For the reaction $A + B \rightarrow \text{product}$,the rate of reaction is $3.6 \times 10^{-2} \ mol \ dm^{-3} \ sec^{-1}$. When $[A] = 0.2 \ mol \ dm^{-3}$ and $[B] = 0.1 \ mol \ dm^{-3}$,calculate the rate constant of the reaction if the reaction is first order in $A$ and second order in $B$.

  • A
    $3.6 \ mol^{-2} \ dm^6 \ sec^{-1}$
  • B
    $1.8 \ mol^{-2} \ dm^6 \ sec^{-1}$
  • C
    $18 \ mol^{-2} \ dm^6 \ sec^{-1}$
  • D
    $36 \ mol^{-2} \ dm^6 \ sec^{-1}$

Explore More

Similar Questions

For an $n^{th}$ order reaction,the rate law is given by $\frac{dx}{dt} = K[A]^n$. Which parameter can be determined from the logarithmic plot?

The rate constants for the following reactions are:
Reaction $1$: $A \xrightarrow{\text{catalyst } 1} P_1, k_1 = 1 \ s^{-1}$
Reaction $2$: $A \xrightarrow{\text{catalyst } 2} P_2, k_2 = 0.1 \ L \ mol^{-1} \ s^{-1}$
Reaction $3$: $A \xrightarrow{\text{catalyst } 3} P_3, k_3 = 0.01 \ L^2 \ mol^{-2} \ s^{-1}$
What is the correct relation between the rates of the reactions at $[A] = 1 \ M$?

For a chemical reaction,which of the following can never be a fraction?

For the following reaction: $NO_{2(g)} + CO_{(g)} \to NO_{(g)} + CO_{2(g)}$,the rate law is: $\text{Rate} = k [NO_2]^2$. If $0.1 \ mol$ of gaseous carbon monoxide is added at constant temperature to the reaction mixture,which of the following statements is true?

The rate equation for the reaction $2 A + B \longrightarrow$ products is $\text{rate} = k[A][B]^2$. If $k$ at $T \ K$ is $5.0 \times 10^{-6} \ mol^{-2} \ L^2 \ s^{-1}$,the initial rate of the reaction,when $[A] = 0.05 \ mol \ L^{-1}$ and $[B] = 0.1 \ mol \ L^{-1}$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo