JEE Main · 2009 · 12 Apr 2009 · Offline · Chemical Kinetics

The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of

99%

of the chemical reaction will be (

lo g 2 = 0.301

) :

Answer: (C)

Solution

For a first-order reaction the rate constant follows from the half-life as

k = \frac{0.693}{t _{1/2}} = \frac{0.693}{6.93} = 0.1 mi n^{- 1}

. Ninety-nine per cent completion leaves one hundredth of the reactant, so with

t = \frac{2.303}{k} lo g \frac{[ A _{0} ]}{[ A ]} = \frac{2.303}{0.1} lo g 100

the time comes out as

23.03 \times 2

, which is 46.06 minutes.

JEE Main · 2009 · 12 Apr 2009 · Offline · Single correct option