`Delta H ` for the reaction `H-C-= N(g) + 2H_(2)(g) rarr H - underset(H) underset(|)overset(H ) overset(|)(C) - overset(H) overset(|)(N) - H (g)`
is `- 150 kJ `. Calculate the bond energy of`C -= N`bond.
[ Given bond energies of `C-H= 414 kJ mol^(-1)`, `H-H= 435kJ mol^(-1)`, `C-N = 293kJ mol^(-1)`, `N-H =396 kJ mol^(-1)`]

Explanation

To calculate the bond energy of the `C≡N` bond for the given reaction, we can use the following steps:

### Step 1: Write the reaction and identify the bonds
The reaction is:
\[ H-C \equiv N(g) + 2H_2(g) \rightarrow H_3C-NH_2(g) \]

### Step 2: Identify the bonds in the reactants and products
- **Reactants:**
- 1 `C≡N` bond
- 2 `H-H` bonds (from `2H_2`)

- **Products:**
- 3 `C-H` bonds (from `H_3C`)
- 2 `N-H` bonds (from `NH_2`)

### Step 3: Write the equation for ΔH in terms of bond energies
The change in enthalpy (ΔH) can be expressed as:
\[ \Delta H = \text{(Sum of bond energies of reactants)} - \text{(Sum of bond energies of products)} \]

Given:
- ΔH = -150 kJ
- Bond energies:
- `C-H` = 414 kJ/mol
- `H-H` = 435 kJ/mol
- `N-H` = 396 kJ/mol
- Let the bond energy of `C≡N` = x kJ/mol

### Step 4: Write the equation using bond energies
Using the information from the bonds:
\[ -150 = [x + 2 \times 435] - [3 \times 414 + 2 \times 396] \]

### Step 5: Simplify the equation
Calculating the bond energies:
- For reactants:
\[ x + 870 \quad \text{(since } 2 \times 435 = 870\text{)} \]

- For products:
\[ 3 \times 414 + 2 \times 396 = 1242 + 792 = 2034 \]

Putting it all together:
\[ -150 = [x + 870] - [2034] \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ -150 = x + 870 - 2034 \]
\[ -150 = x - 1164 \]
\[ x = -150 + 1164 \]
\[ x = 1014 \text{ kJ/mol} \]

### Step 7: Conclusion
The bond energy of the `C≡N` bond is:
\[ \text{Bond energy of } C≡N = 1014 \text{ kJ/mol} \]