The rate of the reaction becomes twice when t | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

$r_{1}=k[A]^{n}$

$r_{2}=2 r_{1}=k[8 A]^{n}$

$	herefore \frac{2 r_{1}}{r_{1}}=(8)^{n}$

$(2)^{1}=(2)^{3 n}$

$	herefore 3 n=1$

$n=\frac{

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

$r_{1}=k[A]^{n}$

$r_{2}=2 r_{1}=k[8 A]^{n}$

$	herefore \frac{2 r_{1}}{r_{1}}=(8)^{n}$

$(2)^{1}=(2)^{3 n}$

$	herefore 3 n=1$

$n=\frac{1}{3}$

The rate of the reaction becomes twice when the concentration of reactant becomes $8$ times then the order of the reaction is

Similar Questions

An elementary reaction between $A$ and $B$ is a second order reaction. Which of the following rate equations must be correct?

$A(g) + 2B(g) \to $ product is a elementary reaction, which of the following is incorrect.

Velocity constant $K$ of a reaction is affected by

The rate law of the reaction $A + 2B \to $Product is given by $\frac{{d[dB]}}{{dt}} = k[{B^2}]$. If $ A$ is taken in excess, the order of the reaction will be

The rate law for a reaction between the substances $ A $ and $B$ is given by, rate $= k{[A]^n}{[B]^m}$. On doubling the concentration of $A$ and halving the concentration of $B$, the ratio of the new rate to the earlier rate of the reaction will be as