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

The half life of decomposition of $N _2O _5$ is a first order reaction represented by
$N _2O _5\, \rightarrow\, N _2O _4\, =\, 1/2O _2$
After 15 min the volume of $O _2$ produced is $9mL$ and at the end of the reaction $35 mL$. The rate constant is equal to :

  1. $\displaystyle \frac{1}{15}\, log\frac{35}{26}$
  2. $\displaystyle \frac{1}{15}\log\frac{44}{26}$
  3. $\displaystyle \frac{1}{15}\, log\frac{35}{36}$

  4. None of the above

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle k\, =\, \frac{2.303}{t}\, log\, \frac{V _{\infty}}{V _{\infty}\, -\, V _t}$

$\displaystyle =\, \frac{1}{t}\, log _e\, \frac{V _{\infty}}{V _{\infty}\, -\, V _t}$

$\displaystyle \frac{1}{15}\, log _e\, \frac{35mL}{(35\, -\, 9)\, mL}\, =\, \frac{1}{15}\, log _e\, \frac{35}{26}$

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

The rate constant $k$, for the reaction
${N} _{2}{O} _{5}(g) \longrightarrow 2{NO} _{2}(g)+\cfrac{1}{2}{O} _{2}(g)$
is $1.3\times {10}^{-2}{s}^{-1}$. Which equation given below describes the change of $[{N} _{2}{O} _{5}]$ with time?
${[{N} _{2}{O} _{5}]} _{0}$ and ${[{N} _{2}{O} _{5}]} _{t}$ correspond to concentration of ${N} _{2}{O} _{5}$ initially and at time $t$.

  1. ${[{N} _{2}{O} _{5}]} _{t}={[{N} _{2}{O} _{5}]} _{0}+kt$
  2. ${[{N} _{2}{O} _{5}]} _{0}={[{N} _{2}{O} _{5}]} _{t}{e}^{kt}$
  3. $\log{{[{N} _{2}{O} _{5}]} _{t}}=\log{{[{N} _{2}{O} _{5}]} _{0}}+kt$
  4. $\ln{\cfrac{{[{N} _{2}{O} _{5}]} _{0}}{{[{N} _{2}{O} _{5}]} _{t}}}=kt$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the unit of rate constant is ${sec}^{-1}$, the reaction is first order reaction. 

${N} _{2}{O} _{5}(g) \longrightarrow 2{NO} _{2}(g)+\cfrac{1}{2}{O} _{2}(g)$
$k{t}=\ln{\cfrac{a}{(a-x)}}$ 
$kt=\ln{\cfrac { { [{ N } _{ 2 }{ O } _{ 5 }] } _{ 0 } }{ { [{ N } _{ 2 }{ O } _{ 5 }] } _{ t } } }$

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

For the reaction, $2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2}$ the rate of reaction is:

  1. $\cfrac{1}{2}\cfrac{d}{dt}[{N} _{2}{O} _{5}]$
  2. $2\cfrac{d}{dt}[{N} _{2}{O} _{5}]$
  3. $\cfrac{1}{2}\cfrac{d}{dt}[{NO} _{2}]$
  4. $4\cfrac{d}{dt}[{NO} _{2}]$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the reaction,  $\displaystyle 2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2} $  the rate of reaction is  $\displaystyle \cfrac{1}{2}\cfrac{d}{dt}[{NO} _{2}]$


 Rate of reaction $\displaystyle -\cfrac{1}{2}\cfrac{d[{N} _{2}{O} _{5}]}{dt}=\cfrac{1}{4}\cfrac{d[{NO} _{2}]}{dt}$

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,

$2{N} _{2}{O} _{5}\longrightarrow 4{NO} _{2}+{O} _{2}$ is $3.0\times {10}^{-4}{s}^{-1}$.

