Q.
In the reversible reaction
The rate of disappearance of ${NO}_{2}$ is equal to
A) $\frac{2k_{1}}{k_{2}}\left\lbrack {NO}_{2} \right\rbrack^{2}$
B) $2k_{1}\left\lbrack {NO}_{2} \right\rbrack^{2} - 2k_{2}\left\lbrack N_{2}O_{4} \right\rbrack$
C) $2k_{2}\left\lbrack {NO}_{2} \right\rbrack^{2} - k_{2}\left\lbrack N_{2}O_{4} \right\rbrack$
D) $\left( 2k_{1} - k_{2} \right)\left\lbrack {NO}_{2} \right\rbrack$
Explanation
(b)
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$Rate = - \frac{1}{2}\frac{d\left\lbrack {NO}_{2} \right\rbrack}{dt}$
$\ = k_{1}\left\lbrack {NO}_{2} \right\rbrack^{2} - k_{2}\left\lbrack N_{2}O_{4} \right\rbrack$
$\therefore$ Rate of disappearance of ${NO}_{2}$
$ie,\ - \frac{d\left\lbrack {NO}_{2} \right\rbrack}{dt} = 2k_{1}\left\lbrack {NO}_{2} \right\rbrack^{2} - 2k_{2}\left\lbrack N_{2}O_{4} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.