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Step-by-step Solution
Step 1: Identify the Reaction
The given reaction is:
$H_2C = CH_2\,(g) + H_2\,(g) \to H_3C - CH_3\,(g)$
We want to find the enthalpy change $ \Delta H $ for this process.
Step 2: List the Given Bond Energies
The bond energies (in kJ/mol) at 298 K are provided as follows:
$C - H = 414$
$C - C = 347$
$C = C = 615$
$H - H = 435$
Step 3: Determine Bonds Broken and Formed
In the reactants, the double bond $C = C$ is present, and there is one $H - H$ bond. We can also count the total number of $C - H$ bonds based on the structure of $H_2C=CH_2$: each carbon has two hydrogens, so there are 4 C - H bonds in ethylene ($H_2C=CH_2$) plus 1 H - H bond in $H_2$.
In the products, we have ethane ($H_3C - CH_3$), which has a single $C - C$ bond and a total of 6 C - H bonds (3 on each carbon).
Hence:
Bonds broken (Reactants):
1 $C = C$ bond
1 $H - H$ bond
(Note that the total 4 $C - H$ bonds in ethylene remain on both sides, but we will see them factor out when we do the difference.)
Bonds formed (Products):
1 $C - C$ bond
2 additional $C - H$ bonds (to convert the double bond into single bonds with extra hydrogens), for a total difference of 6 $C - H$ bonds in ethane minus the 4 $C - H$ bonds in ethylene.
Step 4: Apply the Bond Energy Formula
The general expression for change in enthalpy using bond energies is:
$ \Delta H = \text{(Sum of bond energies of bonds broken)} \;-\; \text{(Sum of bond energies of bonds formed)} $
So for this reaction:
$
\Delta H = [\,1 (C = C) + 1 (H - H)\,] \;-\; [\,1 (C - C) + 2 (C - H)\,].
$
Step 5: Substitute the Numerical Values
Using the given bond energies:
$
\Delta H = [\,615 + 435\,] - [\,347 + 2\times 414\,].
$
$
\Delta H = 1050 - (347 + 828).
$
$
\Delta H = 1050 - 1175.
$
$
\Delta H = -125 \,\text{kJ}.
$
Step 6: Conclude the Answer
The negative sign indicates an exothermic reaction. Therefore, the enthalpy change for the given reaction at 298 K is $-125$ kJ.
Final Answer: $-125$ kJ
