Q.The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are –890.3 kJ mol$^{-1}$, –393.5 kJ mol$^{-1}$ and –285.8 kJ mol$^{-1}$ respectively. Enthalpy of formation of $CH_4(g)$ will be

Concept: Hess's Law and Formation Enthalpy from Combustion Data

The enthalpy of formation of $\mathrm{CH_4(g)}$ is defined for the reaction:

We use the given combustion enthalpies and apply Hess's Law.

Step 1: Write the combustion reactions with their enthalpies:

• $\mathrm{CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)}$, $\Delta H_c^\circ = -890.3 \text{ kJ mol}^{-1}$
• $\mathrm{C(graphite) + O_2(g) \rightarrow CO_2(g)}$, $\Delta H_c^\circ = -393.5 \text{ kJ mol}^{-1}$
• $\mathrm{H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l)}$, $\Delta H_c^\circ = -285.8 \text{ kJ mol}^{-1}$

Step 2: To obtain the formation reaction, reverse the combustion of methane and add the combustion of graphite and twice the combustion of hydrogen: …

The enthalpy of formation is defined as the enthalpy change when one mole of a compound is formed from its constituent elements in their standard states. For methane, that means:

We aren't given this directly. Instead, we have combustion data—the enthalpy changes when substances burn in oxygen. The strategy is to use Hess's law: enthalpy is a state function, so we can add and subtract reactions algebraically to construct the target equation.

Step-by-step construction

1. Write out the combustion reactions with their given enthalpies.

For methane:

For graphite (carbon):

For dihydrogen (hydrogen):

2. Identify what we need on each side of the formation equation.

We want $\text{C(graphite)} + 2\text{H}_2(g)$ on the left (reactants) and $\text{CH}_4(g)$ on the right (product).

3. Reverse the methane combustion to get $\text{CH}_4(g)$ as a product.

Reversing equation (1):

When you reverse a reaction, the sign of $\Delta H$ flips.

4. Add the combustion of graphite to supply $\text{C(graphite)}$ and remove $\text{CO}_2$.

Use equation (2) as written:

5. Add twice the combustion of hydrogen to supply $2\text{H}_2(g)$ and remove $2\text{H}_2\text{O}$.

Multiply equation (3) by 2:

6. Sum the three manipulated equations. …