Respuesta :
Answer:
The rate law is k = [tex]k\frac{[O3]^{2} }{[O2]}\\[/tex]
Explanation:
From the mechanism is necessary to derive the rate law from the elementary steps and explain the effects of [O2] on the rate
The first step is a reversible reaction. Assuming dynamic equilibrium is achieved, the rate of the forward reaction is equal to the rate of the backward reaction
rate(forward) = rate(backward)
k1 [O3] = k-1 [O] [O2]
[O] is not part of the rate law, so we need to express [O] in terms of other species
[O] = [tex]\frac{k1 [O3]}{k-1[O2]}[/tex]
from the second step
rate = k2[O] [O3]
substituting [O] from the first step
rate = [tex]k2 \frac{k1 [O3] [O3]}{k-1[O2]} = \frac{k2k1 [O3]^{2} }{k-1[O2]}\\[/tex]
k = [tex]\frac{k2k1}{k-1}\\[/tex]
The final rate law is then
k = [tex]k\frac{[O3]^{2} }{[O2]}\\[/tex]
So, as the concentration os O2 increase the rate decrease. Also from the first step of the mechanism we can se that O2 can react to O to form back the reactant O3 resulting in decreased reaction rate.
When concentration of oxygen will increase, the reaction rate decrease. The final rate law is
[tex]\bold {K = k\dfrac {[O_3]^2}{O_2}}[/tex]
Since, the given reaction is at dynamic equilibrium,
Reaction rate for the forward and backward reaction will be equal.
k1 [O3] = k-1 [O] [O2].......1
[O] is not part of the rate law, so express [O] in terms of other species,
[tex]\bold {[O] = \dfrac {k1[O_3]}{{k-1[O_2]}}}[/tex].........2
From the second step
rate = k2[O] [O3]
Substitute [O] from the first step,
[tex]\bold {K = K_2 \dfrac {K_1 [O_3]^2}{k-1 [O_2]}}\\\\\bold {K = \dfrac {k_2k_1}{k-1}}[/tex]
The final rate law is then
[tex]\bold {K = k\dfrac {[O_3]^2}{O_2}}[/tex]
Therefore, when concentration of oxygen will increase, the reaction rate decrease.
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