If at 298 K the bond energies of `C-H,C-C,C=CandH-H` bonds are respectively 414, 347, 615 and 435 kJ `mol^(-1)`, the value of enthalpy change for the reaction given below at 298 K will be
`H_(2)C=CH_(2)(g)+H_(2)(g)toH_(3)C-CH_(3)(g)`

Options

A. `-250` KJ
B. `+125` KJ
C. `-125` KJ
D. `+250` KJ

Answer

To calculate 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.

### Step 1: Identify the bonds in the reactants and products
- **Reactants:**
- Ethylene (H₂C=CH₂) has:
- 1 C=C bond
- 4 C-H bonds (2 C-H bonds for each carbon)
- Hydrogen (H₂) has:
- 1 H-H bond

- **Products:**
- Ethane (H₃C-CH₃) has:
- 6 C-H bonds (3 C-H bonds for each carbon)
- 1 C-C bond

### Step 2: Write down the bond energies
- 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 bond energy for reactants
- For H₂C=CH₂:
- 1 C=C bond: 615 kJ/mol
- 4 C-H bonds: \(4 \times 414 = 1656 \, \text{kJ/mol}\)

- For H₂:
- 1 H-H bond: 435 kJ/mol

**Total bond energy of reactants:**
\[ \text{Total (Reactants)} = 615 + 1656 + 435 = 2706 \, \text{kJ/mol} \]

### Step 4: Calculate the total bond energy for products
- For H₃C-CH₃:
- 6 C-H bonds: \(6 \times 414 = 2484 \, \text{kJ/mol}\)
- 1 C-C bond: 347 kJ/mol

**Total bond energy of products:**
\[ \text{Total (Products)} = 2484 + 347 = 2831 \, \text{kJ/mol} \]

### Step 5: Calculate the enthalpy change (ΔH)
Using the formula:
\[ \Delta H = \text{Total bond energy of reactants} - \text{Total bond energy of products} \]
\[ \Delta H = 2706 - 2831 = -125 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy change (ΔH) for the reaction at 298 K is \(-125 \, \text{kJ/mol}\).

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