From the following bond energies :-
`H-H "bond energy" = 420 KJmol^(-1)`
`C-=C "bond energy" = 601 KJmol^(-1)`
`C-C "bond energy"=340KJmol^(-1)`
`C-H "bond energy" = 425 KJ mol^(-1)`
Enthalpy for the reaction :-
`C_(2)H_(2)+2H_(2)toC_(2)H_(6)`

Options

A. `-599 KJ mol^(-1)`
B. `-580 KJ mol^(-1)`
C. `-625 KJ mol^(-1)`
D. `-325 KJ mol^(-1)`

Explanation

To calculate the enthalpy change for the reaction \( C_2H_2 + 2H_2 \rightarrow C_2H_6 \), we will follow these steps:

### Step 1: Identify the bonds in the reactants and products
- **Reactants**:
- In \( C_2H_2 \) (ethyne), there is 1 C≡C bond and 2 C-H bonds.
- In \( 2H_2 \), there are 2 H-H bonds.

- **Products**:
- In \( C_2H_6 \) (ethane), there is 1 C-C bond and 6 C-H bonds.

### Step 2: Write the bond energy equation
The enthalpy change (\( \Delta H \)) for the reaction can be calculated using the formula:
\[ \Delta H = \text{(Total bond energy of reactants)} - \text{(Total bond energy of products)} \]

### Step 3: Calculate the total bond energy of the reactants
- For \( C_2H_2 \):
- 1 C≡C bond: \( 601 \, \text{kJ/mol} \)
- 2 C-H bonds: \( 2 \times 425 \, \text{kJ/mol} = 850 \, \text{kJ/mol} \)

- For \( 2H_2 \):
- 2 H-H bonds: \( 2 \times 420 \, \text{kJ/mol} = 840 \, \text{kJ/mol} \)

**Total bond energy of reactants**:
\[ 601 + 850 + 840 = 2291 \, \text{kJ/mol} \]

### Step 4: Calculate the total bond energy of the products
- For \( C_2H_6 \):
- 1 C-C bond: \( 340 \, \text{kJ/mol} \)
- 6 C-H bonds: \( 6 \times 425 \, \text{kJ/mol} = 2550 \, \text{kJ/mol} \)

**Total bond energy of products**:
\[ 340 + 2550 = 2890 \, \text{kJ/mol} \]

### Step 5: Calculate the enthalpy change
Now, substituting the total bond energies into the enthalpy change equation:
\[ \Delta H = 2291 \, \text{kJ/mol} - 2890 \, \text{kJ/mol} \]
\[ \Delta H = -599 \, \text{kJ/mol} \]

### Final Answer
The enthalpy change for the reaction \( C_2H_2 + 2H_2 \rightarrow C_2H_6 \) is \( -599 \, \text{kJ/mol} \).

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