JEE Main 2021 · Chemistry · Chemical Kinetics · hard

The gas-phase decomposition A(g)→2B(g)+C(g) is first order in A. Starting with pure A at 0.40 atm in a rigid vessel at constant temperature, the…

Numerical answer

The gas-phase decomposition

\mathrm{A}(g)\to2\mathrm{B}(g)+\mathrm{C}(g)

is first order in A. Starting with pure A at

0.40\ \mathrm{atm}

in a rigid vessel at constant temperature, the total pressure is

0.60\ \mathrm{atm}

after

30

minutes. The time, in minutes, needed for

90\%

of A to decompose is (nearest integer;

\ln10=2.303

\ln\frac43=0.2877

)

Show answer and solution

Answer: 240

If pressure of A falls by

x

, then

P_t=(P_0-x)+2x+x=P_0+2x

. So

0.60=0.40+2x

x=0.10\ \mathrm{atm}

and

p_A=0.30\ \mathrm{atm}

. First order:

k=\dfrac{1}{30}\ln\dfrac{0.40}{0.30}=\dfrac{0.2877}{30}=9.59\times10^{-3}\ \mathrm{min^{-1}}

. For

90\%

decomposition,

t=\dfrac1k\ln\dfrac{1}{0.1}=\dfrac{2.303}{9.59\times10^{-3}}=240\ \mathrm{min}

. Equivalent shortcut:

t_{90}=30\times\dfrac{\ln10}{\ln(4/3)}=30\times8.0=240

Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake