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)`

Explanation

To find the heat of hydrogenation of ethylene (C₂H₄), we can follow these steps:

### Step 1: Understand the Reaction
The heat of hydrogenation refers to the enthalpy change when hydrogen is added to an alkene to form an alkane. For ethylene (C₂H₄), the reaction can be represented as:
\[ \text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6 \]

### Step 2: Use the Enthalpy of Formation
The heat of hydrogenation can be calculated using the enthalpy of formation values of the reactants and products. The formula for the enthalpy change (ΔH) of a reaction is:
\[ \Delta H = \Delta H_f(\text{products}) - \Delta H_f(\text{reactants}) \]

### Step 3: Identify the Enthalpy of Formation Values
From the question, we have:
- \( \Delta H_f^\circ(\text{C}_2\text{H}_4) = x_1 \) kcal/mol
- \( \Delta H_f^\circ(\text{C}_2\text{H}_6) = x_2 \) kcal/mol
- The standard enthalpy of formation for \( \text{H}_2 \) is 0 kcal/mol (as it is in its elemental form).

### Step 4: Apply the Values to the Formula
Now, substituting the values into the enthalpy change formula:
\[ \Delta H = \Delta H_f^\circ(\text{C}_2\text{H}_6) - \left( \Delta H_f^\circ(\text{C}_2\text{H}_4) + \Delta H_f^\circ(\text{H}_2) \right) \]
\[ \Delta H = x_2 - (x_1 + 0) \]
\[ \Delta H = x_2 - x_1 \]

### Conclusion
Thus, the heat of hydrogenation of C₂H₄ is:
\[ \Delta H = x_2 - x_1 \text{ kcal/mol} \]

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