Q. The chemical reaction $2O_{3} \rightarrow 3O_{2}$ proceeds as follows $O_{3} \rightleftharpoons O_{2} + O\$(fast) $O + O_{3} \rightarrow 2O_{2}\$(slow) The rate law expression should be
A) $r = k\left\lbrack O_{3} \right\rbrack^{2}$
B) $r = k\left\lbrack O_{3} \right\rbrack^{2}\ \left\lbrack O_{2} \right\rbrack^{- 1}$
C) $r = k\left\lbrack O_{3} \right\rbrack\left\lbrack O_{2} \right\rbrack$
D) Unpredictable
Explanation
(b)
Rate depends upon the slowest step. Hence, from equation
$O + O_{3} \rightarrow 2O_{2}$
$r = k\left\lbrack O_{3} \right\rbrack\lbrack O\rbrack$
And from equation $O_{3} \rightleftharpoons O_{2} + O$
$K_{eq} = \frac{\left\lbrack O_{2} \right\rbrack\lbrack O\rbrack}{\left\lbrack O_{3} \right\rbrack}$
$\ \lbrack O\rbrack = \frac{K_{eq}\left\lbrack O_{3} \right\rbrack}{\left\lbrack O_{2} \right\rbrack}$
$\therefore r = k\left\lbrack O_{3} \right\rbrack\frac{K_{eq}\left\lbrack O_{3} \right\rbrack}{\left\lbrack O_{2} \right\rbrack}$
$= k^{'}\left\lbrack O_{3} \right\rbrack^{2}\left\lbrack O_{2} \right\rbrack^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.