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 chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

Which of the following statement is/are correct ?

  1. The rate of the reaction involving the conversion of ortho-hydrogen to parahydrogen is $\displaystyle -\, \frac{d[H _2]}{dt}\, =\, k[H _2]^{3/2}$
  2. The rate of the reaction involving the thermal decomposition of acetaldehyde is $k[CH _3CHO]^{3/2}$
  3. In the formation of phosgene gas from CO and $Cl _2$, the rate of the reaction is $k[CO][Cl _2]^{1/2}$
  4. In the decomposition of $H _2O _2$, the rate of the reaction is $k[H _2O _2]$.
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

(A) The rate of the reaction involving the conversion of ortho-hydrogen to parahydrogen is $\displaystyle -\, \frac{d[H _2]}{dt}\, =\, k[H _2]^{3/2}$
The order of the reaction is 1.5.
(B) The rate of the reaction involving the thermal decomposition of acetaldehyde is $k[CH _3CHO]^{3/2}$
The order of the reaction is1.5.
(C) In the formation of phosgene gas from CO and $Cl _2$, the rate of the reaction is $k[CO][Cl _2]^{1/2}$
The order of the reaction is 1.5.
(D) In the decomposition of $H _2O _2$, the rate of the reaction is $k[H _2O _2]$.
The order of the reaction is 1.

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

The inversion of a sugar follows first-order rate equation which can be followed by noting the change in the rotation of the plane of polarization of light in the polarimeter. If $r _{\propto},\, r _{\zeta}$ and $r _0$ are the rotations at $t\, =\, \propto$, t = t, and t = 0, then the first order reaction can be written as:

  1. $\displaystyle k\, =\, \frac{1}{t}\, log\, \frac{r _{1}\, -\, r _{\propto}}{r _{0}\, -\, r _{\propto}}$
  2. $\displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{0}\, -\, r _{\propto}}{r _{1}\, -\, r _{\propto}}$
  3. $\displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{\propto}\, -\, r _{0}}{r _{\propto}\, -\, r _{1}}$
  4. $\displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{\propto}\, -\, r _{1}}{r _{\propto}\, -\, r _{0}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a first order reaction $\displaystyle A \rightarrow P$, the expression for the rate constant is
$\displaystyle \displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{ a}{a-x}$
Here, A is reactant, P is product, a is the initial concentration of A and $a-x$ is the concentration of A at time t.

The inversion of a sugar follows first order rate equation which is given below.
$\displaystyle \displaystyle k\, =\, \frac{1}{t}\, ln\, \frac{r _{0}\, -\, r _{\infty}}{r _{t}\, -\, r _{\infty}}$
Here, $\displaystyle a = r _{0}\, -\, r _{\infty}$ and $\displaystyle a-x = r _{t}\, -\, r _{\infty}$

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

Inversion of a sugar folllows first order rate equation which can be followed by nothing the change in rotation of the plane of polarization of light in the polarimeter. If $r _{\infty,}:r _t:and:r _0$ are the rotations at
 $t\,=\,\infty,t\,=\,t:and:t\,=\,0,$ then, first order reaction can be written as:

  1. $\;k\,=\,\displaystyle\frac{1}{t}log\displaystyle\frac{r _t-r _{\infty}}{r _0-r _{\infty}}$
  2. $\;k\,=\,\displaystyle\frac{1}{t}\,ln\displaystyle\frac{r _0-r _{\infty}}{r _t-r _0}$
  3. $\;k\,=\,\displaystyle\frac{1}{t}\,ln\displaystyle\frac{r _{\infty}-r _0}{r _{\infty}-r _t}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\underset{d-Sucrose}{C _{12}H _{22}O _{11}}+H _2O\xrightarrow{H+}\underset{d-Glucose}{C _6H _{12}O _6}+\underset{l-Fructose}{C _6H _{12}O _6}$


Initially               a                Excess                  0                0 
After time t        a-x            Constant                x                x
At infinity           0               Constant               a                 a
If $r _0,r _t$ and $r _{\infty}$ be the observed angle of rotations of the sample at zero times, times $t$ and infinity respectively, and $k _1,k _2$ and $k _3$ proportionate in terms of sucrose,glucose and fructose, respectively.
Then,
$r _0=k _1a$
$r _t=k _1(a-x)+k _2x+k _3x$
$r _{\infty}=k _2a+k _3a$
From these equations it can be shown that
$\dfrac{a}{a-x}=\dfrac{r _0-r _{\infty}}{r _t-r _{\infty}}$
So, the expression for the rate constant rate of this reaction in terms of the optical rotational data may be 
put as $k=\dfrac{2.303}{t}\log \dfrac{r _0-r _{\infty}}{r _t-r _{\infty}}$

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

In the following reaction $2H _2O _2\rightarrow2H _2O+O _2$ rate of formation of $O _2$ is 3.6 M min$^{-1}$.


