JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Chemical Kinetics

For a first order reaction, the time required for completion of

90%

reaction is '

x

' times the half life of the reaction. The value of '

x

' is
(Given:

ln 10 = 2.303

and

lo g 2 = 0.3010

)

Answer: (C)

Solution

For a first-order reaction, the rate constant

k = \frac{l n 2}{t _{1/2}}

.

For 90% completion:

k = \frac{1}{t _{0.90}} ln \frac{A _{0}}{0.1 A _{0}} = \frac{l n 10}{t _{0.90}}

Given

t_{0.90} = x \cdot t_{1/2}

\frac{l n 2}{t _{1/2}} = \frac{l n 10}{x \cdot t _{1/2}}

x = \frac{l n 10}{l n 2} = \frac{2.303}{2.303 \times 0.3010} = 3.32

JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Single correct option