The rate of a gaseous reaction is given by the expression $K [A] [B]$. If the volume of the reaction vessel is suddenly reduced to $1/4$ of the initial volume,the reaction rate relating to the original rate will be

Consider the following reaction: $A \longrightarrow \text{Products}$. This reaction is completed in $100 \ min$. The rate constant of this reaction at $t_1 = 10 \ min$ is $10^{-2} \ min^{-1}$. What is the rate constant (in $min^{-1}$) at $t_2 = 20 \ min$?

What is the order of a reaction which has a rate expression $\text{rate} = K[A]^{3/2}[B]^{-1}$?

For a gaseous reaction $2A + B \rightarrow C + D$,the rate of reaction is given by $Rate = K[A][B]$. If the volume of the container is reduced to $1/4$ of its original volume,what will be the ratio of the new rate to the original rate?

For a reaction $2A + B \to \text{Products}$,doubling the initial concentration of both the reactants increases the rate by a factor of $8$,and doubling the concentration of $B$ alone doubles the rate. The rate law for the reaction is

For the following graphs,choose the correct option regarding the order of reaction: