For given first order reaction, the reactant reduces to 1/4th its initial value in 10 min. The rate constant of the reaction is

Question

For given first order reaction, the reactant reduces to 1/4th its initial value in 10 min. The rate constant of the reaction is (A) 0.1386 text mi textn- 1 (B) 0.0693 text mi textn- 1 (C) 0.1386 text mol textL- 1min- 1  (D) 0.0693 text mol textL- 1min- 1

Tardigrade Admin · Accepted Answer

For the first order reaction, k=(2.303/t)log (a/a - x) Given, t=10 text min a=a a-x=(a/4) k=(2.303/10)log (a/(a)4) =(2.303 × 2/10)log 2 =0.1386 text mi textn- 1

Q. For given first order reaction, the reactant reduces to 1/4th its initial value in 10 min. The rate constant of the reaction is

$0.1386 mi n^{- 1}$

$0.0693 mi n^{- 1}$

$0.1386 mol L^{- 1} mi n^{- 1}$

$0.0693 mol L^{- 1} mi n^{- 1}$

For the first order reaction,

$k = \frac{2.303}{t} l o g \frac{a}{a - x}$

Given, $t = 10 min$

$a = a$

$a - x = \frac{a}{4}$

$k = \frac{2.303}{10} l o g \frac{a}{\frac{a}{4}}$

$= \frac{2.303 \times 2}{10} l o g 2$

$= 0.1386 mi n^{- 1}$

Q. For given first order reaction, the reactant reduces to 1/4th its initial value in 10 min. The rate constant of the reaction is

0.1386 min−1

0.0693 min−1

0.1386 mol L−1min−1

0.0693 mol L−1min−1

For the first order reaction, k=t2.303​loga−xa​ Given, t=10 min a=a a−x=4a​ k=102.303​log4a​a​ =102.303×2​log2 =0.1386 min−1

$0.1386 mi n^{- 1}$

$0.0693 mi n^{- 1}$

$0.1386 mol L^{- 1} mi n^{- 1}$

$0.0693 mol L^{- 1} mi n^{- 1}$

For the first order reaction,

$k = \frac{2.303}{t} l o g \frac{a}{a - x}$

Given, $t = 10 min$

$a = a$

$a - x = \frac{a}{4}$

$k = \frac{2.303}{10} l o g \frac{a}{\frac{a}{4}}$

$= \frac{2.303 \times 2}{10} l o g 2$

$= 0.1386 mi n^{- 1}$