The time required for completion of $93.75 %$ of a first order reaction is $x$ minutes. The half-life of it (in minutes) is

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$x / 8$

$x / 3$

$x / 4$

$x / 2$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Conc at time, t,

$A = A_{0} - \frac{93.75}{100} A_{0} = \frac{6.25}{100} A_{0}$

$A_{0} \to$ Initial conc $k \to$ rate constant ,time,

$t = x$ min using integrated rate law for 1^st order reaction

$k = \frac{2.303}{t} \log \frac{A_{0}}{A}$

$= \frac{2.303}{x} \log \frac{100 A_{0}}{6.25 A_{0}}$

so, $k = \frac{2.303}{x} \log \frac{100}{6.25}$

$= \frac{2.303 x 1.20}{x}$

$\Rightarrow k = \frac{2.303 x 1.20}{x} \min^{- 1}$

We know,

$t_{1 / 2} = \frac{0.693}{k}$ $= \frac{0.693 x}{2.303 x 1.20}$

So, $t_{1 / 2} = \frac{x}{4} \min$

Watch 3-min video & get full concept clarity