Question

In the reaction $A\to B$, if the concentration of A is increased by four times, the rate of the reaction becomes double. The order of the reaction is:

• A. $0$
• B. $1$
• C. $1/2$
• D. $2$

Answer 1

The correct option is C $1/2$

Rate of reaction $=k[A{]}^n$, where $n$ the is order.
$r_1=k[A_1{]}^n$
$r_2=k[A_2{]}^n$
$\frac{r_2}{r_1}={\left[\frac{\left[A_2\right]}{\left[A_1\right]}\right]}^n\Rightarrow \frac{2}{1}={\left[\frac{4}{1}\right]}^n$
$\Rightarrow [4{]}^n=2$
So, $n=\frac{1}{2}$