How do exothermic reactions occur?

Apr 22, 2018

$| E n t h a l p h {y}_{B o n \mathrm{dF}} |$>$| E n t h a l p h {y}_{B o n \mathrm{dB}} |$

Explanation:

In any reaction, there is two stages. The Bond Breaking and Bond Forming stage which occurs after bond breaking.

Lets take ${N}_{2} + 3 {H}_{2} = N {H}_{3}$ as reaction for example. This reaction ,common in the Haber Process, is exothermic. Why? Simply , energy is first used to break the bonds of N≡N in ${N}_{2}$ and $H = H$ in ${H}_{2}$. Let this energy 'absorbed' be $| E n t h a l p h {y}_{B o n \mathrm{dB}} |$ .

After that, free $N$ atoms and free $H$ atoms collide to form $N - H$ single bonds in newly formed $N {H}_{3}$. Let this energy 'released' be $| E n t h a l p h {y}_{B o n \mathrm{dF}} |$ .

Since in this reaction, the energy evolved(released) is larger than the energy absorbed to break the bonds, there is a net release of energy which results in the products having a lower energy content than the reactants as shown in the graph.

aka $| E n t h a l p h {y}_{B o n \mathrm{dF}} |$ > $| E n t h a l p h {y}_{B o n \mathrm{dB}} |$