Question

For a chemical reaction,

A $⟶$ Products

the rate of reaction doubles when the concentration of 'A' is increased by a factor of 4, the order of reaction is :

• A. 2
• B. 0.5
• C. 4
• D. 1

Answer 1

The correct option is B 0.5

For nth order reaction, the rate of the reaction is

$r=k[A{]}^n$ ....(1)
The rate of reaction doubles when the concentration of 'A' is increased by a factor of 4.

$r^{\prime }=k[A{]}^{\prime n}$ .....(2)
Now,

$r^{\prime }=2r$
$[A{]}^{\prime }=4[A]$
Hence, equation (2) becomes

$2r=k\left(4\right[A]{)}^n$ ......(3)
Divide equation (3) with equation (1)

$\frac{2r}{r}=\frac{k\left(4\right[A]{)}^n}{k[A{]}^n}$

$2=4^n$
Take natural logarithm on both sides

$ln2=nln4$
$n=0.5$

$\therefore$ Order of the reaction is 0.5.