# Question #495b7

Feb 22, 2017

$\Delta H = + 2034 k J m o {l}^{-} 1$ or $\Delta H = + 726 k J m o {l}^{-} 1$

#### Explanation:

If the reaction in question is exactly as you have written the answer should be +2034 kj per mol. But when hydrogenating ethene generally the reaction is: ${H}_{2} + {C}_{2} {H}_{4} = {C}_{2} {H}_{6}$ and then the answer would be + 726 kJ so take the answer that fits best with your question.

To calculate this we must use Bond energies (or Bond enthalpies). My values were taken from my course's data booklet and I advise to check your data booklet too to see that it corresponds.

The answer is found by comparing energies from "Breaking and making bonds".

Bonds broken= 4 H-H bonds, 4 C-H bonds, , 1 C-C double bond.
$\Delta H = 4 \left(436\right) + 4 \left(414\right) + 614 = + 4014 k J m o {l}^{1}$ (alt 1 H-H bond , then the answer becomes $436 + 4 \left(414\right) + 614 = + 2706 k J m o {l}^{-} 1$
(as breaking is Endothermic, Delta H values are pos)
Bonds formed= 6 C-H bonds, 1 C-C bond
$6 \left(- 414\right) + \left(- 346\right) = - 1980 k J m o {l}^{-} 1$
(As making is Exothermic, Delta H values are neg.)
Then the total enthalpy(energy) is:
$\Delta H = + 4014 - 1980 = + 2034 k J m o {l}^{-} 1$
or
$\Delta H = + 2706 - 1980 = + 726 k J m o {l}^{-} 1$

This picture also shows the Lewis structure of all molecules involved.