Calculate the enthalpy of hydrogenation :
`C_(2)H_(5)(g) + H_(2)(g) to C_(2)H_(6)(g)`
Given that the bond enthalpies of H - H , C = C, C- C and C - H bonds are 433, 615, 347 and 413 kJ `mol^(-1)` respectively.

Explanation

To calculate the enthalpy of hydrogenation for the reaction:

\[ C_2H_4(g) + H_2(g) \rightarrow C_2H_6(g) \]

we will use the bond enthalpy values provided. The bond enthalpies given are:
- H-H: 433 kJ/mol
- C=C: 615 kJ/mol
- C-C: 347 kJ/mol
- C-H: 413 kJ/mol

### Step-by-Step Solution:

1. **Identify the bonds in the reactants and products:**
- In the reactant \( C_2H_4 \) (ethene), we have:
- 1 C=C bond
- 4 C-H bonds
- In the reactant \( H_2 \), we have:
- 1 H-H bond
- In the product \( C_2H_6 \) (ethane), we have:
- 1 C-C bond
- 6 C-H bonds

2. **Calculate the total bond enthalpy for the reactants:**
- The total bond enthalpy for the reactants is calculated as follows:
\[ \text{Total bond enthalpy (reactants)} = \text{C=C} + 4 \times \text{C-H} + \text{H-H} \]
- Substituting the values:
\[ = 615 \, \text{kJ/mol} + 4 \times 413 \, \text{kJ/mol} + 433 \, \text{kJ/mol} \]
- Calculate:
\[ = 615 + 1652 + 433 = 2700 \, \text{kJ/mol} \]

3. **Calculate the total bond enthalpy for the products:**
- The total bond enthalpy for the products is calculated as follows:
\[ \text{Total bond enthalpy (products)} = 6 \times \text{C-H} + \text{C-C} \]
- Substituting the values:
\[ = 6 \times 413 \, \text{kJ/mol} + 347 \, \text{kJ/mol} \]
- Calculate:
\[ = 2478 + 347 = 2825 \, \text{kJ/mol} \]

4. **Calculate the enthalpy change for the reaction:**
- The enthalpy change (\( \Delta H \)) for the reaction is given by:
\[ \Delta H = \text{Total bond enthalpy (reactants)} - \text{Total bond enthalpy (products)} \]
- Substituting the values:
\[ \Delta H = 2700 \, \text{kJ/mol} - 2825 \, \text{kJ/mol} \]
- Calculate:
\[ \Delta H = -125 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of hydrogenation for the reaction is:
\[ \Delta H = -125 \, \text{kJ/mol} \]