$A$ study of chemical kinetics of the reaction $A + B \to$ Products,gave the following data at $25 \ ^oC$.
$Exp. \ No.$ $[A]$ $[B]$ $Rate$
$1.$ $1.0$ $0.15$ $4.2 \times 10^{-6}$
$2.$ $2.0$ $0.15$ $8.4 \times 10^{-6}$
$3.$ $1.0$ $0.20$ $5.6 \times 10^{-6}$

Find out the rate law.

  • A
    $r = k[A]$
  • B
    $r = k[B]$
  • C
    $r = k[A][B]$
  • D
    $r = k[A][B]^2$

Explore More

Similar Questions

For a zero order reaction,will the molecularity be equal to zero? Explain.

For the reaction $2A + B \rightarrow C + D$,select the correct rate law based on the following data:
$1$. $[A] = 0.1, [B] = 0.1, \text{Initial Rate} = 7.5 \times 10^{-3}$
$2$. $[A] = 0.3, [B] = 0.2, \text{Initial Rate} = 9.0 \times 10^{-2}$
$3$. $[A] = 0.3, [B] = 0.4, \text{Initial Rate} = 3.6 \times 10^{-1}$
$4$. $[A] = 0.4, [B] = 0.1, \text{Initial Rate} = 3.0 \times 10^{-2}$

If the relation between half-life time $(t_{1/2})$ and initial concentration $[R]_0$ is given by $t_{1/2} \propto \frac{1}{[R]_0^{n-1}}$,what is the order of the reaction?

The data for the reaction $A + B \to C$ is given below. The rate law corresponding to the above data is:
$Exp.$ $[A]_0$ $[B]_0$ Initial rate
$(1)$ $0.012$ $0.035$ $0.10$
$(2)$ $0.024$ $0.070$ $0.80$
$(3)$ $0.024$ $0.035$ $0.10$
$(4)$ $0.012$ $0.070$ $0.80$

The rate law for the reaction between substances $A$ and $B$ is given by $\text{Rate} = k[A]^n[B]^m$. If the concentration of $A$ is doubled and the concentration of $B$ is halved,what is the ratio of the new rate to the initial rate?

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