1. Write the reaction

The given reaction is hydrogenation of ethylene to ethane:

$\mathrm{C_2H_4 + H_2 \rightarrow C_2H_6}$

2. Use bond energies

Enthalpy change using bond energies is:

$\Delta H = \sum (\text{bond energies of bonds broken}) - \sum (\text{bond energies of bonds formed})$

3. Identify bonds broken and formed

In the reaction:

• One $\mathrm{C=C}$ bond in ethylene becomes one $\mathrm{C-C}$ bond in ethane.
• One $\mathrm{H-H}$ bond is broken.
• Two new $\mathrm{C-H}$ bonds are formed.

So,

Bonds broken:

• $1 \times \mathrm{C=C} = 615\,\text{kJ}$
• $1 \times \mathrm{H-H} = 435\,\text{kJ}$

Total energy of bonds broken:

$615 + 435 = 1050\,\text{kJ}$

Bonds formed:

• $1 \times \mathrm{C-C} = 347\,\text{kJ}$
• $2 \times \mathrm{C-H} = 2 \times 414 = 828\,\text{kJ}$

Total energy of bonds formed:

$347 + 828 = 1175\,\text{kJ}$

4. Calculate enthalpy change

$\Delta H = 1050 - 1175 = -125\,\text{kJ}$

5. Final integer

Since the question asks:

"The enthalpy ... is $-\_\_\_\_\_\_\_\_\_\_\mathrm{kJ}$"

the required integer is:

$125$

6. Comparison with stored answer

Stored correct answer = $125$

This matches our derived answer.