The rate constant for the reaction A longrigh | vedclass

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

Question

(C) The units of the rate constant $k = 2 	imes 10^{-4} \ L \ mol^{-1} \ min^{-1}$ indicate that the reaction is of second order.The rate law is $Rat

Vedclass · Accepted Answer

$0.5 \ M$

Vedclass · Answer

(C) The units of the rate constant $k = 2 	imes 10^{-4} \ L \ mol^{-1} \ min^{-1}$ indicate that the reaction is of second order.The rate law is $Rate = k[A]^2$.First,convert the rate constant to $sec^{-1}$ units: $k = (2 	imes 10^{-4} \ L \ mol^{-1} \ min^{-1}) / 60 = (1 / 30) 	imes 10^{-4} \ L \ mol^{-1} \ sec^{-1}$.Given $Rate = (1 / 12) 	imes 10^{-5} \ M \ sec^{-1}$,we have $(1 / 12) 	imes 10^{-5} = (1 / 30) 	imes 10^{-4} 	imes [A]^2$.Simplifying: $[A]^2 = [(1 / 12) 	imes 10^{-5}] / [(1 / 30) 	imes 10^{-4}] = (30 / 12) 	imes 10^{-1} = 2.5 	imes 0.1 = 0.25$.Thus,$[A] = \sqrt{0.25} = 0.5 \ M$.

The rate constant for the reaction $A \longrightarrow B$ is $2 \times 10^{-4} \ L \ mol^{-1} \ min^{-1}$. The concentration of $A$ at which the rate of the reaction is $(1 / 12) \times 10^{-5} \ M \ sec^{-1}$ is :-

Similar Questions

How can you determine the rate law of the following reaction?
$2 \, NO \, (g) + O_2 \, (g) \to 2 \, NO_2 \, (g)$

The rate constant for the reaction,$2N_2O_5 \to 4NO_2 + O_2$ is $3.0 \times 10^{-4} \ s^{-1}$. If the reaction starts with $1.0 \ mol \ L^{-1}$ of $N_2O_5$,calculate the rate of formation of $NO_2$ at the moment when the concentration of $O_2$ is $0.1 \ mol \ L^{-1}$.

The rate constant of which of the following reactions is independent of the concentration of the reactants?

The dimension of the rate constant for a second-order reaction involves:

For the reaction,$2N_2O_5 \to 4NO_2 + O_2$,the rate equation can be expressed in two ways $-\frac{d[N_2O_5]}{dt} = k[N_2O_5]$ and $+\frac{d[NO_2]}{dt} = k'[N_2O_5]$. $k$ and $k'$ are related as:

Vedclass Test Series

Exam Paper Generator

Online Exam Module