# Question bca43

Jul 5, 2016

Look at the enthalpy change of the reaction ($\Delta H$)
$\Delta H$ is positive so the reaction is endothermic.

#### Explanation:

Note: Please do not be apprehensive about the long explanation. I have tried to explain the steps I have taken as simply as possible and in full. Feel free to ask further questions if you do not understand any part of the answer.

To figure out whether a reaction is exothermic or endothermic, you need to know whether it releases or absorbs energy (endothermic absorbs energy, exothermic releases energy).

To find the amount of energy that is released or absorbed by the reaction you need to find the enthalpy change ($\Delta H$) of the reaction. Enthalpy, $H$, is the energy contained within the molecules of the reactants and of the products.

$\Delta H = \Sigma E \left(\text{Bonds Formed")-SigmaE("Bonds Broken}\right)$

If $\Delta H$ is negative, the enthalpy of the products is smaller than the enthalpy of the reactants, so energy is released and the reaction is exothermic and if $\Delta H$ is positive, the enthalpy of the products is larger than the enthalpy of the reactants, so energy is absorbed and the reaction is endothermic.

To find $\Sigma E \left(\text{Bonds Broken}\right)$ and $\Sigma E \left(\text{Bonds Formed}\right)$
you need to add up all of the bond energies of the bonds in the reactants and products. I used this table to get the bond energies of various types of covalent bonds, but you will probably be provided with your own. Since the values given are averages they may be different from different sources.

In the reactants there are 2 $\text{C=O}$ double bonds and no others.
SigmaE("Bonds Broken")=(2xxBE("C=O"))
$B E \left(\text{C=O}\right) = 799 k J m o {l}^{-} 1$ so
$\Sigma E \left(\text{Bonds Broken}\right) = \left(2 \times 799 k J m o {l}^{-} 1\right) = 1598 k J m o {l}^{-} 1$

In the products there are 2 $\text{C"-="O}$ triple bonds and no others.
SigmaE("Bonds Formed")=(2xxBE("C"-="O"))#
$B E \left(\text{C"-="O}\right) = 1072 k J m o {l}^{-} 1$ so
$\Sigma E \left(\text{Bonds Formed}\right) = \left(2 \times 1072 k J m o {l}^{-} 1\right) = 2144 k J m o {l}^{-} 1$

$\Delta H = 1 - 2144 k J m o {l}^{-} 1 - 1598 k J m o {l}^{=} 546 k J m o {l}^{-} 1$

$\Delta H$ is negative so the reaction is endothermic.