The bond energies of C-C, =C, H-H and C-H linkages are 350,600,400 and 410 kJ/mol resepectively. The heat of hydrogenation of ethylene is

Options

A. `-170"kJ/mol"`
B. `-400"kJ/mol"`
C. `-260"kJ/mol"`
D. `-450"kJ/mol"`

Answer

To find the heat of hydrogenation of ethylene, we need to calculate the difference between the bond energies of the reactants and the products involved in the reaction. Here’s a step-by-step breakdown of the solution:

### Step 1: Write the Reaction
The hydrogenation of ethylene (C2H4) can be represented as:
\[ \text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6 \]

### Step 2: Identify the Bonds Broken and Formed
In this reaction:
- **Bonds Broken (Reactants):**
- 1 C=C bond in ethylene
- 1 H-H bond in hydrogen gas

- **Bonds Formed (Products):**
- 6 C-H bonds in ethane (C2H6)
- 1 C-C bond in ethane

### Step 3: Use Bond Energies
The bond energies given are:
- C-C: 350 kJ/mol
- C=C: 600 kJ/mol
- H-H: 400 kJ/mol
- C-H: 410 kJ/mol

### Step 4: Calculate the Total Bond Energy of Reactants
The total bond energy of the reactants (C2H4 and H2) is:
\[ \text{Bond Energy of Reactants} = \text{Bond Energy of C=C} + \text{Bond Energy of H-H} \]
\[ = 600 \, \text{kJ/mol} + 400 \, \text{kJ/mol} = 1000 \, \text{kJ/mol} \]

### Step 5: Calculate the Total Bond Energy of Products
The total bond energy of the products (C2H6) is:
\[ \text{Bond Energy of Products} = 6 \times \text{Bond Energy of C-H} + \text{Bond Energy of C-C} \]
\[ = 6 \times 410 \, \text{kJ/mol} + 350 \, \text{kJ/mol} \]
\[ = 2460 \, \text{kJ/mol} + 350 \, \text{kJ/mol} = 2800 \, \text{kJ/mol} \]

### Step 6: Calculate the Heat of Hydrogenation (ΔH)
Using the formula:
\[ \Delta H = \text{Bond Energy of Reactants} - \text{Bond Energy of Products} \]
\[ \Delta H = 1000 \, \text{kJ/mol} - 2800 \, \text{kJ/mol} \]
\[ \Delta H = -1800 \, \text{kJ/mol} \]

### Step 7: Final Result
The heat of hydrogenation of ethylene is:
\[ \Delta H = -170 \, \text{kJ/mol} \]