Determine C-C and C-H bond enthalpy (in `kJ//mol`)
Given : `Delta_(f)H^(@)(C_(2)H_(6),g)=-85kJ//mol," "Delta_(f)H^(@)(C_(3)H_(8),g)=-104kJ//mol`
`Delta_("sub")H^(@)(C,s)=718kJ//mol," "B.E.(H-H)=436kJ//mol`

Options

A. 414345
B. 345414
C. `287404.5`
D. None of these

Explanation

To determine the C-C and C-H bond enthalpy in kJ/mol, we will use the given data and apply Hess's law. Here’s a step-by-step solution:

### Step 1: Write the formation equations for C2H6 and C3H8

1. **C2H6 Formation:**
\[ 2C(s) + 3H_2(g) \rightarrow C_2H_6(g) \quad \Delta_f H^\circ = -85 \text{ kJ/mol} \]

2. **C3H8 Formation:**
\[ 3C(s) + 4H_2(g) \rightarrow C_3H_8(g) \quad \Delta_f H^\circ = -104 \text{ kJ/mol} \]

### Step 2: Write the sublimation and bond dissociation equations

- **Sublimation of Carbon:**
\[ C(s) \rightarrow C(g) \quad \Delta_{sub} H^\circ = 718 \text{ kJ/mol} \]

- **Bond Enthalpy of H-H:**
\[ H_2(g) \rightarrow 2H(g) \quad B.E.(H-H) = 436 \text{ kJ/mol} \]

### Step 3: Set up the equations using Hess's law

For **C2H6**:
\[ \Delta_f H^\circ = \Delta H_{sublimation} + \Delta H_{bond} - \Delta H_{formation} \]
\[ -85 = 2 \times 718 + 3 \times 436 - (x + 6y) \]
Where:
- \(x\) = C-C bond enthalpy
- \(y\) = C-H bond enthalpy

This simplifies to:
\[ -85 = 1436 + 1308 - (x + 6y) \]
\[ x + 6y = 2829 \quad \text{(Equation 1)} \]

For **C3H8**:
\[ \Delta_f H^\circ = \Delta H_{sublimation} + \Delta H_{bond} - \Delta H_{formation} \]
\[ -104 = 3 \times 718 + 4 \times 436 - (2x + 8y) \]
This simplifies to:
\[ -104 = 2154 + 1744 - (2x + 8y) \]
\[ 2x + 8y = 4002 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations

From Equation 1:
\[ x + 6y = 2829 \]

From Equation 2:
\[ 2x + 8y = 4002 \]

Now, we can multiply Equation 1 by 2:
\[ 2x + 12y = 5658 \quad \text{(Equation 3)} \]

Now subtract Equation 2 from Equation 3:
\[ (2x + 12y) - (2x + 8y) = 5658 - 4002 \]
\[ 4y = 1656 \]
\[ y = 414 \text{ kJ/mol} \]

### Step 5: Substitute \(y\) back to find \(x\)

Using \(y = 414\) in Equation 1:
\[ x + 6(414) = 2829 \]
\[ x + 2484 = 2829 \]
\[ x = 345 \text{ kJ/mol} \]

### Final Results
- C-C bond enthalpy (\(x\)) = 345 kJ/mol
- C-H bond enthalpy (\(y\)) = 414 kJ/mol