The rate constant of a first order reaction is 0.0693 min -1. What is the time (in minutes) required for reducing an initial concentration of text20 textmol textL-1 to text2 text.5 textmol textL-1 ?

Question

The rate constant of a first order reaction is 0.0693 min -1. What is the time (in minutes) required for reducing an initial concentration of text20 textmol textL-1 to text2 text.5 textmol textL-1 ? (A) 40 (B) 30 (C) 20 (D) 10

Tardigrade Admin · Accepted Answer

Given, k=0.0693 min -1 t=? a=20 mol L-1 a-x=2.5 mol L-1 k=(2.303/t) log (a/a-x) t=(2.303/k) log (20/2.5) t=(2.303/0.0693) log 8 =33.23× 0.9030 min =30 min

Q. The rate constant of a first order reaction is $0.0693 min^{- 1} .$ What is the time (in minutes) required for reducing an initial concentration of $20 mol L^{- 1}$ to $2 .5 mol L^{- 1}$ ?

40

30

20

10

Given, $k = 0.0693 min^{- 1}$ $t = ?$ $a = 20 m o l L^{- 1}$ $a - x = 2.5 m o l L^{- 1}$ $k = \frac{2.303}{t} lo g \frac{a}{a - x}$ $t = \frac{2.303}{k} lo g \frac{20}{2.5}$ $t = \frac{2.303}{0.0693} lo g 8$ $= 33.23 \times 0.9030 min$ $= 30 min$

Q. The rate constant of a first order reaction is 0.0693min−1. What is the time (in minutes) required for reducing an initial concentration of 20molL−1 to 2.5molL−1 ?

40

30

20

10

Given, k=0.0693min−1 t=? a=20molL−1 a−x=2.5molL−1 k=t2.303​loga−xa​ t=k2.303​log2.520​ t=0.06932.303​log8 =33.23×0.9030min =30min

Q. The rate constant of a first order reaction is $0.0693 min^{- 1} .$ What is the time (in minutes) required for reducing an initial concentration of $20 mol L^{- 1}$ to $2 .5 mol L^{- 1}$ ?

Given, $k = 0.0693 min^{- 1}$ $t = ?$ $a = 20 m o l L^{- 1}$ $a - x = 2.5 m o l L^{- 1}$ $k = \frac{2.303}{t} lo g \frac{a}{a - x}$ $t = \frac{2.303}{k} lo g \frac{20}{2.5}$ $t = \frac{2.303}{0.0693} lo g 8$ $= 33.23 \times 0.9030 min$ $= 30 min$