If `Delta_(f)H^(@)(C_(2)H_(4))` and `Delta_(f)H^(@)(C_(2)H_(6))` are `x_(1)` and `x_(2)` kcal `mol^(-1)`, then heat of hydrogenation of `C_(2)H_(4)` is :

Options

A. `x_(1)+x_(2)`
B. `x_(1)-x_(2)`
C. `x_(2)-x_(1)`
D. `x_(1)+2x_(2)`

Answer

To find the heat of hydrogenation of ethylene (C₂H₄), we can use the standard heats of formation of the reactants and products involved in the reaction.

### Step-by-Step Solution:

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

2. **Understand the Concept of Heat of Formation**: The standard heat of formation (\( \Delta_f H^\circ \)) of a compound is the heat change that results when one mole of the compound is formed from its elements in their standard states.

3. **Write the Heat of Hydrogenation Formula**: The heat of hydrogenation can be calculated using the standard heats of formation of the products and reactants:
\[ \text{Heat of Hydrogenation} = \Delta_f H^\circ (\text{C}_2\text{H}_6) - \Delta_f H^\circ (\text{C}_2\text{H}_4) \]

4. **Substitute the Values**: Let \( \Delta_f H^\circ (\text{C}_2\text{H}_4) = x_1 \) and \( \Delta_f H^\circ (\text{C}_2\text{H}_6) = x_2 \). Therefore, the heat of hydrogenation can be expressed as:
\[ \text{Heat of Hydrogenation} = x_2 - x_1 \text{ kcal/mol} \]

5. **Final Answer**: The heat of hydrogenation of C₂H₄ is:
\[ x_2 - x_1 \text{ kcal/mol} \]