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 :

Correct answer: C. (X+Y+2 Z)/(4) For methane formation: C(g) + 2H_2(g) arrow CH_4(g) . Using enthalpy balance: Δ_f H° = Bonds broken - Bonds formed . Breaking: sublimation of C (Y) and 2 H-H bonds (2Z); Forming: 4 C-H bonds. Energy equation: -X = Y + 2Z - 4B where B is C-H bond enthalpy. Solving: 4B = X + Y + 2Z , so B = (X + Y + 2Z)/(4) .

JEE Main 202621 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 21 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Main 2026 · 21 January · Evening Shift · Chemistry · Thermodynamics (C)

Consider the following data: Δ_ f H^ (methane, g )=-X kJ mol ^ -1...

Consider the following data:

\Delta_{f} \mathrm{H}^{\ominus}

(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 :

Options

A $\frac{-X+Y+Z}{4}$
B $\frac{X+Y+4 Z}{2}$
C $\frac{X+Y+2 Z}{4}$
D $X+Y+Z$

Correct answer

C. $\frac{X+Y+2 Z}{4}$

Step-by-step solution

For methane formation:

\text{C(g)} + 2\text{H}_2(g) \rightarrow \text{CH}_4(g)

.
Using enthalpy balance:

\Delta_f H° = \text{Bonds broken} - \text{Bonds formed}

.
Breaking: sublimation of C (Y) and 2 H-H bonds (2Z); Forming: 4 C-H bonds.
Energy equation:

-X = Y + 2Z - 4B

where B is C-H bond enthalpy.
Solving:

4B = X + Y + 2Z

, so

B = \frac{X + Y + 2Z}{4}

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