`Delta H_(1)C_(2)H_(4) = 12.5 kcal`
Heat of atomisation of C = 171 Kcal
Bond energy of `H_(2)` = `104.3 kcal`
Bond energy C - H = 99.3 kcal
What is C = C bond energy?

Options

A. `140.9 kcal`
B. `49 kcal`
C. `40 kcal`
D. `76 kcal`

Answer

To find the bond energy of the carbon-carbon double bond (C=C) in ethylene (C2H4), we can use the given data and apply the concept of bond energies and the enthalpy of formation. Here’s a step-by-step solution:

### Step 1: Write the formation reaction of C2H4
The formation of ethylene (C2H4) can be represented as:
\[ 2C + 2H_2 \rightarrow C_2H_4 \]

### Step 2: Identify the bond energies and given values
- Heat of formation of C2H4, \( \Delta H_{f} = 12.5 \, \text{kcal} \)
- Heat of atomization of carbon, \( \Delta H_{atomization} = 171 \, \text{kcal} \) (for 1 mole of C)
- Bond energy of H2, \( BE(H_2) = 104.3 \, \text{kcal} \) (for 1 mole of H2)
- Bond energy of C-H bond, \( BE(C-H) = 99.3 \, \text{kcal} \)

### Step 3: Write the equation for the enthalpy change
Using the formula for the enthalpy change of formation:
\[ \Delta H_{f} = \text{(Bond energy of reactants)} - \text{(Bond energy of products)} \]

### Step 4: Calculate the bond energies of the reactants
For the reactants:
- 2 atoms of carbon: \( 2 \times 171 \, \text{kcal} = 342 \, \text{kcal} \)
- 2 molecules of hydrogen: \( 2 \times 104.3 \, \text{kcal} = 208.6 \, \text{kcal} \)

Total bond energy of reactants:
\[ BE_{reactants} = 342 + 208.6 = 550.6 \, \text{kcal} \]

### Step 5: Calculate the bond energies of the products
For the products (C2H4):
- There are 4 C-H bonds: \( 4 \times 99.3 \, \text{kcal} = 397.2 \, \text{kcal} \)
- Let the bond energy of the C=C bond be \( x \).

Total bond energy of products:
\[ BE_{products} = 397.2 + x \]

### Step 6: Set up the equation
Now, substituting into the enthalpy change equation:
\[ 12.5 = 550.6 - (397.2 + x) \]

### Step 7: Solve for x
Rearranging the equation:
\[ 12.5 = 550.6 - 397.2 - x \]
\[ 12.5 = 153.4 - x \]
\[ x = 153.4 - 12.5 \]
\[ x = 140.9 \, \text{kcal} \]

### Conclusion
The bond energy of the carbon-carbon double bond (C=C) in ethylene is:
\[ \text{Bond energy of C=C} = 140.9 \, \text{kcal} \]

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