For the first order reaction half-life is 14 sec, the time required for the initial concentration to reduce to 1/8 of its value is

Question

For the first order reaction half-life is 14 sec, the time required for the initial concentration to reduce to 1/8 of its value is (A) (14)3s (B) 28 s (C) 42 s (D) (14)2s

Tardigrade Admin · Accepted Answer

N=N0 ×((1/2))n (1/8) N0=N0 ×((1/2))n n=3 T=n × t1 / 2 =3 × 14=42 s

Q. For the first order reaction half-life is $14 sec$ , the time required for the initial concentration to reduce to $1/8$ of its value is

$(14)^{3} s$

$28 s$

$42 s$

$(14)^{2} s$

$N = N_{0} \times (\frac{1}{2})^{n}$
$\frac{1}{8} N_{0} = N_{0} \times (\frac{1}{2})^{n}$
$n = 3$
$T = n \times t_{1/2}$
$= 3 \times 14 = 42 s$

Q. For the first order reaction half-life is 14sec, the time required for the initial concentration to reduce to 1/8 of its value is

(14)3s

28s

42s

(14)2s