Chemistry

Chemical Kinetics

256 Questions

Chemical kinetics involves the study of chemical reaction rates and the factors affecting them, such as temperature and concentration. This topic covers rate laws, half-life, and zero, first, and second order reactions. It is a crucial part of the chemistry syllabus for various competitive examinations.

Reaction rate parametersFirst order kineticsZero and second orderHalf-life of reactionRate constant units

Chemical Kinetics Questions

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

In the reaction A + 2B $\longrightarrow $ 2C + D. if the concentration of A is increased four times and B is decreased to half of its initial concentration then the rate becomes:

  1. twice

  2. half

  3. unchanged

  4. one fourth of the rate
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given reaction is $A+2B\longrightarrow2C+O$

Rate law is given by:-
$Rate=[A][B]^2$             $- (i)$

Now, if the concentration of $A$ is increased $4$ times & concentration of $B$ is increased $1/2$ of the initial concentration. Then,

$(Rate) _{New}=[4A][B/2]^2$
$=4[A] \cfrac {[B]^2}{4}$
$\Rightarrow (Rate) _{New}= [A] [B]^{2}$       $- (ii)$

$(i)$ & $(ii)\Rightarrow$  Rate is unchanged

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

If $n _A$ and $n _B$ are the number of moles at any instant in the reaction : $2A _{(g)} \rightarrow 3B _(g)$ carried out in a vessel of $V\ L$, the rate of the reaction at that instant is given by ?

  1. $- \frac{1}{2} \frac{dn _A}{dt} = \frac{1}{3} \frac{dn _B}{dt}$
  2. $- \frac{1}{V} \frac{dn _A}{dt} = \frac{1}{V} \frac{dn _B}{dt}$
  3. $- \frac{1}{2V} \frac{dn _A}{dt} = \frac{1}{3V} \frac{dn _B}{dt}$
  4. $- \frac{1}{V} \frac{n _A}{t} = \frac{1}{V} \frac{n _B}{t}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
If $n _A$ and $n _B$ are the number of moles at any instant in the reaction : $2A _{(g)} \rightarrow 3B _(g)$ carried out in a vessel of VL, the rate of the reaction at that instant is given by

$  \displaystyle  - \frac{1}{2} \frac{d[A]}{dt} =+ \frac{1}{3} \frac{d[B]}{dt}$

$ \displaystyle  - \frac{1}{2V} \frac{dn _A}{dt} =+ \frac{1}{3V} \frac{dn _B}{dt}$

Note: 
$  \displaystyle  [A]= \frac{n _A}{V} $
$  \displaystyle  [B]= \frac{n _B}{V} $
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

The decomposition of ${N} _{2}{O} _{5}$ in ${CCl} _{4}$ solution at 320 K takes place as ${2N} _{2}{O} _{5}\rightarrow{4NO} _{2}+{O} _{2}$; On the bases of given data order and the rate constant of the reaction is :
$\begin{matrix}Time\ in\ mitues&10&15&20&25&\infty\Valume of {O} _{2}&6.30&8.95&11.40&13.50&34.75\end{matrix}$
evolved (in mL)

  1. $1,0.198$ ${min}^{-1}$
  2. $3/2, 0.0198$ ${M}^{-1/2}$ ${min}^{-1}$
  3. $0, 0.0198$ $ {M}$ $ {min}^{-1}$
  4. $1, 0.0198$ $ {min}^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Decomposition of N2O5 is a known first-order reaction. The rate constant can be determined from the time-volume data using the first-order integrated rate equation.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

Consider the reaction  : 
$2H _2(g) + 2NO(g) \rightarrow\  N _2(g) + 2H _2O(g)$
The rate law for this reaction is :
$Rate = k[H _2][NO]^2$
Under what conditions could these steps represent the mechanism?
Step 1 : $2NO(g) \rightleftharpoons  N _2O _2(g)$
Step 2 : $N _2O _2  + H _2 \rightarrow\ N _2O + H _2O$
Step 3 : $N _2O + H _2 \rightarrow\ H _2O + N _2$

  1. These steps can never satisfy the rate law

  2. Step 1 should be the slowest step

  3. Step 2 should be the slowest step

  4. Step 3 should be the slowest step
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given reaction is:-

$2H _2(g)+2NO(g)\longrightarrow N _2(g)+2H _2O(g)$

The given rate law is:-
$Rate=K [H _2][NO]^2$

The rate of the chemical reaction is determined by the slowest step. So, in the slowest step we should have $2$ molecules of $NO$ and $1$ molecule of $H _2$ because the rate of the reaction is determined by that.

