Calculate the enthalpy of hydrogenation of`C_(2)H_(2)(g) ` to `C_(2)H_(4)(g)`.
( Given bond energies: `C-H = 414.0 kJ mol^(-1)`,
`C-=C = 827.6 kJ mol^(-1)`,
`C=C= 606 .0 kJ mol^(-1)`,
`H-H = 430 .5 kJ mol^(-1))`

Explanation

To calculate the enthalpy of hydrogenation of \( C_2H_2(g) \) to \( C_2H_4(g) \), we will use the bond energies provided. The reaction can be represented as follows:

\[ C_2H_2(g) + H_2(g) \rightarrow C_2H_4(g) \]

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

**Reactants:**
- In \( C_2H_2 \) (ethyne), there is one carbon-carbon triple bond and two carbon-hydrogen single bonds.
- In \( H_2 \), there is one hydrogen-hydrogen single bond.

**Products:**
- In \( C_2H_4 \) (ethene), there is one carbon-carbon double bond and four carbon-hydrogen single bonds.

### Step 2: Write down the bond energies

Given bond energies:
- \( C-H = 414.0 \, kJ/mol \)
- \( C \equiv C = 827.6 \, kJ/mol \)
- \( C=C = 606.0 \, kJ/mol \)
- \( H-H = 430.5 \, kJ/mol \)

### Step 3: Calculate the total bond energy of the reactants

For the reactants \( C_2H_2 + H_2 \):
- Bond energies:
- 1 triple bond \( C \equiv C = 827.6 \, kJ/mol \)
- 2 single bonds \( C-H = 2 \times 414.0 \, kJ/mol = 828.0 \, kJ/mol \)
- 1 single bond \( H-H = 430.5 \, kJ/mol \)

Total bond energy of reactants:
\[ BE_{reactants} = 827.6 + 828.0 + 430.5 = 2086.1 \, kJ/mol \]

### Step 4: Calculate the total bond energy of the products

For the products \( C_2H_4 \):
- Bond energies:
- 1 double bond \( C=C = 606.0 \, kJ/mol \)
- 4 single bonds \( C-H = 4 \times 414.0 \, kJ/mol = 1656.0 \, kJ/mol \)

Total bond energy of products:
\[ BE_{products} = 606.0 + 1656.0 = 2262.0 \, kJ/mol \]

### Step 5: Calculate the change in enthalpy (\( \Delta H \))

Using the formula:
\[ \Delta H = BE_{reactants} - BE_{products} \]
Substituting the values:
\[ \Delta H = 2086.1 - 2262.0 = -175.9 \, kJ/mol \]

### Final Answer
The enthalpy of hydrogenation of \( C_2H_2(g) \) to \( C_2H_4(g) \) is:
\[ \Delta H = -175.9 \, kJ/mol \]