Calculate `Delta H ^(@)` for the reaction `CH_(2)= CH_(2) + 3O_(2) rarr 2CO_(2)+ 2H_(2)O`
Given that the average bond energies of the different bonds are `:`
`{:("Bond",C-H,O=O,C=O,O-H,C=C),("Bond energy " ( kJ mol^(-1)),414,499,724,460,619):}`

Explanation

To calculate the standard enthalpy change (ΔH°) for the reaction:

\[ \text{CH}_2= \text{CH}_2 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O} \]

we will use the bond energies provided. The formula to calculate ΔH° is:

\[ \Delta H^\circ = \text{Sum of bond energies of reactants} - \text{Sum of bond energies of products} \]

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

**Reactants:**
- Ethylene (CH₂=CH₂):
- 4 C-H bonds
- 1 C=C bond
- Oxygen (O₂):
- 3 O=O bonds

**Products:**
- Carbon Dioxide (CO₂):
- 2 CO₂ molecules → 4 C=O bonds
- Water (H₂O):
- 2 H₂O molecules → 4 O-H bonds

### Step 2: Calculate the total bond energies for reactants

**Bond Energies:**
- C-H: 414 kJ/mol
- C=C: 619 kJ/mol
- O=O: 499 kJ/mol

**Total bond energy for reactants:**
- For CH₂=CH₂:
- C-H: \(4 \times 414 = 1656 \, \text{kJ/mol}\)
- C=C: \(1 \times 619 = 619 \, \text{kJ/mol}\)
- For 3 O₂:
- O=O: \(3 \times 499 = 1497 \, \text{kJ/mol}\)

**Total bond energy of reactants:**
\[ \text{Total} = 1656 + 619 + 1497 = 3772 \, \text{kJ/mol} \]

### Step 3: Calculate the total bond energies for products

**Bond Energies:**
- C=O: 724 kJ/mol
- O-H: 460 kJ/mol

**Total bond energy for products:**
- For 2 CO₂:
- C=O: \(4 \times 724 = 2896 \, \text{kJ/mol}\)
- For 2 H₂O:
- O-H: \(4 \times 460 = 1840 \, \text{kJ/mol}\)

**Total bond energy of products:**
\[ \text{Total} = 2896 + 1840 = 4736 \, \text{kJ/mol} \]

### Step 4: Calculate ΔH°

Now, we can calculate ΔH° using the formula:

\[ \Delta H^\circ = \text{Sum of bond energies of reactants} - \text{Sum of bond energies of products} \]
\[ \Delta H^\circ = 3772 \, \text{kJ/mol} - 4736 \, \text{kJ/mol} \]
\[ \Delta H^\circ = -964 \, \text{kJ/mol} \]

### Final Answer:
The standard enthalpy change (ΔH°) for the reaction is:

\[ \Delta H^\circ = -964 \, \text{kJ/mol} \]

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