So, I. $2NO(g)+H _2(g)\longrightarrow N _2(g)+H _2O _2$ (slow)
      II. $H _2O _2+H _2(g)\longrightarrow 2H _2O(g)$ (fast)

This could be the mechanism of the reaction as given by rate law.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

In a first order reaction, the concentration of reactant, decrease from 0.8 M to 0.4 M in 15 minutes. The time taken for concentration to change from 0.1 M to 0.025 M is:

  1. 7.5 minutes

  2. 15 minutes

  3. 30 minutes

  4. 60 minutes
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Its a 1st order reaction,


$k = \dfrac{2.303}{t} log \dfrac{[A]}{[A - x]}$

So,
$k = \dfrac{2.303}{15} log \dfrac{[0.8]}{[0.4]}$

In the 2nd Case,
$k = \dfrac{2.303}{{t}^{1}} log \dfrac{[0.1]}{[0.025]}$

On substituting the value of k, We get
$t = 30\space min$

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

The decomposition of $N _{2}O _{5}$ in $CCI _{4}$ solution at 320 K takes place as
$2N _{2}O _{5} \rightarrow 4NO _{2} + O _{2}$; On the bases of given data order and the rate constant of the reaction is :

Time in minutes 10 15 20 25 $\infty$
Volume of $O _{2}$ evolved (in mL) 6.30 8.95 11.40 13.50 34.75
  1. 1,0.198 $min^{-1}$
  2. 3/2, 0.0198 $M^{-1/2} min^{-1}$
  3. 0,0.198 $M^{-1/2} min^{-1}$
  4. 1,0.0198 $min^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

In a reaction $2X \rightarrow Y$, the concentration of $X$ decreases from $3.0$ moles/ litre to $2.0\ moles/ litre$ in $5$ minutes. The rate of reaction is :

  1. $0.1\ mol\ L^{-1} min^{-1}$
  2. $5\ mol\ L^{-1} min^{-1}$
  3. $1\ mol\ L^{-1} min^{-1}$
  4. $0.5\ mol\ L^{-1} min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Rate = -\dfrac {1}{2} \dfrac {\triangle [X]}{\triangle t}$
$= -\dfrac {1}{2} \dfrac {(3 - 2)}{5} = -0.1\ mol\ L^{-1} min^{-1}$
Negative sign signifies the decrease in concentration.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

The rate law for a reaction, $A + B \rightarrow C + D$ is given by the expression $k[A]$. The rate of reaction will be:

  1. doubled on doubling the concentration of $B$
  2. halved on reducing the concentration of $A$ to half

  3. decreased on increasing the temperature of the reaction

  4. unaffected by any change in concentration of temperature
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rate of reaction- The speed at which a chemical reaction proceeds, It is often expressed in terms of either the concentration (amount per unit volume) of a product that is formed in a unit of time or the concentration of a reactant that is consumed in a unit of time.

other terms of expression are produced or consumed $\dfrac{mol}{time}$ and in case of gas we can use pressure term also.
So the correct option is $[B]$

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

Which of the following expressions is correct for the rate of reaction given below?


$5Br^{-} _{(aq)} + BrO _{3(aq)}^{-} + 6H^{+} _{(aq)} \rightarrow 3Br _{2(aq)} + 3H _{2}O _{(l)}$

  1. $\dfrac {\triangle [Br^{-}]}{\triangle t} = 5\dfrac {\triangle [H^{+}]}{\triangle t}$
  2. $\dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {6}{5}\dfrac {\triangle [H^{+}]}{\triangle t}$
  3. $\dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {5}{6}\dfrac {\triangle [H^{+}]}{\triangle t}$
  4. $\dfrac {\triangle [Br^{-}]}{\triangle t} = 6\dfrac {\triangle [H^{+}]}{\triangle t}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the reaction,


$5Br^{-} _{(aq)} + BrO _{3(aq)}^{-} + 6 _{(aq)}^{+} \rightarrow 3Br _{2(aq)} + 3H _{2}O _{(l)}$


Rate of disappearance $= \dfrac {-1}{5}\dfrac {\triangle [Br^{-}]}{\triangle t} = -\dfrac {\triangle [BrO _{3}^{-}]}{\triangle t} = \dfrac {-1}{6} \dfrac {\triangle [H^{+}]}{\triangle t}$

$ \dfrac {\triangle [Br^{-}]}{\triangle t} = \dfrac {5}{6} \dfrac {\triangle [H^{+}]}{\triangle t}$.

Hence, the correct answer is option $\text{C}$.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

The rate of a gaseous reaction is given by the expression $k[A]^{2}[B]^{3}$. The volume of the reaction vessel is reduced to one half of the initial volume. What will be the reaction rate as compared to the original rate $a$?

  1. $\dfrac {1}{8}a$
  2. $\dfrac {1}{2}a$
  3. $2a$
  4. $32a$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Rate = k[A]^{2} [B]^{3} = a$


When volume is reduced to one half then conc. of reactants will be doubled.

$Rate = k[2A]^{2}[2B]^{2}$

         $=32 k[A]^{2} [B]^{3} = 32a$.