(a) What is rate of formation of $H _2O\ ?$        
(b) What is rate of disappearance of $H _2O _2$?

  1. (i) $7.2$ mol litre$^{-1}$ min$^{-1},$ (ii) $7.2$ mol litre$^{-1}$ min$^{-1}$
  2. (i) $3.6$ mol litre$^{-1}$ min$^{-1},$ (ii) $3.6$ mol litre$^{-1}$ min$^{-1}$
  3. (i) $14.4$ mol litre$^{-1}$ min$^{-1},$ (ii) $14.4$ mol litre$^{-1}$ min$^{-1}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ (a)2H _{2}O _{2}\rightarrow 2H _{2}O+O _{2} $

We know, $ \dfrac{-1}{2}\dfrac{d[H _{2}O _{2}]}{dt} = \dfrac{-1}{2}\dfrac{d[H _{2}O]}{dt} = \dfrac{d[O _{2}]}{dt} $

$ \dfrac{d[H _{2}O]}{dt} = 2\dfrac{d[O _{2}]}{dt} = 2\times 3.6\,M\,min^{-1} $

$ \dfrac{d[H _{2}O]}{dt} = 7.2\,M\,min^{-1} $

$(b) \dfrac{-d[H _{2}O _{2}]}{dt} = \dfrac{2}{2}\dfrac{d[H _{2}O]}{dt} = \dfrac{-d[H _{2}O _{2}]}{dt} = 7.2\,M\,min^{-1} $ 

Option A is correct.
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

Derive an expression for the Rate (k) of reaction :
$2N _{2}O _{5}(g)\rightarrow 4NO _{2}(g)+O _{2}(g)$


With the help of following mechanism:

$N _{2}O _{5}\overset{K _a}{\rightarrow}NO _{2}+NO _{3}$
$NO _{3}+NO _{2}\overset{K _{-a}}{\rightarrow}N _{2}O _{5}$
$NO _{2}+NO _{3}\overset{K _b}{\rightarrow}NO _{2}+O _{2}+NO$
$NO+NO _{3}\overset{K _c}{\rightarrow}2NO _{2}$

  1. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+2k _{b}}[N _{2}O _{5}]$
  2. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}-2k _{b}}[N _{2}O _{5}]$
  3. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+k _{b}}[N _{2}O _{5}]$
  4. $\displaystyle Rate=\frac{k _{a}\times k _{b}}{2k _{-a}-2k _{b}}[N _{2}O _{5}]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Rate $\displaystyle = k _b[NO _2][NO _3] $ .....(1)

But $\displaystyle \dfrac {[NO _2][NO _3]}{[N _2O _5]}=  \dfrac {K _a}{K _{-a} + 2k _b}$

Hence $\displaystyle [NO _2][NO _3]  =\dfrac {K _a}{K _{-a}+2k _b} [N _2O _5]$......(2)

Substitute equation (2) in equation (1):

$\displaystyle \displaystyle Rate=\frac{k _{a}\times k _{b}}{k _{-a}+2k _{b}}[N _{2}O _{5}]$

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

The rate constant for the reaction, ${ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 2N{ O } _{ 2 }\left( g \right) +\dfrac { 1 }{ 2 } { O } _{ 2 }\left( g \right) $, is $2.3\times { 10 }^{ -2 }\ { sec }^{ -1 }$. Which equation given below describes the change of $\left[ { N } _{ 2 }{ O } _{ 5 } \right] $ with time, ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }$ and ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }$ corresponds to concentration of ${ N } _{ 2 }{ O } _{ 5 }$ initially and time $t$ respectively?

