Question

Given the reaction $A\to C$, it is found that the rate of the reaction doubles when the concentration of $A$ is increased four times, the order of the reaction is:

• A. 1
• B. 4
• C. 1/2
• D. 2

Answer 1

The correct option is C 1/2

For the give reaction ; $A\to C$
Rate = r = $k[A{]}^n$........(i) ; where n = order of reaction.
Also given that, rate doubles when concentration of A becomes 4 times
Therefore new rate = 2r = $k[4A{]}^n$.......(ii)
On dividing (i) by (ii) we get:

$\frac{r}{2r}=\frac{k[A{]}^n}{k[4A{]}^n}$

$\therefore \frac{1}{2}=[\frac{1}{4}{]}^n$

$\therefore [\frac{1}{2}{]}^1=[\frac{1}{2}{]}^{2n}$

2n = 1 & n = $\frac{1}{2}$