For a certain reaction of order n, the time for half change, t1 / 2, is given by t1 / 2=((2-√2)/k) × C01 / 2 where k is constant and C 0 is the initial concentration. What is n ?

Question

For a certain reaction of order n, the time for half change, t1 / 2, is given by t1 / 2=((2-√2)/k) × C01 / 2 where k is constant and C 0 is the initial concentration. What is n ? (A) 1 (B) 2 (C) 0 (D) 0.5

Tardigrade Admin · Accepted Answer

t 1 / 2=((2-√2/ K )) C 01 / 2 For n text th order reaction t 1 / 2 propto( C 0)1- text order 1- n =(1/2) So, n =0.5.

Q. For a certain reaction of order $n$ , the time for half change, $t_{1/2}$ , is given by $t_{1/2} = \frac{( 2 - 2 )}{k} \times C_{0}^{1/2}$ where $k$ is constant and $C_{0}$ is the initial concentration. What is $n$ ?

1

2

0

0.5

$t_{1/2} = (\frac{2 - 2}{K}) C_{0}^{1/2}$
For $n^{th}$ order reaction
$t_{1/2} \propto (C_{0})^{1 - order}$
$1 - n = \frac{1}{2}$
So, $n = 0.5$ .

Q. For a certain reaction of order n, the time for half change, t1/2​, is given by t1/2​=k(2−2​)​×C01/2​ where k is constant and C0​ is the initial concentration. What is n ?

1

2

0

0.5

t1/2​=(K2−2​​)C01/2​ For nth order reaction t1/2​∝(C0​)1− order 1−n=21​ So, n=0.5.

Q. For a certain reaction of order $n$ , the time for half change, $t_{1/2}$ , is given by $t_{1/2} = \frac{( 2 - 2 )}{k} \times C_{0}^{1/2}$ where $k$ is constant and $C_{0}$ is the initial concentration. What is $n$ ?

$t_{1/2} = (\frac{2 - 2}{K}) C_{0}^{1/2}$
For $n^{th}$ order reaction
$t_{1/2} \propto (C_{0})^{1 - order}$
$1 - n = \frac{1}{2}$
So, $n = 0.5$ .