Calculate the enthalpy of the following reaction:
`H_(2)C =CH_(2)(g) +H_(2)(g) rarr CH_(3)-CH_(3)(g)`
The bond energies of `C-H, C-C,C=C`, and `H-H` are `99,83,147`,and `104kcal` respectively.

Explanation

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

\[ \text{H}_2\text{C}=\text{CH}_2(g) + \text{H}_2(g) \rightarrow \text{CH}_3\text{CH}_3(g) \]

we will use the bond energies provided for the relevant bonds involved in the reaction.

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

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