  1. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }{ e }^{ kt }$
  2. $\log _{ e }{ \dfrac { { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 } }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t } } } =kt$
  3. $\log _{ 10 }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t } } =\log _{ 10 }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 } } -kt$
  4. ${ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right] } _{ 0 }+kt$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

${ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }{ e }^{ -kt }\ { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 }={ \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t }{ e }^{ kt }\ \ln { \left( \cfrac { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 } }{ { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t } }  \right)  } ={ e }^{ kt }\ \ln { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ t } } =\ln { { \left[ { N } _{ 2 }{ O } _{ 5 } \right]  } _{ 0 } } -kt$

Multiple choice enzymes and catalysts catalysis adsorption and colloids surface chemistry chemistry adsorption theory of heterogeneous catalysis

In presence of a catalyst, the activation energy is lowered by 3 kcal at $27^o C$. Hence, the rate of reaction will increase by:

  1. 32 times

  2. 243 times

  3. 3 times

  4. 48 times
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$K={Ae}^{\dfrac{-Ea}{RT}}$ 

So Ratio, ${\dfrac{K _2}{K _1}}={e^\dfrac{(Ea _2-Ea _1)}{RT}}$
$\dfrac{K _2}{K _1}$=$e^{\dfrac{(3\times1000)}{(8.314\times300)}}$=$e^{{3000}/{2494}}$ = $e^{1.2}$ = $3.3$
Nearly 3 times.

Multiple choice chemistry nitrogen and sulfur nitrogen gas component of air - nitrogen nitrogen

In the reaction of $2NO+{ O } _{ 2 }\rightarrow 2{ NO } _{ 2 }$, if the rate of disappearance of ${ O } _{ 2 }$ is $16gm$ ${ min }^{ -1 }$, then the rate of apperance of ${ NO } _{ 2 }$ is:

  1. $90\ gm\ { min }^{ -1 }$
  2. $46\ gm\ { min }^{ -1 }$
  3. $28\ gm\ { min }^{ -1 }$
  4. $32\ gm\ { min }^{ -1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$2NO+{ O } _{ 2 }\rightarrow 2N{ O } _{ 2 }$
$-\cfrac { 1 }{ 2 } \cfrac { d\left[ NO \right]  }{ dt } =\cfrac { -d\left[ { O } _{ 2 } \right]  }{ dt } =\cfrac { 1 }{ 2 } \cfrac { d\left[ N{ O } _{ 2 } \right]  }{ dt } $
$\cfrac { -d\left[ { O } _{ 2 } \right]  }{ dt } =16ℊ{ min }^{ -1 }=\cfrac { 16 }{ 16 } mol{ min }^{ -1 }=1mol{ min }^{ -1 }$
$\cfrac { d\left[ N{ O } _{ 2 } \right]  }{ dt } =2\times \cfrac { -d\left[ { O } _{ 2 } \right]  }{ dt } =2\times 1=2mol/min$
No. of moles$=\cfrac { weight }{ mol.wt } $
$2\times $Mol. wt of $N{ O } _{ 2 }=$wt
$2\times 46=$ wt $\Rightarrow $Weight of $N{ O } _{ 2 }=92ℊ$
Rate of appearance of $N{ O } _{ 2 }=92ℊ/min\approx 90ℊ/min$
Therefore, option $A$ is correct.
Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Rate constant in case of first order reaction is :

  1. Inversely proportional to the concentration units

  2. Independent of concentration units

  3. Directly proportional to concentration units

  4. Inversely proportional to the square of concentration units
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For First order of reaction,
$Rate= k [A]$,


$k = \cfrac {mol/L}{sec\times {mol/L}}=sec^{-1}$

Option B is correct.

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Fill in the blanks by choosing the correct option;
Order of the reaction is the $X$ of the powers to which concentration terms are raised in experimentally determined rate equation. The unit of first order rate constant is $Y$. The unit of first order rate constant when concentration is measured in terms of pressure and time in minutes is $Z$.

  1. $X\rightarrow product, Y\rightarrow mol\ L^{-1} time^{-1}, Z\rightarrow atm\ min^{-1}$
  2. $X\rightarrow sum, Y\rightarrow L\ mol^{-1}time^{-1}, Z\rightarrow atm\ min^{-1}$
  3. $X\rightarrow product, Y\rightarrow L\ mol^{-1}, Z\rightarrow atm\ min^{-1}$
  4. $X\rightarrow sum, Y\rightarrow time^{-1}, Z \rightarrow min^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The order of a chemical reaction is defined as the sum of the powers of the concentration of the reactants in the rate equation of that particular chemical reaction.


General formula for the unit of rate constant = mole$^{(1-n)}$ L$^{(n-1)}$ min$^{-1}$.

