For the reaction,$CH_3Br_{(aq)} + OH_{(aq)}^{-} \rightarrow CH_3OH_{(aq)} + Br_{(aq)}^{-}$,the rate law is $\text{rate} = k[CH_3Br][OH^{-}]$. What is the change in the rate of reaction if the concentration of both reactants is doubled?

If the rate of reaction is given by $Rate = K [A]^{3/2} [B]^{-1/2}$,then find the order of the reaction.

For the reaction $A + B \rightarrow C$,select the appropriate rate law based on the following data:
$1$. $[A] = 0.012, [B] = 0.035 \rightarrow \text{Initial Rate} = 0.10$
$2$. $[A] = 0.024, [B] = 0.070 \rightarrow \text{Initial Rate} = 1.6$
$3$. $[A] = 0.024, [B] = 0.035 \rightarrow \text{Initial Rate} = 0.20$
$4$. $[A] = 0.012, [B] = 0.070 \rightarrow \text{Initial Rate} = 0.80$

For a second-order reaction where the initial concentrations of both reactants are equal,it takes $3000 \ s$ for the reaction to be $60\%$ complete. How much time (in $s$) will it take for the reaction to be $20\%$ complete?

When the concentration of a reactant,$A$,in a reaction $A \to \text{Products}$ is doubled,the rate of reaction increases seven times. The order of the reaction is between:

In the sequence of reaction $A$ $\xrightarrow{k_1} X$ $\xrightarrow{k_2} Y$ $\xrightarrow{k_3} Z$ with $k_3 > k_2 > k_1$,then the rate determining step is