Consider the following data: $\Delta_{\mathrm{f}} \mathrm{H}^{\Theta}$(methane, g)$=-\mathrm{X} \mathrm{kJ} \mathrm{mol}^{-1}$ Enthalpy of sublimation of graphite $=\mathrm{Y} \mathrm{kJ} \mathrm{mol}^{-1}$ Dissociation enthalpy of $\mathrm{H}_{2}=\mathrm{Z} \mathrm{kJ} \mathrm{mol}^{-1}$ The bond enthalpy of $\mathrm{C}-\mathrm{H}$ bond is given by:

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Chemical Thermodynamics

Consider the following data:

Δ_{f} H^{Θ}

(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:

Answer: (A)

Forming methane from its elements,

C (s) + 2 H_{2} (g) \to CH_{4} (g)

, can be assembled by atomising the reactants and then making four C-H bonds. Subliming graphite supplies one gaseous carbon atom at a cost

Y

, dissociating the two hydrogen molecules costs

2 Z

, and forming the four C-H bonds releases

4 E_{C - H}

. Hess's law then reads

- X = Y + 2 Z - 4 E_{C - H}

, and rearranging for the bond enthalpy gives

E_{C - H} = \frac{X + Y + 2 Z}{4}

JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Single correct option