Question

$1.386$ hours are required for the disappearance of $75\%$ of a reactant of first-order reaction. What is the rate constant of the reaction?

• A. $3.6\times 10^{-3}{s}^{-1}$
• B. $7.2\times 10^{-3}{s}^{-1}$
• C. $2.7\times 10^{-4}{s}^{-1}$
• D. $1.8\times 10^{-3}{s}^{-1}$

Answer 1

Answer: (C)

Let original amount $N_0=100$
Since, disappeared amount $=75\%$
$\therefore$ left amount, $N=100-75=25$
$\therefore$ Since,$\frac{N_0}{N}=2^n$ where n = no. of half lives
or $\frac{100}{25}$ or $2^2=2^n,$ here $n=2$
$\because =n\times t_{12}1.386$ hours $=2\times t_{1/2}$
$t_{1/2}=0.693-2494.8$ sec
$k=\frac{0.693}{2494.8}=2.7\times 10^{-4}se{c}^{-1}$