Chemistry

Chemical Kinetics

276 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

For the reaction $2A + B + C \rightarrow 2D$. The observed rate law is Rate=$K[A]{ [B] }^{ 2 }$. Correct statements are
a) An increase of cone .of C does not affect the rate
b)Doubling the conc of A doubles the rate
c)Tripling the conc of B increases the rate by 9 times
d)Doubling the conc of C, doubling the rate

  1. a,b,c

  2. b,c,d

  3. d

  4. c,d
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given  rate is $k[A]{ [B] }^{ 2 }$

 It is overall ${ 3 }^{ rd }$ order  reaction with a following  first order  and B following second order.
 (a) True  C is not affecting the rate of reaction.
 (b) True $rate _{ 1 }\quad =k[A]{ [B] }^{ 2 }\ rate _{ 2 }\quad =k[2A]{ [B] }^{ 2 }=2k[A]{ [B] }^{ 2 }\ =2\times rate _{ 1 }$
 (c) True $rate _{ 1 }\quad =k[A]{ [B] }^{ 2 }\ rate _{ 3 }=k[A]{ [3B] }^{ 2 }=9Kk[A]{ [B] }^{ 2 }\ =9\times rate _{ 1 }$
(d)False C does not affected  the rate in any manner.

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

${ SO } _{ 2 }$ react with ${ O } _{ 2 }$ as follows :


 ${ 2SO } _{ 2 }+{ O } _{ 2 }\rightarrow { 2SO } _{ 3}$ 


The rate of disappearance of ${ SO } _{ 2 }$ is $2.4\times { 10 }^{ -4 }$ mol ${ lit }^{ -1 }{ min}^{ -1 }$, then :

  1. Rate of reaction is $1.2\times { 10 }^{ -4 }\quad mole{ lit }^{ -1\quad }{ min }^{ -1 }$
  2. Rate of appearance of ${ SO } _{ 3 }$ is $2.4\times { 10 }^{ -4 } { mole\quad lit }^{ -1 }min^{ -1 }$
  3. Rate of disappearance of ${ O } _{ 2 }$ is $1.2\times { 10 }^{ -4 } { mole\quad lit }^{ -1 }min^{ -1 }$
  4. Rate of reaction is twice the rate of disappearance of ${ SO } _{ 2 }$
Reveal answer Fill a bubble to check yourself
B,C,D Correct answer
Multiple choice chemistry rates of reaction introduction to rate of reaction rate of chemical reaction rate of reaction

Rate of formation of $SO _{3}$ according to the reaction $2SO _{2}+O _{2} \rightarrow 2SO _{3}$ is $1.6 \times 10^{-3}\ kg\ min^{-1}$ Hence rate at which $SO _{2}$ reacts is :-

  1. $1.6 \times 10^{-3}\ kg\ min^{-1}$
  2. $8.0 \times 10^{-4}\ kg\ min^{-1}$
  3. $3.2 \times 10^{-3}\ kg\ min^{-1}$
  4. $1.28\times 10^{-3}\ kg\ min^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to the balanced chemical equation 2SO2 + O2 -> 2SO3, the stoichiometric coefficients for SO2 and SO3 are both 2, meaning they react and form in a 1:1 molar ratio. Since their molar masses are different, we account for them: rate of SO2 consumption = rate of SO3 formation * (Molar Mass of SO2 / Molar Mass of SO3). Molar mass of SO2 is 64 g/mol and SO3 is 80 g/mol. Thus, rate = 1.6 * 10^-3 * (64 / 80) = 1.28 * 10^-3 kg min^-1.

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

$C _{4}H _{8}\rightarrow 2C _{2}H _{4}$; rate constant $=2.303\times 10^{4}\sec^{-1}$, After what time the molar ratio of $\dfrac{C _{2}H _{4}}{C _{4}H _{8}}$ attain the value $1$

  1. $176\ sec$
  2. $3522\ sec$
  3. $1661\ sec$
  4. $1761\ sec$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the reaction C4H8 -> 2C2H4, if the ratio [C2H4]/[C4H8] = 1, then [C2H4] = [C4H8]. Using the integrated rate law for a first-order reaction, we solve for t.

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

On increasing the pressure three fold, the rate of reaction of ${ 2H } _{ 2 }{ S }$ + ${ O } _{ 2 }$ $\rightarrow $ products would increase

  1. 3 times

  2. 9 times

  3. 12 times

  4. 27 times
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Rate$=$ ${K _{{p^2}}}{H _2}S \times P{O _2} = x$ $($rate$)$


On increasing the pressure three fold :


Rate$=$ $K{\left( {3{P _{{H _2}S}}} \right)^2}\left( {3P{O _2}} \right)$

$=$ $K \times 9{P^2} _{{H _2}S} \times 3P{O _2}$

$=$ $K \times 27 \times {P _{{H _2}S}} \times P{O _2} = 27$

the rate will increases $27$ times 

Hence, option $(D)$ is correct answer.