So, the correct option is $D$

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

In a reaction, $2X \rightarrow Y$, the concentration of $X$ decreases from $0.50\ M$ to $0.38\ M$ in $10\ min$. What is the rate of reaction in $M\ s^{-1}$ during this interval?

  1. $2\times 10^{-4}$
  2. $4\times 10^{-2}$
  3. $2\times 10^{-2}$
  4. $1\times 10^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rate of reaction $= \dfrac {\triangle [X]}{\triangle t}$


$\triangle [X] = X _{i} - X _{f} = 0.50 - 0.38 = 0.12\ M$

$Rate = \dfrac {0.12}{10\times 60} = 2\times 10^{-4} M\ s^{-1}$.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

The rate equation for a reaction is r =  $K[A]^{\circ}[B]^3$. Which of the following statements are true?

  1. Doubling the concentration of B quadruples the rate of reaction
  2. The units of rate constant are mole$^{-2} L^2 S^{-1}$

  3. The plot of concentration of A Vs time is parallel to the time axis
  4. If the volume of the reaction vessel is decreased to $\frac{1}{3}$, the rate of reaction is $ \frac{1^th}{27}$ of the original rate
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$r= K[A]^{o} $ $[B]^3$

The reaction is zero order with respect $A$  and order with respect to $B$ is 3. The overall order being 3, the units of the rate constant are mole $^{-2}$l$^2$ s$^{-1}$. The rate of reaction does not change with a change in concentration of $A$. Therefore, the plot of concentration of $A$ vs time is parallel to a time axis. If the volume of the reaction vessel is decreased to $\frac{1}{3}$, the rate of reaction is 27 times of the original rate.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

The reaction $A(g)+2B(g)\rightarrow C(g)+D(g)$ is an elementary process. In an experiment, the initial partial pressure of $A$ and $B$ are $P _A=0.6$ and $P _B=0.8$ atm when $P _C=0.2$ atm the rate of reaction relative to the initial rate is:

  1. $\dfrac{1}{48}$
  2. $\dfrac{1}{24}$
  3. $\dfrac{9}{16}$
  4. $\dfrac{1}{16}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$ \implies \space \space \space A(g) \space \space \space \space \space \space+ \space \space \space \space \space \space 2B(g) \rightarrow \space \space \space \space \space \space \space \space C(g) + D(g) $
$ At \space t=0 \space0.6 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0.8 \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space 0 \space \space \space \space \space \space \space \space \space \space \space \space \space 0 $
$ At \space t=t \space (0.6-0.2) \space \space \space \space (0.8 - 2 \times 0.2) \space \space 0.2 \space \space \space \space \space \space \space \space \space \space0.2$

$ Rate _i = [0.6][0.8]^2$
$ Rate _t = [0.4][0.4]^2$
$ \dfrac{Rate _t}{Rate _i} = \dfrac{1}{6}$

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

Two gases A and B are filled in a container. The experimental rate law for the reaction for the reaction between them has been found to be $Rate = k [A]^2 [B]$. Predict the effect on the rate of the reaction when pressure is doubled?

  1. The rate is doubled

  2. The rate becomes four times

  3. The rate becomes six times

  4. The rate becomes eight times
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

If, $Rate=kx[A]^2[B]^1$

$order=3$.

If pressure is increased by factor of $2$, then rate will be increased by factor of $2^3=8$.

$\therefore $ Rate becomes eight times.

Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

For the reaction A + B $\rightarrow$ products, it is observed that :-
(a) on doubling the initial concentration of A only, the rate of reaction is also doubled and 
(b) on doubling the initial concentrations of both A and B, there is a change by a factor of 8 in the rate of the reaction.

  1. rate = k[A][B]
  2. rate = $k[A]^2$[B]
  3. rate = k[A]$[B]^2$
  4. rate = k$[A]^2[B]^2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The given reaction is $A+B\longrightarrow$ Product

Let us suppose the rate law of this reaction is:-
$Rate=K[A]^a[B]^b$
where $K$ is a rate constant.
$a$ and $b$ are order of the reaction with respect to the reactants $A$ and $B$ respectively.
Given that,
When [A] is doubled, the rate of the reaction is also doubled, so the reaction is first order $w.r.t. A$ and hence $a=1$
When $[A],[B]$ is doubled, the rate of reaction becomes $8$ times. Now,
$(Rate) _{new}=K [2A]^1[2B]^b$          $- (ii)$
$Rate=K[A]^1[B]^b$          $-(iii)$
Now, $\because$ New rate of reaction is $8$ times, so dividing $(ii)$ by $(iii)$ :-
$\Rightarrow 8=2.2^b$
$2^3=2^{1+b}$
Equating the exponents:-
$\Rightarrow 3=1+b\Rightarrow b=2$
So, order of reaction $w.r.t$ to $B$ is $2$
So, $Rate=K[A][B]^2$