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.

$\ce{C2H4 + H2O -&gt; CH3CH2OH}$

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?
Tabulation table such as the ones you reference, describe 'typical' bonds when using generic groups such as R.  You are right in your discomfort that the numbers should be somewhat different, but that would require a thermochemical measurement for every known molecule, which is difficult to achieve in practice.

Calculations such as the one you are performing are 'estimates', and while good ones, they are never as credible as an actual experiment.