For 1$^{st}$ order reaction, 
n = 1
Therefore, unit of rate constant $=$ min$^{-1}$

The unit of rate constant in terms of pressure and time:
  mol $^{( 1-n)}$  L $^{ (n-1)}$ min$^{ -1 }$ or  atm$^{( 1-n) }$ min $^{ -1 }$


For 1$^{st}$ order reaction, 
n = 1
Therefore, the unit is min$^{-1}.$

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

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Match the rate law given in column I with the dimensions of rate constant given in column II and mark the appropriate choice.

Column I Column II
(A) $Rate = k[NH _{3}]^{0}$ (i) $mol\ L^{-1} s^{-1}$
(B) $Rate = k[H _{2}O _{2}][I^{-}]$ (ii) $L\ mol^{-1} s^{-1}$
(C) $Rate = k[CH _{3}CHO]^{3/2}$ (iii) $s^{-1}$
(D) $Rate = k[C _{2}H _{5}Cl]$ (iv) $L^{1/2} mol^{-1/2} s^{-1}$
  1. $(A)\rightarrow (iv), (B) \rightarrow (iii), (C)\rightarrow (ii), (D) \rightarrow (i)$
  2. $(A)\rightarrow (i), (B) \rightarrow (ii), (C)\rightarrow (iii), (D) \rightarrow (iv)$
  3. $(A)\rightarrow (ii), (B) \rightarrow (i), (C)\rightarrow (iv), (D) \rightarrow (iii)$
  4. $(A)\rightarrow (i), (B) \rightarrow (ii), (C)\rightarrow (iv), (D) \rightarrow (iii)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
(A) $Rate= k[NH _3]^0$
It is zero order reaction.
$\therefore$ Units of rate constant are same as rate.
i.e $ mol\ l^{-1}s^{-1}$

(B) $Rate=k[H _2O _2]^1[I^-]^1$
It is second order reaction as order $=1+1=2$
$\therefore$ Units of Rate constant are $L \ mol^{-1}s^{-1}$

(C) $Rate=k[CH _3CHO]^{3/2}$
It is fractional order reaction with order $= \cfrac 32$
$\therefore$ Units of rate constant are $L^{\cfrac 12}mol^{-\cfrac 12}s^{-1}$

(D) $Rate=k[C _2H _5Cl]^{-1}$
It is first order reaction. So units of rate constant are $s^{-1}$

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

For the second order reaction, concentration $(x)$ of the product at time $t$ starting with initial concentration $[A] _0$ is:

  1. $\dfrac{kt[A _0]^2}{1 + kt[A _0]}$
  2. $\dfrac{k + [A _0]^2}{1 + kt}$
  3. $\dfrac{1 + kt[A _0]^2}{k + [A _0]^2}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A reaction said to be a second order when the overall order is $2$. The rate of second order reaction may be proportional to one concentration squared.

$R=K[A]^2$
For rate proportional to single concentration squared, the time dependance of concentration is given by
$\cfrac{1}{[A]}=\cfrac{1}{A _0}+Kt
Therefore, concentration of product after time $t=\cfrac{kt[A_0]^2}{1+kt[A_0]}$.

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Units of rate constant of a first order reaction is :

  1. $mole.lit^{-1}$
  2. $lit. mole$
  3. $mole. sec^{-1}$
  4. $sec^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A $\rightarrow$ product

For first order reaction, rate is dependent on single reactant A for example, rate = k[A]
$k=\frac{rate}{[A]}$

$=\frac{mole}{liter}sec\times \frac{liter}{mole}$
$=sec^{-1}$

Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

Consider the reaction $2A+B$ $\rightarrow$products,when the concentration of a alone was doubled, the half-life of the  reaction did not change.When the concentration of B alone was double,the rate was not altered.The unit of rate constant for this reaction is

  1. $S^{-1}$
  2. $L\ mol^{-1}\ s^{-1}$
  3. $mol\ L^{-1}\ s^{-1}$
  4. $mol^{-2}\ L^{5}\ S^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry chemical kinetics dependence of reaction rate on concentration of reactants order of reactions factors influencing rate of a reaction

For the reaction $A\rightarrow C+D$, the initial concentration of $A$ is $1000 M$. After $10^{2} sec$ concentration of $A$ is $100\ M$. The rate constant of the reaction has the numerical value of $9.0$. What is the unit of the reaction rate constant? 

  1. $M^{-1}s^{-1}$
  2. $Ms^{-1}$
  3. $s^{-1}$
  4. $M^{-1.5}s^{-1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The reaction is zero order because the rate of change of concentration is constant (1000M to 100M in 100s, rate = 9 M/s). The rate constant k for a zero-order reaction has units of concentration per time, which is M s^-1.