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# Rate Law - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

## Quick Facts

• Rate Law is considered one of the most asked concept.

• 61 Questions around this concept.

## Solve by difficulty

Consider the reaction, $\dpi{100} 2A+B\rightarrow$ products. When concentration of $B$ alone was doubled, the half­-life did not change. When the concentration of $\dpi{100} A$ alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

The rate law for a reaction between the substances A and B  is given by rate =  $\dpi{100} k\left [ A \right ]^{n}\left [ B \right ]^{m}.$

On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

For the reaction,$\dpi{100} 3A+2B\rightarrow C+D,$  the differential rate law can be written as :

For a reaction of the type  $\mathrm{aA + bB \rightarrow products; -d[A]/dt}$  is equal to

In a reaction, $\mathrm{X_2 + Y_2 \rightarrow 2XY}$ , the X disappears at

A reaction involving two different reactants can never be

Identify the correct statement for a reaction, A+ 2B→2C

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During the course of a Zero order chemical reaction, the rate of a reaction

The rate of a reaction aA +bB Products, is given by  $\mathrm{k[A]^{n} [B]{^{m}}}$. On doubling the concentration of A and halving the concentration of B the rate of reaction changes by

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For a chemical reaction $\mathrm{ A+2 B \rightarrow C+2 D}$ , the initial rate  $\mathrm{ -\frac{d[A]}{d t} }$ at  $\mathrm{t=0 }$ is experimentally found to be  $\mathrm{2.5 \times 10^{-2} \mathrm{M} \mathrm{s}^{-1} }$
What is the value of $\mathrm{-\frac{d[B]}{d t} }$  at $\mathrm{ t=0 }$ in  $\mathrm{ \mathrm{M} \mathrm{s}^{-1} }$ ?

## Concepts Covered - 2

Rate Law

The rate of any reaction depends upon the concentration of the reactants in rate law equation that is rate law and rate of reaction depend upon the order of the reaction.
Let us consider this reaction:
$\mathrm{xA}+\mathrm{yB} \rightarrow \text { Product }$
Rate law equation for the reaction can be given as:

$\begin{array}{l}{\mathrm{R} \propto[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}} \\\\ {\mathrm{R}=\mathrm{K}[\mathrm{A}]^{\mathrm{p}}[\mathrm{B}]^{\mathrm{q}}}\end{array}$
$\text { Order of reaction }=\mathrm{p}+\mathrm{q}$
Here p, q are experimental quantities which may or may not be equal to the respective stoichiometric coefficients ( x, y).

Unit of Rate Constant

The differential rate expression for nth order reaction is as follows:
$-\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{k}(\mathrm{a}-\mathrm{x})^{\mathrm{n}}$
$\text { or } \quad \mathrm{k}=\frac{\mathrm{d} \mathrm{x}}{(\mathrm{a}-\mathrm{x})^{\mathrm{n}} \mathrm{dt}}=\frac{(\text { concentration })}{(\text { concentration })^{\mathrm{n}} \text { time }}=(\text { conc. })^{1-\mathrm{n}} \text { time }^{-1}$
If concentration be expressed in mole L-1 and time in minutes, then
$\mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{1-\mathrm{n}} \mathrm{min}^{-1}$
$\begin{array}{l}{\text { For zero order reaction, } \mathrm{n}=0 \text { and hence, } \mathrm{k}=\mathrm{mole} \mathrm{L}^{-1} \mathrm{min}^{-1}} \\ {\text { For first order reaction, } \mathrm{n}=1 \text { and hence, }} \\ {\: \mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{0} \min ^{-1}=\min ^{-1}} \\ {\text {For second order reaction, } \mathrm{n}=2 \text { and hence, }} \\ {\: \mathrm{k}=\left(\text { mole } \mathrm{L}^{-1}\right)^{-1} \min ^{-1}=\text { mole }^{-1} \mathrm{L} \min ^{-1}}\end{array}$
The rate constant of a first-order reaction has only time in its unit. It has no concentration term in the unit. This means the numerical value of k first-order reaction is independent of the unit in which concentration is expressed. If the concentration unit is changed, the numerical value of k for a first-order reaction will not change. However, it would change with change in time.

## Study it with Videos

Rate Law
Unit of Rate Constant

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