Calculate the enthalpy change `(Delta H)` of the following reaction
`2C_(2)H_(2)(g)+5O_(2)(g)rarr4CO_(2)(g)+2H_(2)O(g)` given average bond enthalpies of various bonds, i.e., `C-H,C-=C,O=O,C=O,O-H` as `414,814,499,724` and `640 kJ mol^(-1)` respectively.

Explanation

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

\[ 2C_{2}H_{2}(g) + 5O_{2}(g) \rightarrow 4CO_{2}(g) + 2H_{2}O(g) \]

we will follow these steps:

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

1. **Reactants:**
- **C2H2 (ethyne)**: Contains 1 C≡C bond and 2 C-H bonds.
- **O2**: Contains 5 O=O bonds.

2. **Products:**
- **CO2**: Contains 2 C=O bonds (each CO2 molecule has 2 C=O bonds, and there are 4 CO2 molecules).
- **H2O**: Contains 2 O-H bonds (each H2O molecule has 2 O-H bonds, and there are 2 H2O molecules).

### Step 2: Write the bond enthalpy values

- C-H: 414 kJ/mol
- C≡C: 814 kJ/mol
- O=O: 499 kJ/mol
- C=O: 724 kJ/mol
- O-H: 640 kJ/mol

### Step 3: Calculate the total bond enthalpy for reactants

- For **2 C2H2**:
- Bonds: 2 C-H bonds per C2H2 → 2 * 2 = 4 C-H bonds
- 1 C≡C bond per C2H2 → 2 * 1 = 2 C≡C bonds
- Total bond enthalpy for C2H2:
\[ \text{Total for C2H2} = 2 \times 814 + 4 \times 414 \]

- For **5 O2**:
- Bonds: 5 O=O bonds
- Total bond enthalpy for O2:
\[ \text{Total for O2} = 5 \times 499 \]

### Step 4: Calculate the total bond enthalpy for products

- For **4 CO2**:
- Bonds: 2 C=O bonds per CO2 → 4 * 2 = 8 C=O bonds
- Total bond enthalpy for CO2:
\[ \text{Total for CO2} = 8 \times 724 \]

- For **2 H2O**:
- Bonds: 2 O-H bonds per H2O → 2 * 2 = 4 O-H bonds
- Total bond enthalpy for H2O:
\[ \text{Total for H2O} = 4 \times 640 \]

### Step 5: Calculate ΔH

Using the formula:
\[ \Delta H = \text{Total bond enthalpy of reactants} - \text{Total bond enthalpy of products} \]

Substituting the values:
1. Total bond enthalpy of reactants:
\[ \text{Total for C2H2} = 2 \times 814 + 4 \times 414 = 1628 + 1656 = 3284 \text{ kJ/mol} \]
\[ \text{Total for O2} = 5 \times 499 = 2495 \text{ kJ/mol} \]
\[ \text{Total for reactants} = 3284 + 2495 = 5779 \text{ kJ/mol} \]

2. Total bond enthalpy of products:
\[ \text{Total for CO2} = 8 \times 724 = 5792 \text{ kJ/mol} \]
\[ \text{Total for H2O} = 4 \times 640 = 2560 \text{ kJ/mol} \]
\[ \text{Total for products} = 5792 + 2560 = 8352 \text{ kJ/mol} \]

Finally, calculate ΔH:
\[ \Delta H = 5779 - 8352 = -2573 \text{ kJ/mol} \]

### Final Answer:
\[ \Delta H = -2573 \text{ kJ/mol} \]