If bond energies are denoted by `epsilon`, the enthalpy of the reaction:
`CH-=CH(g)+2H_(2)(g)toC_(2)H_(6)(g)` is

Options

A. `E_(C-=C)+2E_(H-H)-E_(C-C)-4E_(C)-H`
B. `E_(C-=C)+2E_(H-H)-E_(C-C)-6E_(C-H)`
C. `E_(C-=C)-2E_(H-H)-E_(C-C)-4E_(C)-H)`
D. `E_(C-=C)+2E_(H-H)-E_(C=C)-4E_(C-H)`

Answer

To calculate the enthalpy change (ΔH) for the reaction:

\[ \text{C}\equiv\text{C}(g) + 2\text{H}_2(g) \rightarrow \text{C}_2\text{H}_6(g) \]

we will follow these steps:

### Step 1: Identify the bonds broken in the reactants
In the reactants, we have:
- One C≡C (triple bond) bond
- Two H-H (single bond) bonds from the two H₂ molecules

### Step 2: Identify the bonds formed in the products
In the products, we have:
- One C-C (single bond) bond
- Six C-H (single bond) bonds in C₂H₆

### Step 3: Write the equation for ΔH
The enthalpy change for the reaction can be calculated using the formula:

\[ \Delta H = \text{(Sum of bond energies of bonds broken)} - \text{(Sum of bond energies of bonds formed)} \]

### Step 4: Write the bond energy terms
From the previous steps, we can express the bond energies:
- Bonds broken:
- 1 × E(C≡C)
- 2 × E(H-H)

- Bonds formed:
- 1 × E(C-C)
- 6 × E(C-H)

### Step 5: Substitute the bond energies into the ΔH equation
Now we can substitute these into our ΔH equation:

\[ \Delta H = \left( E(C \equiv C) + 2 \cdot E(H-H) \right) - \left( E(C-C) + 6 \cdot E(C-H) \right) \]

### Step 6: Simplify the equation
This simplifies to:

\[ \Delta H = E(C \equiv C) + 2 \cdot E(H-H) - E(C-C) - 6 \cdot E(C-H) \]

### Final Result
Thus, the enthalpy change for the reaction is:

\[ \Delta H = E(C \equiv C) + 2 \cdot E(H-H) - E(C-C) - 6 \cdot E(C-H) \]