# How do you calculate the energy change of reaction for the following reaction?

## EDIT: This is nearly impossible unless we are talking about the change in enthalpy of the reaction. We would otherwise have to calculate the ground-state energies of each molecule by solving the Schrodinger equation three times. - Truong-Son

Apr 3, 2018

Using bond enthalpies(?)

#### Explanation:

Assuming you meant the ENTHALPY change of the reaction it becomes clearer. As Truong-Son pointed out it would be a hassle to calculate using the Schrodinger equation if we are truly talking about the ENERGY change.

Given that we are talking about Enthalpy changes, we can use bond enthalpies from a table to solve this. I found my bond enthalpies in this booklet, table 11 (Courtesy of Ibchem.com)

We need to determine what bonds are broken and what bonds are formed. Bond breaking is endothermic- we need to put energy into breaking the bond so the value for $\Delta H$ will be positive.

Bond making is exothermic, meaning energy will be released to the surroundings and $\Delta H$ will be negative.

From the diagram's product side, we can see that the Hydrogen gas and the C-O double bond have vanished, so the respective bonds must have been broken in the first step!

Hence:
Breaking a C-O double bond=$\Delta H = + 745 k j m o {l}^{-} 1$*
Breaking an H-H single bond= $\Delta H = + 436 k j m o {l}^{-} 1$

*(Not the value in the booklet as some pointed out that the value in the booklet was too high)

If we wanted to be thorough, we could compare all the bonds on both the product and reactant side, but here we can see that there is no change in the Methyl $\left(- C {H}_{3}\right)$ groups so the "breaking and making" would cancel out, mathematically.

Anyway, on the product side, we now have the central carbon single bonded to a hydrogen, an oxygen and in turn that oxygen is bonded to a hydrogen. We have 3 new bonds that were not present in the reactant step.

We have formed the following bonds:

Forming a C-H single bond= $\Delta H = - 414 k j m o {l}^{-} 1$
Forming an O-H single bond=$\Delta H = - 463 k j m o {l}^{-}$
Forming a C-O single bond= $\Delta H = - 358 k j m o {l}^{-} 1$

So the total enthalpy change should be all of these enthalpy changes summed up.

$745 + 436 + \left(- 414\right) + \left(- 463\right) + \left(- 358\right) = - 54$

$\Delta H = - 54 k j m o {l}^{-} 1$
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