For a given reaction $t_{1/2} = \frac{1}{k \cdot a}$,the order of reaction will be:

  • A
    $0$
  • B
    $1$
  • C
    $2$
  • D
    $3$

Explore More

Similar Questions

The rate constant for the reaction $2N_2O_5 \rightarrow 4NO_2 + O_2$ is $3.0 \times 10^{-5} \text{ s}^{-1}$. If the rate of reaction is $2.4 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$, then the concentration of $N_2O_5$ in $\text{mol L}^{-1}$ is:

Higher order $(> 3)$ reactions are rare due to:

If doubling the concentration of a reactant $A$ increases the rate $4$ times and tripling the concentration of $A$ increases the rate $9$ times,the rate is proportional to

The initial rates of decrease of $I_2$ in acetone-iodine reaction catalysed by $H^{+}$ are given in the table.
ExperimentInitial $[I_2]$ $(mol \ L^{-1})$Initial $[H^{+}]$ $(mol \ L^{-1})$Initial $[CH_3COCH_3]$ $(mol \ L^{-1})$Initial rate $(mol \ L^{-1} \ s^{-1})$
$1$$0.01$$0.1$$0.1$$0.096$
$2$$0.01$$0.2$$0.1$$0.192$
$3$$0.02$$0.2$$0.1$$0.192$
$4$$0.01$$0.2$$0.2$$0.384$

The order with respect to $I_2, H^{+}$,acetone and total order of the reaction respectively are:

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

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