JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Chemical Kinetics

For a first order reaction

A \to B

, the rate constant,

k = 5.5 \times 1 0^{- 14} s^{- 1}

. The time required for

67%

completion of reaction is

x \times 1 0^{- 1}

times the half life of reaction. The value of

x

is ______.
(Nearest integer)
(Given:

lo g 3 = 0.4771

)

Answer: 16

Solution

For a first-order reaction, the time for a given fraction completion is:

t = \frac{1}{k} ln (\frac{1}{1 - f})

For 67% completion (

f = 0.67 = \frac{2}{3}

t_{67%} = \frac{1}{k} ln (\frac{1}{1 - \frac{2}{3}}) = \frac{1}{k} ln (3)

The half-life is:

t_{1/2} = \frac{ln 2}{k}

Taking the ratio:

\frac{t _{67%}}{t _{1/2}} = \frac{ln 3}{ln 2} = \frac{lo g 3}{lo g 2} = \frac{0.4771}{0.301} = 1.585

Given

t_{67%} = x \times 1 0^{- 1} \times t_{1/2}

x \times 1 0^{- 1} = 1.585

x = 15.85 \approx 16

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Integer answer