# Enthalpy and Bond Dissociation Energies

I'm not sure if I understand the reasoning behind this particular question.

The sigma and pi bonds of $\ce{C=C}$ have a combined bond dissociation energy of 632 kJ/mol. Using this information, predict whether the following reaction is exothermic or endothermic.

$\ce{C2H4 + H2O -> CH3CH2OH}$

My attempt:

The bonds being broken are the $\ce{C=C}$ pi bond and the $\ce{H-OH}$ bond. I realize there are two bonds being formed, but I am unable to correctly identify the second one. One is $\ce{CH_3CH_2-OH}$. I think the other should be $\ce{CH_2-H}$. The textbook says this bond is $\ce{CH_2H-R}$. I think R is referring to the side group, which varies (and in this case, is $\ce{CH_2-OH}$?), but the enthalpy value for breaking the $\ce{CH_2H-R}$ is listed as 410 kJ in the solutions. According to the table in my book, the value for a $\ce{CH_3CH_2-H}$ is 410 kJ/mol. I do not understand how these two bonds are equivalent. Can someone please explain this?