Q. For the reaction system $2NO(g) + O_{2}(g) \rightarrow 2{NO}_{2}(g)$ if the volume of the reaction vessel is reduced to one-third of its original volume, what will be the order of the reaction?
A) Diminished to one fourth of its initial value
B) Diminished to one twenty seven of its initial value
C) Increase to twenty seven times of its initial value
D) Increase to four times of its initial value
Explanation
(c)
The rate of reaction is
$rate = k\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack$
When the volume is reduced to $\frac{1}{3},$ the concentration of each reactant is increased by 3 times
${rate}^{'} = k\lbrack 3NO\rbrack^{2}\left\lbrack 3O_{2} \right\rbrack$
$= 27k\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack$
$\frac{rate'}{rate} = \frac{27k\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack}{k\lbrack NO\rbrack^{2}\left\lbrack O_{2} \right\rbrack}$
${rate}^{'} = 27\ rate$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.