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

Which of the following is not a valid way to describe the rate of the following reaction?
$A + B + C \rightarrow D + E$

  1. $\dfrac {-\triangle [A]}{\triangle t}$
  2. $\dfrac {-\triangle [B]}{\triangle t}$
  3. $\dfrac {-\triangle [C]}{\triangle t}$
  4. $\dfrac {-\triangle [D]}{\triangle t}$
  5. $\dfrac {-\triangle [E]}{\triangle t}$
Reveal answer Fill a bubble to check yourself
D,E Correct answer
Explanation

$A+B+C\longrightarrow D+E$

 Rate of reaction is defined as the change in concentration of reactant or product to time.
 Rate $ (R)=\cfrac { -d[A] }{ dt } =\cfrac { -d[B] }{ dt } =\cfrac { -d[C] }{ dt } =\cfrac { d[D] }{ dt } =\cfrac { d[E] }{ dt } $
Therefore, (D)  &  (E) i.e. $ \cfrac { -d[D] }{ dt } & \quad \cfrac { -d[E] }{ dt } $ respectively are not valid ways of describing the rate of the following reaction.

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

The rate constant of the relation $ A \rightarrow B $ is $ 0.6 \times 10^{-3} $ mole per second. If the concentration of $B$ after $20$ minutes is :

  1. $0.36$ M
  2. $0.72$ M
  3. $1.08$ M
  4. $3.60$ M
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By unit of rate constant it is clear,that the $reaction$ is zero order. 

$\therefore \left [ B \right ]=k\times t$
         $=0.72M$

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

The rate law for a reaction between the substances $A$ and $B$ is given by rate$=k{ \left[ A \right]  }^{ n }{ \left[ B \right]  }^{ m }$. On doubling the concentration of $A$ and having the concentration of $B$ halved, the ratio of the new rate to the earlier rate of the reaction will be as:

  1. $\cfrac { 1 }{ { 2 }^{ m+n } } $
  2. $(m+n)$
  3. $(n-m)$
  4. ${2}^{(n-m)}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that

                    $R=K[A]^n[B]^m$
after doubling the concentration of $A$ and concentration of $B$ is halfed 
$R^1=K[2A]^n[\dfrac{B}{2}]^m$
$R^1 =(2)^{n-m} R$
$ \dfrac{R^1}{R}= \dfrac{2^{n-m}}{1}$

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

The rate equation for the reaction $2A+B \rightarrow C$ is found to be rate = $k[A] [B]$. The correct statement in relation to this reaction is that the :

  1. units of $k$ must be$\ mol^{-1} L$ $s^{-1}$.
  2. $t _{1/2}$ is constant

  3. rate of formation of C is twice the rate of disappearance of A
  4. value of $k$ is independent of the initial concentration of A and B
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The given reaction is $2A+B\longrightarrow C$

The given rate equation is:-
$rate=K[A] [B]$

The unit of rate is $mol L^{-1} s^{-1}$
Unit of $[A]= mol L^{-1}$
Unit of $[B]= mol L^{-1}$

Unit of $K$=$\cfrac {mol L^{-1} s^{-1}}{mol L^{-1} mol L^{-1}}$
$=mol^{-1} L$ $s^{-1}$.

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 in volving this reaction. The initial pressure of A and B are $P _A=0.6$ atm $P _B=0.8$atm respectively when $P _C=0.2$ atm, the rate of reaction relative to the initial rate is:

  1. $\displaystyle\frac{1}{6}$
  2. $\displaystyle\frac{1}{12}$
  3. $\displaystyle\frac{1}{36}$
  4. $\displaystyle\frac{1}{18}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
        $A(g) + 2(B) \rightarrow C(g) + D(g)$
 t = 0  0.6         0.8              0     0
 at t     0.6-x      0.8-x          x       x
since this the elementary reaction
rate,r = $K[B]^2 [A]$
now $r _i = k (0.6)(0.8)^2 = 0.38K$
when $P _i = x - 0.2$ atm
when $P _A= 0.6-x =0.4$ atm
when $P _B= 0.8 - 2x =0.4$ atm
$r _f = K(0.4) (0.4)^2 = 0.064K$
$r _1/r _2 = 0.064/0.384 = 1/6$
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
$t = 30\space min$