The half-life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/litre and concentration at some instant is 0.01 mol/litre, then t is

Question

The half-life of a first order reaction is 10 minutes. If initial amount is 0.08 mol/litre and concentration at some instant is 0.01 mol/litre, then t is (A) 10 minutes (B) 30 minutes (C) 20 minutes (D) 40 minutes

Tardigrade Admin · Accepted Answer

k=(0.693/t1/2)=(0.693/10) t =(2.303/k) log (0.08/0.01)=(2.303/0.693)log 8=30 minutes Alternatively, Amount left after n half-lives,0.01 =(0.08/2n) 2n =(0.08/0.01)=8 Then, total time = 3 ×10 = 30 minutes

Q. The half-life of a first order reaction is $10$ minutes. If initial amount is $0.08 m o l / l i t re$ and concentration at some instant is $0.01 m o l / l i t re$ , then t is

10 minutes

30 minutes

20 minutes

40 minutes

$k = \frac{0.693}{t _{1/2}} = \frac{0.693}{10}$
$t = \frac{2.303}{k} l o g \frac{0.08}{0.01} = \frac{2.303}{0.693} l o g 8 = 30 min u t es$
Alternatively,
Amount left after $n$ half-lives, $0.01 = \frac{0.08}{2 ^{n}}$
$2^{n} = \frac{0.08}{0.01} = 8$
Then, total time = $3 \times 10 = 30 min u t es$

Q. The half-life of a first order reaction is 10 minutes. If initial amount is 0.08mol/litre and concentration at some instant is 0.01mol/litre, then t is