For a chemical reaction, A → B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order in terms of A for this reaction is

Question

For a chemical reaction, A → B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order in terms of A for this reaction is (A) two (B) one (C) half (D) zero

Tardigrade Admin · Accepted Answer

rate (r1)=k[A]n (where, n is order of reaction) dots (i) 2 r1=k[4 A]n dots(ii) On solving equation (i) and (ii), n=(1/2)

Q. For a chemical reaction, $A \to B$ , it is found that the rate of reaction doubles when the concentration of $A$ is increased four times. The order in terms of $A$ for this reaction is

two

one

half

zero

rate $(r_{1}) = k [A]^{n}$ (where, $n$ is order of reaction) $\dots (i)$

$2 r_{1} = k [4 A]^{n}$ $\dots (ii)$

On solving equation (i) and (ii), $n = \frac{1}{2}$

Q. For a chemical reaction, A→B, it is found that the rate of reaction doubles when the concentration of A is increased four times. The order in terms of A for this reaction is

two

one

half

zero

rate (r1​)=k[A]n (where, n is order of reaction) …(i) 2r1​=k[4A]n …(ii) On solving equation (i) and (ii), n=21​

Q. For a chemical reaction, $A \to B$ , it is found that the rate of reaction doubles when the concentration of $A$ is increased four times. The order in terms of $A$ for this reaction is

rate $(r_{1}) = k [A]^{n}$ (where, $n$ is order of reaction) $\dots (i)$

$2 r_{1} = k [4 A]^{n}$ $\dots (ii)$

On solving equation (i) and (ii), $n = \frac{1}{2}$