From the following data :
Enthalpy of formation of `CH_(3)CN`=87.86 kJ/mol
Enthalpy of formation of `C_(2)H_(6)=-` 83.68 kJ/mol
Enthalpy of sublimation of graphite =719.65 kJ/mol
enthalpy of dissociation of nitrogen =945.58 kJ/mol
Enthalpy of dissociation of `H_(2) =` 435.14 kJ/mol
`C-H` bond enthalpy =414.22 kJ/mol
Calculate the bond enthalpy of (i) `C-C` , (ii) `C equiv N`

Explanation

To calculate the bond enthalpy of (i) `C-C` and (ii) `C≡N`, we will use the given enthalpy values and apply Hess's law. Here’s a step-by-step breakdown of the calculations:

### Step 1: Write the Reaction for Formation of `CH₃CN`
The formation of `CH₃CN` can be represented as:
\[ 2C (s) + \frac{3}{2} H_2 (g) + \frac{1}{2} N_2 (g) \rightarrow CH_3CN (l) \]

### Step 2: Write the Enthalpy Change for the Reaction
According to Hess's law, the enthalpy change for the formation of `CH₃CN` can be expressed as:
\[ \Delta H_f = \Delta H_{sublimation} + \Delta H_{bond\ dissociation} - \Delta H_{bonds\ formed} \]

### Step 3: Identify the Components
1. **Enthalpy of formation of `CH₃CN`**: \( \Delta H_f = 87.86 \, \text{kJ/mol} \)
2. **Enthalpy of sublimation of graphite (to form carbon atoms)**: \( \Delta H_{sublimation} = 719.65 \, \text{kJ/mol} \)
3. **Enthalpy of dissociation of \( H_2 \)**: \( \Delta H_{H-H} = 435.14 \, \text{kJ/mol} \)
4. **Enthalpy of dissociation of \( N_2 \)**: \( \Delta H_{N-N} = 945.58 \, \text{kJ/mol} \)
5. **C-H bond enthalpy**: \( \Delta H_{C-H} = 414.22 \, \text{kJ/mol} \)

### Step 4: Set Up the Equation
The total bond enthalpy for the bonds broken and formed can be expressed as:
\[ 87.86 = 719.65 + \frac{3}{2} \times 435.14 + \frac{1}{2} \times 945.58 - (3 \times 414.22 + x + y) \]

Where:
- \( x \) is the bond enthalpy of `C-C`
- \( y \) is the bond enthalpy of `C≡N`

### Step 5: Calculate the Total Energy for Bonds Broken
Calculating the total energy for bonds broken:
1. For \( H_2 \): \( \frac{3}{2} \times 435.14 = 652.71 \, \text{kJ/mol} \)
2. For \( N_2 \): \( \frac{1}{2} \times 945.58 = 472.79 \, \text{kJ/mol} \)

Total energy for bonds broken:
\[ 719.65 + 652.71 + 472.79 = 1845.15 \, \text{kJ/mol} \]

### Step 6: Calculate Total Energy for Bonds Formed
The total energy for bonds formed includes:
- 3 C-H bonds: \( 3 \times 414.22 = 1242.66 \, \text{kJ/mol} \)
- 1 C-C bond: \( x \)
- 1 C≡N bond: \( y \)

### Step 7: Rearranging the Equation
Now we can rearrange the equation:
\[ 87.86 = 1845.15 - (1242.66 + x + y) \]
\[ x + y = 1845.15 - 87.86 - 1242.66 \]
\[ x + y = 514.63 \]

### Step 8: Solve for Individual Bond Enthalpies
To find the individual bond enthalpies, we need additional information or assumptions. However, if we assume a typical value for `C-C` bond enthalpy (e.g., around 348 kJ/mol), we can solve for `C≡N`:

Assuming \( x = 348 \):
\[ 348 + y = 514.63 \]
\[ y = 514.63 - 348 \]
\[ y = 166.63 \, \text{kJ/mol} \]

### Final Results
1. **Bond enthalpy of `C-C`**: \( 348 \, \text{kJ/mol} \) (assumed)
2. **Bond enthalpy of `C≡N`**: \( 166.63 \, \text{kJ/mol} \)