From the following bond energies
`H-H` bond energy `431.37 kJ mol^(-1)`
`C=C` bond energy `606.10 kJ mol^(-1)`
`C-C` bond energy `336.49 kJ mol^(-1)`
`C-H` bond energy `410.5 kJ mol^(-1)`
Enthalpy for the reaction
`overset(H)overset(|)underset(H)underset(|)(C )=overset(H)overset(|)underset(H)underset(|)(C )+H-HtoH-overset(H)overset(|)underset(H)underset(|)(C )-overset(H)overset(|)underset(H)underset(|)(C )-H`
will be

Options

A. `-243.6 kJ mol^(-1)`
B. `-120.0 kJ mol^(-1)`
C. `553.0 kJ mol^(-1)`
D. `1523.6 kJ mol^(-1)`

Answer

To calculate the enthalpy change (ΔH) for the given reaction, we will use the bond energies provided. The reaction can be represented as follows:

\[ \text{C}_2\text{H}_4 + \text{H}_2 \rightarrow \text{C}_2\text{H}_6 \]

### Step 1: Identify the bonds in the reactants and products

- **Reactants:**
- Ethene (C2H4) has:
- 1 C=C bond
- 4 C-H bonds
- Hydrogen (H2) has:
- 1 H-H bond

- **Products:**
- Ethane (C2H6) has:
- 6 C-H bonds
- 1 C-C bond

### Step 2: Write the bond energy equation

The enthalpy change for the reaction can be calculated using the formula:

\[ \Delta H = \text{(Bond energies of reactants)} - \text{(Bond energies of products)} \]

### Step 3: Calculate the bond energies for reactants

- For Ethene (C2H4):
- C=C bond energy: 606.10 kJ/mol
- C-H bonds: 4 × 410.5 kJ/mol = 1642 kJ/mol

- For Hydrogen (H2):
- H-H bond energy: 431.37 kJ/mol

Total bond energy of reactants:
\[ \text{Total (Reactants)} = 606.10 + 1642 + 431.37 = 2679.47 \text{ kJ/mol} \]

### Step 4: Calculate the bond energies for products

- For Ethane (C2H6):
- C-C bond energy: 336.49 kJ/mol
- C-H bonds: 6 × 410.5 kJ/mol = 2463 kJ/mol

Total bond energy of products:
\[ \text{Total (Products)} = 336.49 + 2463 = 2799.49 \text{ kJ/mol} \]

### Step 5: Calculate ΔH

Now, substituting the values into the ΔH equation:

\[ \Delta H = \text{Total (Reactants)} - \text{Total (Products)} \]
\[ \Delta H = 2679.47 - 2799.49 = -120.02 \text{ kJ/mol} \]

### Conclusion

The enthalpy change (ΔH) for the reaction is approximately:
\[ \Delta H \approx -120 \text{ kJ/mol} \]

### Final Answer

The enthalpy for the reaction is approximately **-120 kJ/mol**.

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