Explain why the average rate of reaction decreases with the concentration of the reactant using an example.
(N/A) The rate of a chemical reaction depends on the concentration of the reactants. As the reaction proceeds,the concentration of reactants decreases,which leads to a decrease in the frequency of effective collisions between reactant molecules,thereby decreasing the rate of reaction.
Example: The reaction of $C_{4}H_{9}Cl$ with $OH^{-}$ ions:
$C_{4}H_{9}Cl + OH^{-} \rightarrow C_{4}H_{9}OH + Cl^{-}$
The following table shows the average rate of reaction at different time intervals:
As time increases,the concentration of the reactant decreases,causing the average rate of reaction to decrease.
Example: The reaction of $C_{4}H_{9}Cl$ with $OH^{-}$ ions:
$C_{4}H_{9}Cl + OH^{-} \rightarrow C_{4}H_{9}OH + Cl^{-}$
The following table shows the average rate of reaction at different time intervals:
| $t_{1(s)} - t_{2(s)}$ | $r_{av} = -\frac{\Delta [R]}{\Delta t} \ (mol \ L^{-1} \ s^{-1})$ |
| $0 - 50 \ s$ | $1.9 \times 10^{-4}$ |
| $50 - 100 \ s$ | $1.7 \times 10^{-4}$ |
| $100 - 150 \ s$ | $1.58 \times 10^{-4}$ |
| $200 - 300 \ s$ | $1.22 \times 10^{-4}$ |
As time increases,the concentration of the reactant decreases,causing the average rate of reaction to decrease.
Explore More
Similar Questions
The rate of reaction is expressed as
$+\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt}$
The reaction is:
$+\frac{1}{2}\frac{d[C]}{dt} = -\frac{1}{3}\frac{d[D]}{dt} = +\frac{1}{4}\frac{d[A]}{dt} = -\frac{d[B]}{dt}$
The reaction is:
Medium
View SolutionWhich are the units of rate of reaction?
Medium
View SolutionWhat is the average rate of reaction when the change in concentration of product is $0.05 \ M$ in $20 \ s$ (in $M \ s^{-1}$)?
If the instantaneous rate of reaction is given as $-\frac{1}{a} \frac{d[A]}{dt} = -\frac{1}{b} \frac{d[B]}{dt} = \frac{1}{c} \frac{d[C]}{dt} = \frac{1}{d} \frac{d[D]}{dt}$,then the reaction is represented as:
For the reaction,$3 I_{(aq)}^{-} + S_2 O_{8_{(aq)}}^{2-} \longrightarrow I_{3_{(aq)}}^{-} + 2 SO_{4_{(aq)}}^{2-}$,the rate of formation of $SO_4^{2-}$ is $0.022 \ mol \ dm^{-3} \ sec^{-1}$. What is the rate of formation of $I_{3_{(aq)}}^{-}$?
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo