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
$k=\frac{0.693}{2494.8}=2.7\times10^{-4}sec^{-1}$