If a `298K` the bond energies of `C-H,C-C,C=C` and `H-H` bonds are respectivly `414, 347, 615KJmol^(-1)` , the vlaue of enthalpy change for the reaction
`H_(2)C=CH_(2)(g)+H_(2)(g)+H_(2)(g)rarrH_(3)C-CH_(3)(g)` at `298K` will be

Options

A. `+124kJ`
B. `-125kJ`
C. `+250kJ`
D. `-250kJ`

Answer

To find the enthalpy change (ΔH) for the reaction:

\[ \text{H}_2\text{C}=\text{CH}_2(g) + \text{H}_2(g) \rightarrow \text{H}_3\text{C}-\text{CH}_3(g) \]

we will use the bond energies provided and apply the formula for enthalpy change based on bond energies.

### Step 1: Identify the bonds broken and formed

1. **Reactants**:
- In ethylene (\( \text{H}_2\text{C}=\text{CH}_2 \)):
- 4 C-H bonds
- 1 C=C bond
- In hydrogen (\( \text{H}_2 \)):
- 1 H-H bond

Total bonds broken:
- 4 C-H
- 1 C=C
- 1 H-H

2. **Products**:
- In ethane (\( \text{H}_3\text{C}-\text{CH}_3 \)):
- 6 C-H bonds
- 1 C-C bond

Total bonds formed:
- 6 C-H
- 1 C-C

### Step 2: Write the bond energy values

- C-H bond energy = 414 kJ/mol
- C-C bond energy = 347 kJ/mol
- C=C bond energy = 615 kJ/mol
- H-H bond energy = 435 kJ/mol

### Step 3: Calculate the total energy of bonds broken

Total energy of bonds broken:
\[ \text{Energy}_{\text{broken}} = (4 \times 414) + (1 \times 615) + (1 \times 435) \]

Calculating:
\[ = 1656 + 615 + 435 = 2706 \text{ kJ/mol} \]

### Step 4: Calculate the total energy of bonds formed

Total energy of bonds formed:
\[ \text{Energy}_{\text{formed}} = (6 \times 414) + (1 \times 347) \]

Calculating:
\[ = 2484 + 347 = 2831 \text{ kJ/mol} \]

### Step 5: Calculate the enthalpy change (ΔH)

Using the formula:
\[ \Delta H = \text{Energy}_{\text{broken}} - \text{Energy}_{\text{formed}} \]
\[ \Delta H = 2706 - 2831 = -125 \text{ kJ/mol} \]

### Final Answer:
The enthalpy change for the reaction at 298 K is:
\[ \Delta H = -125 \text{ kJ/mol} \]

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