JEE Main · 2005 · 08 May 2005 · Offline · Chemical Kinetics

t_{1/4}

can be taken as the time taken for the concentration of a reactant to drop to

\frac{3}{4}

of its initial value. If the rate constant for a first order reaction is K , the

t_{1/4}

can be written as

Answer: (B)

Solution

For a first-order reaction the integrated law is

ln \frac{[ A ] _{0}}{[ A ]} = K t

. When the concentration has fallen to three-quarters of its start,

[A] = \frac{3}{4} [A]_{0}

, so

ln \frac{[ A ] _{0}}{\frac{3}{4} [ A ] _{0}} = ln \frac{4}{3} = K t_{1/4}

. Since

ln \frac{4}{3} \approx 0.29

, the time works out to

t_{1/4} = \frac{0.29}{K}

JEE Main · 2005 · 08 May 2005 · Offline · Single correct option