The bond dissociation energy depends upon the nature of the bond and nature of the molecule. If any molecule more than 1 bonds of similar nature are present then the bond energy reported is the average bond energy.
Determine `C-C and C-H` bond enthalpy (in kJ/mol). Given: `Delta_(f)H^(0) (C_(2)H_(6),g)= -85kJ//mol, Delta_(f) H^(0) (C_(3)H_(8), g)= -104kJ//mole, Delta_("sub")H^(0) (C,s)= 718kJ//mol`, B.E. (H-H)= 436 kJ/mol,

Options

A. 414345
B. 345414
C. 287405.5
D. None of these

Answer

To determine the bond enthalpy of C-C and C-H bonds in ethane (C₂H₆) and propane (C₃H₈), we will use the given standard enthalpy of formation values and the bond enthalpy of hydrogen.

### Step-by-Step Solution:

1. **Write the Formation Reactions**:
- For ethane (C₂H₆):
\[ 2C(s) + 3H_2(g) \rightarrow C_2H_6(g) \quad \Delta_f H^\circ = -85 \text{ kJ/mol} \]
- For propane (C₃H₈):
\[ 3C(s) + 4H_2(g) \rightarrow C_3H_8(g) \quad \Delta_f H^\circ = -104 \text{ kJ/mol} \]

2. **Use the Given Data**:
- Sublimation enthalpy of carbon:
\[ \Delta_{sub} H^\circ (C, s) = 718 \text{ kJ/mol} \]
- Bond dissociation energy of hydrogen:
\[ B.E. (H-H) = 436 \text{ kJ/mol} \]

3. **Set Up the Equations**:
- For ethane:
\[ 2 \times 718 + 3 \times 436 - x - 6y = -85 \]
Simplifying gives:
\[ 1436 + 1308 - x - 6y = -85 \]
\[ -x - 6y = -85 - 2744 \]
\[ x + 6y = 2859 \quad \text{(Equation 1)} \]

- For propane:
\[ 3 \times 718 + 4 \times 436 - 3x - 8y = -104 \]
Simplifying gives:
\[ 2154 + 1744 - 3x - 8y = -104 \]
\[ -3x - 8y = -104 - 3898 \]
\[ 3x + 8y = 4002 \quad \text{(Equation 2)} \]

4. **Solve the Equations**:
- From Equation 1:
\[ x + 6y = 2859 \]
- From Equation 2:
\[ 3x + 8y = 4002 \]

Multiply Equation 1 by 3:
\[ 3x + 18y = 8577 \quad \text{(Equation 3)} \]

Now subtract Equation 2 from Equation 3:
\[ (3x + 18y) - (3x + 8y) = 8577 - 4002 \]
\[ 10y = 4575 \]
\[ y = 457.5 \text{ kJ/mol} \]

Substitute \(y\) back into Equation 1:
\[ x + 6(457.5) = 2859 \]
\[ x + 2745 = 2859 \]
\[ x = 114 \text{ kJ/mol} \]

5. **Final Results**:
- C-C bond enthalpy \(x = 114 \text{ kJ/mol}\)
- C-H bond enthalpy \(y = 457.5 \text{ kJ/mol}\)