The enthalpy of reaction, `2HC = CH + 5O_(2) rightarrow 4CO_(2) + 2H_(2)O`
If the bond energies of `C-H, C=C, O=O, C=O and O-H` bonds are p,q,r,s,t repectively

Options

A. `[8s + 4t] - [4p + q + 5r]`
B. `[4p + 2q + 5r] - [8s + 4t]`
C. `[4p + 2q + 5r + 8s + 4t]`
D. `[2p + q + 5r] - [8s + 4t]`

Answer

To find the enthalpy of the reaction given by the equation:

\[ 2HC \equiv CH + 5O_2 \rightarrow 4CO_2 + 2H_2O \]

we will use the bond energies of the bonds involved in the reaction. The bond energies are represented as follows:

- C-H bond energy = \( P \)
- C≡C bond energy = \( Q \)
- O=O bond energy = \( R \)
- C=O bond energy = \( S \)
- O-H bond energy = \( T \)

### Step-by-Step Solution:

1. **Identify Bonds Broken in Reactants:**
- In the reactants \( 2HC \equiv CH + 5O_2 \):
- Each \( HC \equiv CH \) has:
- 2 C-H bonds (from 2HC)
- 1 C≡C bond (from 2HC)
- There are 5 O=O bonds from \( 5O_2 \).
- Total bonds broken:
- C-H: \( 2 \times 2 = 4 \) (since there are 2 HC)
- C≡C: \( 2 \) (from 2HC)
- O=O: \( 5 \)

The total energy required to break these bonds is:
\[ \Delta H_{\text{reactants}} = 4P + 2Q + 5R \]

2. **Identify Bonds Formed in Products:**
- In the products \( 4CO_2 + 2H_2O \):
- Each \( CO_2 \) has:
- 2 C=O bonds (and there are 4 \( CO_2 \))
- Each \( H_2O \) has:
- 2 O-H bonds (and there are 2 \( H_2O \))
- Total bonds formed:
- C=O: \( 4 \times 2 = 8 \)
- O-H: \( 2 \times 2 = 4 \)

The total energy released from forming these bonds is:
\[ \Delta H_{\text{products}} = 8S + 4T \]

3. **Calculate the Enthalpy of Reaction:**
The enthalpy change for the reaction is given by:
\[ \Delta H_{\text{reaction}} = \Delta H_{\text{reactants}} - \Delta H_{\text{products}} \]
Substituting the values we calculated:
\[ \Delta H_{\text{reaction}} = (4P + 2Q + 5R) - (8S + 4T) \]
Simplifying this gives:
\[ \Delta H_{\text{reaction}} = 4P + 2Q + 5R - 8S - 4T \]

### Final Answer:
The enthalpy of the reaction is:
\[ \Delta H_{\text{reaction}} = 4P + 2Q + 5R - 8S - 4T \]