How to calculate the enthalpy change for the formation of ethylene epoxide from ethylene?

Question

I get everything up until the very end. We need to identify the bond enthalpies first: \begin{align} \ce{C=C} &= 612\\ \ce{O=O} &= 248\\ \text{SUM of these}&= 860\\ \end{align}

\begin{align} \ce{C-C} &= 348\\ 2\cdot(\ce{C-O}) &= 720\\ \text{SUM of these} &= 1068\\ \end{align} We know that enthalpy change is = products - reactants, which is $1068 - 860 = 208$, but the problem comes here: I know a minus sign is supposed to go there, but how do we know that? Exothermic reaction? Is it because of the oxygen, suggesting it's a combustion reaction or not?

Answer 1

When you use bond enthalpies to calculate $\Delta H$, you need to subtract the sum of the bond enthalpies for the products from the sum of the bond enthalpies for the reactants. In other words "reactants - products". This is because all bond enthalpies are positive, since energy must be put into the bond to break it. Bond formation releases energy; the minus sign compensates for this. In short, bonds of reactant molecule must break, and new are formed to form product molecules. So, your answer should be negative.

Answer 2

There is one mistake in your calculation: oxirane is not just two standard C-O bonds plus a standard C-C bond. Due to ring strain there is some extra energy in the three-member ring.