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 calculate the enthalpy change (ΔH) for the reaction \( H_2(g) + C_2H_4(g) \rightarrow C_2H_6(g) \), we will use the bond energies provided. The bond energies are as follows:

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

### Step 1: Identify the bonds broken in the reactants
In the reactants \( H_2 \) and \( C_2H_4 \):
- \( H_2 \) has 1 \( H-H \) bond.
- \( C_2H_4 \) has:
- 1 \( C=C \) bond (double bond).
- 4 \( C-H \) bonds.

**Total bond energies for reactants:**
\[ \text{Bonds broken} = 1 \times \text{(H-H)} + 1 \times \text{(C=C)} + 4 \times \text{(C-H)} \]
\[ = 1 \times 103 + 1 \times 145 + 4 \times 99 \]
\[ = 103 + 145 + 396 = 644 \, \text{kcal/mol} \]

### Step 2: Identify the bonds formed in the products
In the product \( C_2H_6 \):
- \( C_2H_6 \) has:
- 1 \( C-C \) bond.
- 6 \( C-H \) bonds.

**Total bond energies for products:**
\[ \text{Bonds formed} = 1 \times \text{(C-C)} + 6 \times \text{(C-H)} \]
\[ = 1 \times 80 + 6 \times 99 \]
\[ = 80 + 594 = 674 \, \text{kcal/mol} \]

### Step 3: Calculate the enthalpy change (ΔH)
Using the formula:
\[ \Delta H = \text{(Bonds broken)} - \text{(Bonds formed)} \]
\[ \Delta H = 644 - 674 = -30 \, \text{kcal/mol} \]

### Final Answer:
The enthalpy change for the reaction \( H_2(g) + C_2H_4(g) \rightarrow C_2H_6(g) \) is \( -30 \, \text{kcal/mol} \).

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