Formation reaction of methane:
\[ C(graphite) + 2H_2(g) \rightarrow CH_4(g) \]
Given:
\[ \Delta_f H^\circ = -X \]
To calculate bond enthalpy, we consider atomization process.
Step 1: Convert graphite to gaseous carbon
\[ C(graphite) \rightarrow C(g) \]
Energy required = Y
Step 2: Break H₂ into atoms
\[ 2H_2 \rightarrow 4H \]
Each H₂ requires Z energy, so:
\[ Energy = 2Z \]
Total energy to form gaseous atoms:
\[ Y + 2Z \]
Step 3: Formation of CH₄ from atoms
\[ C(g) + 4H(g) \rightarrow CH_4(g) \]
This releases energy equal to 4(C–H bond enthalpy).
Let bond enthalpy of C–H = D
Total energy released = 4D
Using Hess’s Law:
\[ (Y + 2Z) - 4D = -X \]
\[ Y + 2Z + X = 4D \]
\[ D = \frac{X + Y + 2Z}{4} \]
Correct Option: A
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.