Time required for 100 per cent completion of a zero order reaction is

Question

Time required for 100 per cent completion of a zero order reaction is (A)  (2k/a)  (B)  (a/2k)  (C)  (a/k)  (D)  ak

Tardigrade Admin · Accepted Answer

For zero order reactions, [A]=-kt+[A]0 where, [A]0= initial concentration =a [A]= remaining concentration =a-a=0 On putting values of [A]0 and [A], we get t=(a/k)

Q. Time required for 100 per cent completion of a zero order reaction is

$\frac{2 k}{a}$

$\frac{a}{2 k}$

$\frac{a}{k}$

$ak$

For zero order reactions, $[A] = - k t + [A]_{0}$ where, $[A]_{0} =$ initial concentration $= a$ $[A] =$ remaining concentration $= a - a = 0$ On putting values of $[A]_{0}$ and $[A],$ we get $t = \frac{a}{k}$

Q. Time required for 100 per cent completion of a zero order reaction is

a2k​

2ka​

ka​

ak

For zero order reactions, [A]=−kt+[A]0​ where, [A]0​= initial concentration =a [A]= remaining concentration =a−a=0 On putting values of [A]0​ and [A], we get t=ka​

$\frac{2 k}{a}$

$\frac{a}{2 k}$

$\frac{a}{k}$

$ak$

For zero order reactions, $[A] = - k t + [A]_{0}$ where, $[A]_{0} =$ initial concentration $= a$ $[A] =$ remaining concentration $= a - a = 0$ On putting values of $[A]_{0}$ and $[A],$ we get $t = \frac{a}{k}$