JEE Main 2026 Jan 21 Shift 2, Chemistry Q6: Hess’s Law and Bond Energy

JEE Main2026Jan 21, Shift 2Chemistry

Consider the following data:

$Δ_{f} H^{⊖}$ (methane, g) = $- X$ kJ mol $^{- 1}$

Enthalpy of sublimation of graphite = $Y$ kJ mol $^{- 1}$

Dissociation enthalpy of $H_{2}$ = $Z$ kJ mol $^{- 1}$

The bond enthalpy of C – H bond is given by:

Solution

Build $C H_{4} (g)$ atom-by-atom from elements in their standard states and equate to the formation enthalpy.

$C (s) + 2 H_{2} (g) \to C H_{4} (g), Δ_{f} H = - X$ .

Decompose via gaseous atoms: sublime C $(+ Y)$ , dissociate 2 mol $H_{2}$ to atoms $(+ 2 Z)$ , then form 4 C–H bonds (releases $4 \times$ B.E.):

$- X = Y + 2 Z - 4 (B . E ._{C - H})$ .

Therefore $B . E ._{C - H} = \frac{X + Y + 2 Z}{4}$ .

Concept behind this question

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