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. `-13.67 kJ`
C. `273.6 kJ`
D. `1.368 kJ`

Answer

To find the heat of hydrogenation of ethene (C₂H₄) to form ethane (C₂H₆), we can use the heats of combustion provided for the substances involved. The heat of hydrogenation can be calculated using the following steps:

### Step-by-Step Solution:

1. **Write the Reaction for Hydrogenation:**
The reaction for the hydrogenation of ethene (C₂H₄) to ethane (C₂H₆) can be written as:
\[ C_2H_4 + H_2 \rightarrow C_2H_6 \]

2. **Identify the Heats of Combustion:**
- The heat of combustion of C₂H₄ is given as \( -1409.5 \, \text{kJ} \).
- The heat of combustion of C₂H₆ is given as \( -1558.3 \, \text{kJ} \).
- The heat of combustion of H₂ is given as \( -285.6 \, \text{kJ} \).

3. **Apply Hess's Law:**
According to Hess's Law, the heat of a reaction can be calculated by taking the difference between the total enthalpy of the products and the total enthalpy of the reactants. Therefore, the heat of hydrogenation (\( \Delta H \)) can be calculated as:
\[ \Delta H = \text{(Heat of combustion of products)} - \text{(Heat of combustion of reactants)} \]

4. **Calculate the Heat of Hydrogenation:**
- The product is C₂H₆, and its heat of combustion is \( -1558.3 \, \text{kJ} \).
- The reactants are C₂H₄ and H₂, so their heats of combustion are \( -1409.5 \, \text{kJ} \) and \( -285.6 \, \text{kJ} \) respectively.
- Plugging in the values:
\[ \Delta H = (-1558.3) - [(-1409.5) + (-285.6)] \]
\[ \Delta H = -1558.3 - (-1409.5 - 285.6) \]
\[ \Delta H = -1558.3 - (-1695.1) \]
\[ \Delta H = -1558.3 + 1695.1 \]
\[ \Delta H = -136.8 \, \text{kJ} \]

5. **Conclusion:**
The heat of hydrogenation of ethene (C₂H₄) to form ethane (C₂H₆) is \( -136.8 \, \text{kJ} \).