The enthalpy change for the reaction `H_(2)(g)+C_(2)H_(4)(g)rarr C_(2)H_(6)(g)` is ……………… The bond energies are, `H - H=103, C-H=99, C-C=80` & `C=C=145 "K cal mol"^(-1)`

Options

A. `-10 "K cal mol"^(-1)`
B. `+10 "K cal mol"^(-1)`
C. `-30 "K cal mol"^(-1)`
D. `+30 "K cal mol"^(-1)`

Explanation

To find the enthalpy change for the reaction

\[ H_2(g) + C_2H_4(g) \rightarrow C_2H_6(g) \]

we can use the bond energies provided in the question. The bond energies are:

- \( H-H = 103 \, \text{kcal/mol} \)
- \( C-H = 99 \, \text{kcal/mol} \)
- \( C-C = 80 \, \text{kcal/mol} \)
- \( C=C = 145 \, \text{kcal/mol} \)

### Step 1: Identify the bonds broken and formed
In the reaction, we start with ethene (\(C_2H_4\)) and hydrogen (\(H_2\)) to form ethane (\(C_2H_6\)).

- **Bonds broken:**
- 1 \(H-H\) bond
- 1 \(C=C\) bond in ethene

- **Bonds formed:**
- 6 \(C-H\) bonds in ethane

### Step 2: Calculate the total energy of bonds broken
The total energy of the bonds broken is the sum of the bond energies of the bonds that are broken:

\[ \text{Energy of bonds broken} = \text{Energy of } H-H + \text{Energy of } C=C \]

Substituting the values:

\[ \text{Energy of bonds broken} = 103 \, \text{kcal/mol} + 145 \, \text{kcal/mol} = 248 \, \text{kcal/mol} \]

### Step 3: Calculate the total energy of bonds formed
The total energy of the bonds formed is the sum of the bond energies of the bonds that are formed:

\[ \text{Energy of bonds formed} = 6 \times \text{Energy of } C-H \]

Substituting the values:

\[ \text{Energy of bonds formed} = 6 \times 99 \, \text{kcal/mol} = 594 \, \text{kcal/mol} \]

### Step 4: Calculate the enthalpy change (\(\Delta H\))
The enthalpy change for the reaction can be calculated using the formula:

\[ \Delta H = \text{Energy of bonds broken} - \text{Energy of bonds formed} \]

Substituting the values we calculated:

\[ \Delta H = 248 \, \text{kcal/mol} - 594 \, \text{kcal/mol} = -346 \, \text{kcal/mol} \]

### Step 5: Conclusion
The enthalpy change for the reaction \( H_2(g) + C_2H_4(g) \rightarrow C_2H_6(g) \) is

\[ \Delta H = -346 \, \text{kcal/mol} \]