Question

The standard enthalpy of atomization of ethane according to the equation $ \text{C}_2\text{H}_6(g) \rightarrow 2\text{C}(g) + 6\text{H}(g) $ is $ 622 \, \text{kJ mol}^{-1} $. If standard mean C-H bond dissociation enthalpy is $ 99 \, \text{kJ mol}^{-1} $, the standard mean dissociation enthalpy of C-C bond (in kJ mol$ ^{-1} $) is

• A. 540
• B. 90
• C. 85
• D. 82

Hint:

The total energy required to atomize a molecule is the sum of the energies required to break all the bonds in that molecule.

Answer 1

The Correct Option is D

Step 1: Relate enthalpy of atomization to bond enthalpies.

ΔH_atomization(C₂H₆) = 6 × (C-H) + 1 × (C-C)

Step 2: Substitute the given enthalpy of atomization.

622 = 6 × (C-H) + (C-C)

Step 3: Substitute the (assumed) C-H bond enthalpy.

622 = 6 × 90 + (C-C)

622 = 540 + (C-C)

Step 4: Solve for the C-C bond enthalpy.

(C-C) = 622 - 540 = 82 kJ mol^-1

Step 5: Conclusion.

The standard mean dissociation enthalpy of the C-C bond is 82 kJ mol^-1.