# Question 8cdfb

Feb 12, 2016

You were given the mechanism, which is (fixing the typo):

2"N"_2"O"_5 stackrel(k_1" ")(=>) 4"NO"_2 + color(red)("O"_2)
$2 {\text{NO"_2 + 1/2"O"_2 stackrel(k_2" ")(=>) "N"_2"O}}_{5}$

The overall reaction therefore is:

$\setminus m a t h b f \left({\text{N"_2"O"_5 stackrel(k_(obs))(->) 2"NO"_2 + 1/2"O}}_{2}\right)$

Notice how:

2"N"_2"O"_5 stackrel(k_1" ")(=>) 4"NO"_2 + color(red)("O"_2)#
$2 \left(2 {\text{NO"_2 + 1/2"O"_2 stackrel(k_2" ")(=>) "N"_2"O}}_{5}\right)$
$\text{-----------------------------------------------}$
$\text{Nothing}$

The second step is half the scale of the first. Hence, the second elementary step's activation energy is half as large. Or:

$\textcolor{b l u e}{{E}_{2} = \frac{1}{2} {E}_{1}}$

Correct me if I'm wrong, but I don't think it's c, because that looks backwards. In that case, I would say ${E}_{1} > {E}_{2}$.