A reaction has a half life of 1 min. The time required for 99.9 % completion of the reaction is . min. (Round off to the Nearest integer) [Use: ln 2=0.69, ln 10=2.3]

Question

A reaction has a half life of 1 min. The time required for 99.9 % completion of the reaction is . min. (Round off to the Nearest integer) [Use: ln 2=0.69, ln 10=2.3] (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

( t 99.9 %/ t 50 %)=((1/ K ) ln (100/0.1)/(1) K ln 2) =( ln 1000/ ln 2) × t 50 % =(3 ln 10/ ln 2) × 1 =(3 × 2.3/0.69)=10

Q. A reaction has a half life of 1 min. The time required for $99.9%$ completion of the reaction is ____. min. (Round off to the Nearest integer)
[Use : $ln 2 = 0.69, ln 10 = 2.3]$

$\frac{t _{99.9%}}{t _{50%}} = \frac{\frac{1}{K} l n \frac{100}{0.1}}{\frac{1}{K} l n 2}$

$= \frac{l n 1000}{l n 2} \times t_{50%}$

$= \frac{3 l n 10}{l n 2} \times 1$

$= \frac{3 \times 2.3}{0.69} = 10$

Q. A reaction has a half life of 1 min. The time required for 99.9% completion of the reaction is ____. min. (Round off to the Nearest integer) [Use : ln2=0.69,ln10=2.3]

t50%​t99.9%​​=K1​ln2K1​ln0.1100​​ =ln2ln1000​×t50%​ =ln23ln10​×1 =0.693×2.3​=10

Q. A reaction has a half life of 1 min. The time required for $99.9%$ completion of the reaction is ____. min. (Round off to the Nearest integer)
[Use : $ln 2 = 0.69, ln 10 = 2.3]$

$\frac{t _{99.9%}}{t _{50%}} = \frac{\frac{1}{K} l n \frac{100}{0.1}}{\frac{1}{K} l n 2}$

$= \frac{l n 1000}{l n 2} \times t_{50%}$

$= \frac{3 l n 10}{l n 2} \times 1$

$= \frac{3 \times 2.3}{0.69} = 10$