The time taken for the completion of 3/4 of a first order reaction is

Question

The time taken for the completion of 3/4 of a first order reaction is (A) (2.303/k) log 3/4 (B) (2 .303/k) log 4 (C) (2.303/k) log 1/4 (D) (2.303/0.75) log k

Tardigrade Admin · Accepted Answer

t3/4=(2.303/k)log (1/1-3/4)=(2.303/k)log 4

Q. The time taken for the completion of 3/4 of a first order reaction is

$(2.303/ k) l o g 3/4$

(2 .303/k) log 4

(2.303/k) log 1/4

(2.303/0.75) log k

$t_{3/4} = \frac{2.303}{k} l o g \frac{1}{1 - 3/4} = \frac{2.303}{k} l o g 4$

Q. The time taken for the completion of 3/4 of a first order reaction is

(2.303/k)log3/4

(2 .303/k) log 4

(2.303/k) log 1/4

(2.303/0.75) log k

t3/4​=k2.303​log1−3/41​=k2.303​log4

$(2.303/ k) l o g 3/4$

$t_{3/4} = \frac{2.303}{k} l o g \frac{1}{1 - 3/4} = \frac{2.303}{k} l o g 4$