The heats of combustion of `C_(2)H_(4), C_(2)H_(6)` and `H_(2)` gases are -1409.5 KJ, -1558.3 KJ and -285.6 KJ respectively. The heat of hydrogenation of ethene is

Options

A. `-136.8 KJ`
B. `-13.68 KJ`
C. `273.6 KJ`
D. `1.368 KJ`

Explanation

To find the heat of hydrogenation of ethene (C₂H₄), we can use the heats of combustion of ethene (C₂H₄), ethane (C₂H₆), and hydrogen (H₂). The heat of hydrogenation can be calculated using the following reaction:

\[ \text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6 \]

### Step-by-step Solution:

1. **Write the heat of combustion values:**
- Heat of combustion of C₂H₄ = -1409.5 kJ
- Heat of combustion of C₂H₆ = -1558.3 kJ
- Heat of combustion of H₂ = -285.6 kJ

2. **Set up the equation for the heat of hydrogenation:**
The heat of hydrogenation (ΔH_hydrogenation) can be calculated using the formula:
\[ \Delta H_{\text{hydrogenation}} = \Delta H_{\text{combustion of products}} - \Delta H_{\text{combustion of reactants}} \]

3. **Identify the products and reactants:**
- Products: C₂H₆ (ethane)
- Reactants: C₂H₄ (ethene) and H₂ (hydrogen)

4. **Substitute the values into the equation:**
\[ \Delta H_{\text{hydrogenation}} = \Delta H_{\text{combustion of C₂H₆}} - (\Delta H_{\text{combustion of C₂H₄}} + \Delta H_{\text{combustion of H₂}}) \]
\[ \Delta H_{\text{hydrogenation}} = (-1558.3 \, \text{kJ}) - [(-1409.5 \, \text{kJ}) + (-285.6 \, \text{kJ})] \]

5. **Calculate the values inside the brackets:**
\[ -1409.5 \, \text{kJ} + (-285.6 \, \text{kJ}) = -1695.1 \, \text{kJ} \]

6. **Now substitute back into the equation:**
\[ \Delta H_{\text{hydrogenation}} = -1558.3 \, \text{kJ} - (-1695.1 \, \text{kJ}) \]
\[ \Delta H_{\text{hydrogenation}} = -1558.3 \, \text{kJ} + 1695.1 \, \text{kJ} \]
\[ \Delta H_{\text{hydrogenation}} = 136.8 \, \text{kJ} \]

7. **Final result:**
The heat of hydrogenation of ethene is **136.8 kJ**.