$A$ substance $S$ undergoes two reactions under similar reaction conditions:
$\mathop S\limits_{(2.0 \ M)} \xrightarrow{K_0} X$ (zero order)
$\mathop S\limits_{(2.0 \ M)} \xrightarrow{K_2} Y$ (second order)
The time taken for the concentration of $S$ to become half is $40 \ s$ and $10 \ s$ for the zero-order and second-order reactions,respectively. What is the value of the ratio $K_0 / K_2$?

  • A
    $1$
  • B
    $0.5$
  • C
    $2$
  • D
    $4$

Explore More

Similar Questions

$A$ person's wound was exposed to some bacteria and then bacteria growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay $(r)$ was found to be proportional to the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the 'before' and 'after' situation of the application of the medicine?
[$Given: N = \text{No. of bacteria}, t = \text{time}$, bacterial growth follows $1^{st}$ order kinetics.]

Under identical reaction conditions,for a reactant with an initial concentration of $1.386 \, M$,the time taken for the concentration to become half is $40 \, s$ for a first-order reaction and $20 \, s$ for a zero-order reaction. The ratio of the rate constant of the first-order reaction $(K_1)$ to the rate constant of the zero-order reaction $(K_0)$ is ......... $mol \, L^{-1}$.

The following graph shows the variation of rate constant $(k)$ with temperature $(T)$. Which graph follows the Arrhenius equation?

Difficult
View Solution

For the following reaction
$2X + Y \xrightarrow{i} P$
the rate of reaction is $\frac{d[P]}{dt} = k[X]$. Two moles of $X$ are mixed with one mole of $Y$ to make $1.0 \ L$ of solution. At $50 \ s$,$0.5 \ mole$ of $Y$ is left in the reaction mixture. The correct statement$(s)$ about the reaction is(are)
(Use: $\ln 2 = 0.693$)
$(A)$ The rate constant,$k$,of the reaction is $13.86 \times 10^{-4} \ s^{-1}$.
$(B)$ Half-life of $X$ is $50 \ s$.
$(C)$ At $50 \ s$,$-\frac{d[X]}{dt} = 13.86 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.
$(D)$ At $100 \ s$,$-\frac{d[Y]}{dt} = 3.46 \times 10^{-3} \ mol \ L^{-1} \ s^{-1}$.

The consecutive reaction $X$ $\longrightarrow Y$ $\longrightarrow Z$ takes place in a closed container. Initially,the container has $A_{0}$ moles of $X$ (and no $Y$ and $Z$). The plot of total moles of the constituents in the container as a function of time will be:

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