Question

If the rate of reaction, $A⟶B$, doubles on increasing the concentration of $A$ by $4$ times, the order of reaction is:

• A. $2$
• B. $1$
• C. $1/2$
• D. $4$

Answer 1

Answer: (C)

$1/2$

$v_1=k[{A]}^m;v_{2}k{\left[A\right]}^m{4}^m=k{\left[A\right]}^m(2^{2m})\frac{v_2}{v_1}=\frac{k[A{]}^m}{k{\left[A\right]}^m}{2}^{2m}=22m=1m=\frac{1}{2}$