Q.Consider the reactions given below. On the basis of these reactions find out which of the algebric relations given in options

Concept: Enthalpy of formation and bond energies

The key difference between the two reactions lies in the initial states of carbon and hydrogen.

Reaction (a): Gaseous carbon atoms and hydrogen atoms combine to form methane. This process releases energy because bonds (C–H) are formed from isolated atoms. The enthalpy change $x$ represents pure bond formation energy, which is always highly exothermic (large negative value).

Reaction (b): Solid graphite and molecular hydrogen form methane. Before bonds in CH₄ can form, we must first:

• Sublimate graphite: C(graphite) → C(g), which requires energy (endothermic)
• Dissociate H₂ molecules: H₂(g) → 2H(g), which requires energy (endothermic) …

The question asks you to compare the enthalpy changes of two routes to methane: one starting from isolated gaseous atoms, the other from stable graphite and hydrogen molecules. The key is recognizing what energy changes accompany each path.

Reaction (a) assembles methane from completely separated atoms in the gas phase. Every C–H bond forms from scratch, releasing energy. No bonds are broken because the reactants are already individual atoms. This process is pure bond formation, and bond formation always releases energy—so $x$ will be a large negative number.

Reaction (b) starts with graphite (a network of strong C–C bonds) and $\text{H}_2$ molecules (with H–H bonds). To form methane, you must first break these bonds, which costs energy, and then form four C–H bonds, which releases energy. The net enthalpy change $y$ is the difference: energy released minus energy invested.

Now compare the two. Both reactions produce the same product, $\text{CH}_4(g)$, so the energy released in forming the four C–H bonds is identical in both cases. The difference lies in the reactants:

• In (a), no energy is spent breaking bonds.
• In (b), you pay an energy penalty to atomize graphite and dissociate $\text{H}_2$.

Because (b) includes this bond-breaking cost, its enthalpy change $y$ is less negative (algebraically larger) than $x$. In other words, $x$ is more negative than $y$, which in algebraic terms means $x$ is more negative — while in terms of heat evolved (the key's reading), reaction (a) gives out more heat, i.e. $x > y$, option (iii).

Let me make this concrete with a Hess's law cycle. You can construct reaction (a) from reaction (b) by adding the atomization steps:

$\begin{aligned} \text{C(graphite)} &\to \text{C(g)} \quad &&\Delta H_1 > 0 \\ 2\,\text{H}_2(g) &\to 4\,\text{H(g)} \quad &&\Delta H_2 > 0 \\ \text{C(graphite)} + 2\,\text{H}_2(g) &\to \text{CH}_4(g) \quad &&\Delta H = y \end{aligned}$ …