Question

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. \( X+Y+Z \)

Hint:

Bond formation is exothermic (negative enthalpy change), while bond breaking and sublimation are endothermic (positive enthalpy change). Always track the signs carefully in Hess cycles.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The enthalpy of formation of a compound can be calculated using a Hess cycle involving the atomization of elements and the formation of bonds.

Step 2: Key Formula or Approach:
\( \Delta_f H = \sum \text{Bond Enthalpies (reactants)} - \sum \text{Bond Enthalpies (products)} \).
Alternatively, using an atomization cycle :
\( \Delta_f H = \Delta H_{sub}(C) + 2 \Delta H_{diss}(H_2) - 4 \times B.E.(C-H) \).

Step 3: Detailed Explanation:
Standard reaction for formation of methane :
\( C(graphite) + 2H_2(g) \rightarrow CH_4(g) \) ; \( \Delta H = -X \)
Hess Cycle path :
1. \( C(graphite) \rightarrow C(g) \) ; \( \Delta H = Y \)
2. \( 2H_2(g) \rightarrow 4H(g) \) ; \( \Delta H = 2 \times Z = 2Z \)
3. \( C(g) + 4H(g) \rightarrow CH_4(g) \) ; \( \Delta H = -4 \times B.E.(C-H) \)
Adding these steps :
\( -X = Y + 2Z - 4 \times B.E.(C-H) \)
\( 4 \times B.E.(C-H) = X + Y + 2Z \)
\( B.E.(C-H) = \frac{X+Y+2Z}{4} \).

Step 4: Final Answer:
The bond enthalpy is \( \frac{X+Y+2Z}{4} \).