For a two step reaction. A \underset{}{\rightleftharpoons} R + B R + C \overset{k_{2}}{\rightarrow} P (Where, R is a reactive intermediate whose concentration is maintained at some low steady state throughout the reaction). If the concentration of C is very high then the order of reaction for formation of "P" is

Engineering

Chemistry

Reaction Mechanism and Molecularity

Question

For a two step reaction.
A $\underset{}{⇌}$ R + B

R + C $\overset{k_{2}}{\to}$ P

(Where, R is a reactive intermediate whose concentration is maintained at some low steady state throughout the reaction). If the concentration of C is very high then the order of reaction for formation of "P" is

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Rate law = $\frac{dp}{dt}$ = K₂ [R] [C]
at steady state $\frac{d [R]}{dt} = 0$

$\frac{d [R]}{dt} =$ K₁ [A] – K₂ [R] [B]– K₂[R] [C] = 0
K₁ [A] = {K₂[B] + K₂[C]} [R]
$[R] = \frac{K_{1} [A]}{K_{2} {[B] + [C]}}$
$Rate = K_{2} \times \frac{K_{1}}{K_{2}} \frac{[A]}{[B] + [C]} \times [C]$

$= \frac{K_{1} [A]}{\frac{[B]}{[C]} + 1}$
where C is very high
Rate = K₁ [A]
$∴$ order of reaction = 1