For a first order reaction, the time required for completion of 90 % reaction is ' x ' times the half life of the reaction. The value of ' x ' is (Given: ln 10=2.303 and log 2=0.3010 )

Question

For a first order reaction, the time required for completion of 90 % reaction is ' x ' times the half life of the reaction. The value of ' x ' is (Given: ln 10=2.303 and log 2=0.3010 ) (A) 1.12 (B) 2.43 (C) 3.32 (D) 33.31

Tardigrade Admin · Accepted Answer

Given t 0.90= t 0.90= xt 1 / 2 First order rate constant K =( ln 2/ t 1 / 2)=(1/ xt 1 / 2) ln ( A 0/ A 0- A 0 × (90)100) ( ln 2/ t 1 / 2)=( ln 10/ xt 1 / 2) x=( ln 10/ ln 2)=(2.303/2.303 × 0.3010)=3.32

Q. For a first order reaction, the time required for completion of $90%$ reaction is ' $x$ ' times the half life of the reaction. The value of ' $x$ ' is
(Given: $ln 10 = 2.303$ and $lo g 2 = 0.3010$ )

$1.12$

$2.43$

$3.32$

$33.31$

Given $t_{0.90} = t_{0.90} = x t_{1/2}$
First order rate constant
$K = \frac{l n 2}{t _{1/2}} = \frac{1}{x t _{1/2}} ln \frac{A _{0}}{A _{0} - A _{0} \times \frac{90}{100}}$
$\frac{l n 2}{t _{1/2}} = \frac{l n 10}{x t _{1/2}}$
$x = \frac{l n 10}{l n 2} = \frac{2.303}{2.303 \times 0.3010} = 3.32$

Q. For a first order reaction, the time required for completion of 90% reaction is ' x ' times the half life of the reaction. The value of ' x ' is (Given: ln10=2.303 and log2=0.3010 )

1.12

2.43

3.32

33.31