Question

In a chemical reaction, $A⟶B$, it is found that the rate of the reaction doubles when the concentration of A is increased four times. The order of the reaction with respect to A is :

• A. $0$
• B. $\frac{1}{2}$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$\frac{1}{2}$

$r_1=k[A{]}^n$

$r_2=2r_1=k[4A{]}^n$

$\frac{2r_1}{r_1}=(4{)}^n$

$\Rightarrow (2{)}^1=(2{)}^{2n}$

$\therefore 2n=1\Rightarrow n=\frac{1}{2}$