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JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Chemical Kinetics

Time required for completion of

99.9%

of a First order reaction is ____ times of half life

(t_{1/2})

of the reaction.

Answer: 10

Solution

For a first-order reaction,

t = \frac{2.303}{k} lo g \frac{a}{a - x}

. At 99.9% completion,

a - x = 0.001 a

, so

t_{99.9%} = \frac{2.303}{k} lo g (1000) = \frac{2.303}{k} \times 3

.

The half-life is

t_{1/2} = \frac{2.303}{k} lo g 2

.

Dividing,

\frac{t _{99.9%}}{t _{1/2}} = \frac{3}{l o g 2} = \frac{3}{0.301} = 10

.

JEE Main · 2024 · 27 Jan 2024 · Shift 2 · Integer answer

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