The rate constant for a first order reaction is 20 min -1. The time required for the initial concentration of the reactant to reduce to its (1/32) level is × 10-2 min. (Nearest integer) (Given: ln 10=2.303 log 2=0.3010 )

Question

The rate constant for a first order reaction is 20 min -1. The time required for the initial concentration of the reactant to reduce to its (1/32) level is × 10-2 min. (Nearest integer) (Given: ln 10=2.303 log 2=0.3010 ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

C =( C o /2 n ) =( C o /32) n =5 t =5 t 1 / 2 =(5 × 0.693/20)=(0.693/4) =0.17325 min =17.325 × 10-2 min

Q. The rate constant for a first order reaction is $20 min^{- 1}$ . The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ___ $\times 1 0^{- 2} min$ . (Nearest integer)
(Given: $ln 10 = 2.303$ $lo g 2 = 0.3010$ )

$C = \frac{C _{o}}{2 ^{n}} = \frac{C _{o}}{32}$
$n = 5$
$t = 5 t_{1/2}$
$= \frac{5 \times 0.693}{20} = \frac{0.693}{4}$
$= 0.17325 min = 17.325 \times 1 0^{- 2} min$

Q. The rate constant for a first order reaction is 20min−1. The time required for the initial concentration of the reactant to reduce to its 321​ level is ___×10−2min. (Nearest integer) (Given: ln10=2.303 log2=0.3010 )

C=2nCo​​=32Co​​ n=5 t=5t1/2​ =205×0.693​=40.693​ =0.17325min=17.325×10−2min

Q. The rate constant for a first order reaction is $20 min^{- 1}$ . The time required for the initial concentration of the reactant to reduce to its $\frac{1}{32}$ level is ___ $\times 1 0^{- 2} min$ . (Nearest integer)
(Given: $ln 10 = 2.303$ $lo g 2 = 0.3010$ )

$C = \frac{C _{o}}{2 ^{n}} = \frac{C _{o}}{32}$
$n = 5$
$t = 5 t_{1/2}$
$= \frac{5 \times 0.693}{20} = \frac{0.693}{4}$
$= 0.17325 min = 17.325 \times 1 0^{- 2} min$