Following reaction takes place in one step: 2NO(g)+O_2(g) 2NO_2(g…

Question

Following reaction takes place in one step:
$\text{2NO(g)+O}_2\text{(g)}\rightarrow\text{2NO}_2\text{(g)}$
How will the rate of the above reaction change if the volume of the reaction vessel is reduced to one-third of its original volume? Will there be any change in the order of the reaction with reduced volume?

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Answer

$\therefore$ Rate $=\text{k[NO]}^2[\text{O}_2]$
Let initially, moles of $NO = a,$ moles of $O_2= b,$ volume of vesse $l= V.$Then
$\text{[NO]}=\frac{\text{a}}{\text{V}}\text{M},[\text{O}_2]=\frac{\text{b}}{\text{V}}\text{M}$
Rate $\text{(r}_1)=\text{k}\bigg(\frac{\text{a}}{\text{V}}\bigg)^2\bigg(\frac{\text{b}}{\text{V}}\bigg)=\text{k}\frac{\text{a}^2\text{b}}{\text{V}^3}\dots(\text{i)}$
Now, new volume $=\frac{\text{V}}{3 }$
$\therefore$ New concentrations:$ [\text{NO]}=\frac{\text{a}}{\frac{\text{V}}{3}}=\frac{3}{\text{V}}$
$[\text{O}_2]= \frac{\text{b}}{\frac{\text{V}}{3}}=\frac{\text{3a}}{\text{V}}$
$\therefore$ New rate $(r_2)= \text{k}\bigg(\frac{\text{3a}}{\text{V}}\bigg)^2\bigg(\frac{3\text{b}}{\text{V}}\bigg)= \frac{27\text{ka}^2\text{b}}{\text{V}^3}\dots \text{(ii)}$
$\therefore \frac{\text{r}_2}{\text{r}_2}=27 \text{ or}\text{ r}_2=27\text { r}_1$,i.e., rate becomes $27$ times.
Thus, there is no effact on the order of reaction.

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