The time taken for 90% of a first order reaction to complete is approximately

Question

The time taken for 90% of a first order reaction to complete is approximately (A) 1.1 times that of half life (B) 2.2 times that of half life (C) 3.3 times that of half life (D) 4.4 times that of half life

Tardigrade Admin · Accepted Answer

t 90 %=(2.303/k) log (a/a-0.9 a) = (2.303/k) log 10 (2.303/k) t1/2=(2.303/k) log(a/a-a/2)=(2.303/k) log 2 =(2.303/k)× 0.3010 t 90 % : t1/2=(1/0.3010)=3.3 i,e., t90 %=3.3 × t1/2

Q. The time taken for 90% of a first order reaction to complete is approximately

1.1 times that of half life

2.2 times that of half life

3.3 times that of half life

4.4 times that of half life

$t 90% = \frac{2.303}{k} lo g \frac{a}{a - 0.9 a} = \frac{2.303}{k} lo g 10$
$\frac{2.303}{k}$
$t_{1/2} = \frac{2.303}{k} lo g \frac{a}{a - a /2} = \frac{2.303}{k} lo g 2$
$= \frac{2.303}{k} \times 0.3010$
$t 90% t_{1/2} = \frac{1}{0.3010} = 3.3$
i,e., $t 90% = 3.3 \times t_{1/2}$

Q. The time taken for 90% of a first order reaction to complete is approximately

1.1 times that of half life

2.2 times that of half life

3.3 times that of half life

4.4 times that of half life

t90%=k2.303​loga−0.9aa​=k2.303​log10 k2.303​ t1/2​=k2.303​loga−a/2a​=k2.303​log2 =k2.303​×0.3010 t90%t1/2​=0.30101​=3.3 i,e.,t90%=3.3×t1/2​

$t 90% = \frac{2.303}{k} lo g \frac{a}{a - 0.9 a} = \frac{2.303}{k} lo g 10$
$\frac{2.303}{k}$
$t_{1/2} = \frac{2.303}{k} lo g \frac{a}{a - a /2} = \frac{2.303}{k} lo g 2$
$= \frac{2.303}{k} \times 0.3010$
$t 90% t_{1/2} = \frac{1}{0.3010} = 3.3$
i,e., $t 90% = 3.3 \times t_{1/2}$