For a reaction scheme. $A \xrightarrow{k_{1}} B \xrightarrow{k_{2}} C$ if the rate of formation of $B$ is set to be zero then the concentration of $B$ is given by

(A) $\left(\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}\right)[\mathrm{A}]$

(B) $\left(\mathrm{k}_{1}-\mathrm{k}_{2}\right)[\mathrm{A}]$

(C) $\mathrm{k}_{1} \mathrm{k}_{2}[\mathrm{~A}]$

(D) $\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right)[\mathrm{A}]$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Chemical Kinetics

For a reaction scheme.

A k_{1} B k_{2} C

if the rate of formation of

B

is set to be zero then the concentration of

B

is given by

Answer: (A)

Setting the rate of formation of the intermediate B to zero is the steady-state approximation. B is produced from A at rate

k_{1} [A]

and consumed into C at rate

k_{2} [B]

, so

\frac{d [ B ]}{d t} = k_{1} [A] - k_{2} [B] = 0

. Solving,

k_{2} [B] = k_{1} [A]

, which gives the steady-state concentration

[B] = \frac{k _{1}}{k _{2}} [A]

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option