JEE Advanced 2018 Paper 2 · Chemistry · Chemical Kinetics · medium

In the consecutive first-order reaction Axrightarrowk1Bxrightarrowk2C with k1=0.20 min^-1 and k2=0.050 min^-1, only A is present at t=0. The maximum…

Numerical answer

In the consecutive first-order reaction

\mathrm{A\xrightarrow{k_1}B\xrightarrow{k_2}C}

with

k_1=0.20\ \mathrm{min^{-1}}

and

k_2=0.050\ \mathrm{min^{-1}}

, only

\mathrm{A}

is present at

t=0

. The maximum concentration of the intermediate

\mathrm{B}

, expressed as a percentage of

[\mathrm{A}]_0

, is ______. Give the answer correct to two decimal places.

Show answer and solution

Answer: 63.00

For

\mathrm{A\rightarrow B\rightarrow C}

[\mathrm{B}]=\dfrac{k_1[\mathrm{A}]_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)

. The maximum occurs when

d[\mathrm{B}]/dt=0

, i.e.

k_1e^{-k_1t}=k_2e^{-k_2t}

, giving

t_{max}=\dfrac{\ln(k_1/k_2)}{k_1-k_2}=\dfrac{\ln4}{0.15}=9.24\ \mathrm{min}

. At that instant

e^{-k_2t}=e^{-0.4621}=0.6300

and

e^{-k_1t}=e^{-1.848}=0.1575

. Therefore

[\mathrm{B}]_{max}/[\mathrm{A}]_0=\dfrac{0.20}{-0.15}(0.1575-0.6300)=0.6300

. The same result follows from the closed form

(k_1/k_2)^{-k_2/(k_1-k_2)}=4^{-1/3}=0.62996

. The maximum is therefore

63.00\%

of the initial concentration of A, reached after

9.24

minutes. After this time B is consumed faster than it is formed, so its concentration falls while C keeps rising.

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