Consider the following data: f H^ ( methane, g) = -X kJ mol^-1 Enthalpy of sublimation of graphite = Y kJ mol^-1 Dissociation enthalpy of H2 = Z kJ mol^-1 The bond enthalpy of C-H bond is given by:

Given: - f H^ ( methane, g) = -X kJ mol^-1 - Enthalpy of sublimation of graphite = Y kJ mol^-1 - Dissociation enthalpy of H2 = Z kJ mol^-1 Find: The bond enthalpy of the C-H bond. Use Hess's law and an atomization cycle. For the standard formation of methane, C( graphite) + 2H2(g) CH4(g) with enthalpy change H = -X Break the formation process into steps shown in the solution: 1. Sublimation of graphite: C( graphite) C(g) H = Y 2. Dissociation of hydrogen: 2H2(g) 4H(g) H = 2Z 3. Bond formation to make methane: C(g) + 4H(g) CH4(g) H = -4 B.E.(C-H) Adding all steps, -X = Y + 2Z - 4 B.E.(C-H) Rearranging, 4 B.E.(C-H) = X + Y + 2Z Therefore, B.E.(C-H) = X+Y+2Z4 Hence, the correct option is B. The hint also matches this sign convention: bond breaking and sublimation are endothermic, while bond formation is exothermic. Therefore, the bond enthalpy is X+Y+2Z4.

Consider the following data: f H^ ( methane, g) = -X kJ mol^-1 En… — Solution | JEE Chemistry

MCQMediumJEE 2026Enthalpies (Bond, Combustion, Formation…)

Consider the following data : $\Delta_f H^\ominus (\text{methane}, g) = -X \; \text{kJ mol}^{-1}$

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

Dissociation enthalpy of $H_2 = Z \; \text{kJ mol}^{-1}$

The bond enthalpy of $C-H$ bond is given by :

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

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given:

$\Delta_f H^\ominus(\text{methane}, g) = -X \; \text{kJ mol}^{-1}$
Enthalpy of sublimation of graphite $= Y \; \text{kJ mol}^{-1}$
Dissociation enthalpy of $H_2 = Z \; \text{kJ mol}^{-1}$

Find: The bond enthalpy of the $C-H$ bond.

Use Hess's law and an atomization cycle.

For the standard formation of methane,

C(\text{graphite}) + 2H_2(g) \rightarrow CH_4(g)

with enthalpy change

\Delta H = -X

Hess Cycle Expansion

Break the formation process into steps shown in the solution:

Sublimation of graphite:

C(\text{graphite}) \rightarrow C(g)

\Delta H = Y

Dissociation of hydrogen:

2H_2(g) \rightarrow 4H(g)

\Delta H = 2Z

Bond formation to make methane:

C(g) + 4H(g) \rightarrow CH_4(g)

\Delta H = -4 \times \text{B.E.}(C-H)

Adding all steps,

-X = Y + 2Z - 4 \times \text{B.E.}(C-H)

Rearranging,

4 \times \text{B.E.}(C-H) = X + Y + 2Z

Therefore,

\text{B.E.}(C-H) = \frac{X+Y+2Z}{4}

Hence, the correct option is B.

The hint also matches this sign convention: bond breaking and sublimation are endothermic, while bond formation is exothermic. Therefore, the bond enthalpy is $\frac{X+Y+2Z}{4}$ .

Common mistakes

Using the wrong sign for the enthalpy of formation of methane. Here $\Delta_f H = -X$ , not $+X$ . Treating it as positive changes the final algebra. Always write the formation reaction first and substitute the sign exactly as given.
Forgetting that dissociation of $2H_2$ produces $4H$ atoms, so the total enthalpy is $2Z$ , not $Z$ or $4Z$ . Count the number of moles of $H_2$ molecules being dissociated before substituting.
Missing the factor of $4$ for methane. Since $CH_4$ contains four $C-H$ bonds, the bond-formation step is $-4 \times \text{B.E.}(C-H)$ . Use the total bond enthalpy first, then divide by $4$ to get one bond.

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