Q13 / 75 · JEE Main 2026 (21 January Shift 2) · Chemistry
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 :
$\Delta_{f} \mathrm{H}^{\ominus}$ (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 :
- A$\frac{-X+Y+Z}{4}$
- B$\frac{X+Y+4 Z}{2}$
- Correct answerC$\frac{X+Y+2 Z}{4}$
- D$X+Y+Z$
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 $B = \frac{X + Y + 2Z}{4}$.
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 $B = \frac{X + Y + 2Z}{4}$.
Hence, the correct answer is option: 'C'.