From the following bond energies: `H-H` bond energy: `431.37KJmol^(-1)`
`C=C` bond energy: `606.10KJmol^(-1)`
`C-C` bond energy: `336.49KJmol^(-1)`
`C-H` bond energy: `410.50KJmol^(-1)`
Enthalpy for the reaction will be:
`overset(H)overset(|)underset(H)underset(|)(C)=overset(H)overset(|)underset(H)underset(|)(C)+H-H to Hoverset(H)overset(|)underset(H)underset(|)(C)-overset(H)overset(|)underset(H)underset(|)(C)-H`

Options

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

Explanation

To calculate the enthalpy change for the given reaction using bond energies, we will follow these steps:

### Step 1: Identify the bonds in the reactants and products
The reaction given is:
\[ \text{C} \equiv \text{C} + \text{H-H} \rightarrow \text{C-H} + \text{C-H} + \text{H} \]

**Reactants:**
1. C=C (1 double bond)
2. H-H (1 single bond)

**Products:**
1. C-H (4 single bonds)
2. C-C (1 single bond)

### Step 2: Write down the bond energies
From the data provided:
- H-H bond energy = 431.37 kJ/mol
- C=C bond energy = 606.10 kJ/mol
- C-C bond energy = 336.49 kJ/mol
- C-H bond energy = 410.50 kJ/mol

### Step 3: Calculate the total bond energy for reactants
Using the bond energies:
- For the reactants:
- 1 C=C bond: \( 1 \times 606.10 \, \text{kJ/mol} \)
- 1 H-H bond: \( 1 \times 431.37 \, \text{kJ/mol} \)

Total bond energy for reactants:
\[ \text{Total BE (reactants)} = 606.10 + 431.37 = 1037.47 \, \text{kJ/mol} \]

### Step 4: Calculate the total bond energy for products
Using the bond energies:
- For the products:
- 1 C-C bond: \( 1 \times 336.49 \, \text{kJ/mol} \)
- 4 C-H bonds: \( 4 \times 410.50 \, \text{kJ/mol} \)

Total bond energy for products:
\[ \text{Total BE (products)} = 336.49 + (4 \times 410.50) = 336.49 + 1642.00 = 1978.49 \, \text{kJ/mol} \]

### Step 5: Apply Hess's Law to find the enthalpy change
According to Hess's Law:
\[ \Delta H = \text{Total BE (reactants)} - \text{Total BE (products)} \]
Substituting the values:
\[ \Delta H = 1037.47 - 1978.49 = -941.02 \, \text{kJ/mol} \]

### Step 6: Conclusion
The enthalpy change for the reaction is:
\[ \Delta H = -941.02 \, \text{kJ/mol} \]