wrongly conclude $\Delta H = \Delta U$. Always check the physical state in the equation.

3. Interpret the sign of $\Delta n_g$:

   Since $\Delta n_g = -2$, the reaction consumes 2 more moles of gas than it produces. At constant pressure, the surroundings do work on the system as the volume decreases. This means the $P\Delta V$ work is negative from the system’s perspective.

4. Apply the relationship:

Substituting $\Delta n_g = -2$:

Since $RT > 0$ (always positive), $2RT$ is a positive quantity. Therefore: …