 If start made with $1.0$ $mol$ ${L}^{-1}$ of ${N} _{2}{O} _{5}$, calculate the rate of formation of ${NO} _{2}$ at the moment of the reaction when concentration of ${O} _{2}$ is $0.1mol$ ${L}^{-1}$ :

  1. $2.7\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
  2. $2.4\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
  3. $4.8\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
  4. $9.6\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Mol$ ${L}^{-1}$ of ${N} _{2}{O} _{5}$ reacted $=2\times 0.1=0.2$

$[{N} _{2}{O} _{5}]$ left $=1.0-0.2=0.8mol$ ${L}^{-1}$
Rate of reaction $=k\times [{N} _{2}{O} _{5}]$
$=3.0\times {10}^{-4}\times 0.8$
$=2.4\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$
Rate of formation of ${NO} _{2}$
$=4\times 2.4\times {10}^{-4}=9.6\times {10}^{-4}mol$ ${L}^{-1}{s}^{-1}$

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

$H _2O _2$ decomposes with first order kinetics in a 3 lit. container. If the pressure developed in 10 min. is 380 mm, the average rate at $27^oC$ is:

  1. $0.01M.min^{-1}$
  2. $0.002M.min^{-1}$
  3. $0.05M.min^{-1}$
  4. $0.06M.min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$t=A.{ e }^{ -kt }\ So,\quad (A-{ A } _{ o })=A.({ e }^{ -kt }-1)\ \therefore 380=A.({ e }^{ -kt }-1)\ { A } _{ o }=\cfrac { 380 }{ { e }^{ -10k }-1 } $
 Otherewise,
$ { P } _{ o }=[{ A } _{ o }]RT\ { P } _{ o }={ [{ A }] } _{ 10 }RT\ \cfrac { { P } _{ 10 }-{ P } _{ o } }{ 7 } =\cfrac { ({ A } _{ 10 }-{ A } _{ o })RT }{ 7 } \ \cfrac { \cfrac { 760 }{ 380 }  }{ 10 } =\vartheta .RT\ \cfrac { 2 }{ 10RT } =\vartheta $
$ \vartheta \sim 0.01{ M. }{ min }^{ -1 }\longrightarrow$ Option (A)

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

The rate constant of the reaction, $2{ H } _{ 2 }{ O } _{ 2 }\left( aq. \right) \rightarrow 2{ H } _{ 2 }O\left( l \right) +{ O } _{ 2 }\left( g \right) $, is $3\times { 10 }^{ -3 }{ min }^{ -1 }$.
At what concentration of ${ H } _{ 2 }{ O } _{ 2 }$, the rate of the reaction will be $2\times { 10 }^{ -4 }M{ s }^{ -1 }$?

  1. $6.67\times { 10 }^{ -3 }\ M$
  2. $2\ M$
  3. $4\ M$
  4. $0.08\ M$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Rate $=k{ \left[ { H } _{ 2 }{ O } _{ 2 } \right]  }^{ 1 }$
$2\times { 10 }^{ -4 }=\dfrac { 3\times { 10 }^{ -3 } }{ 60 } \times \left[ { H } _{ 2 }{ O } _{ 2 } \right] $
$\left[ { H } _{ 2 }{ O } _{ 2 } \right] =4 M$

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

Inversion of a sugar follows first order rate equation which can be followed by noting the change in rotation of the plane of polarisation of light in a 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=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ t }-{ r } _{ \infty } }{ { r } _{ 0 }-{ r } _{ \infty } } } $
  2. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ 0 }-{ r } _{ \infty } }{ { r } _{ t }-{ r } _{ 0 } } } $
  3. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ \infty }-{ r } _{ 0 } }{ { r } _{ \infty }-{ r } _{ t } } } $
  4. $k=\dfrac { 1 }{ t } \log _{ e }{ \dfrac { { r } _{ \infty }-{ r } _{ t } }{ { r } _{ \infty }-{ r } _{ 0 } } } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$({ r } _{ t }-{ r } _{ 0 })=({ r } _{ 0 }-{ r } _{ \infty  }){ e }^{ -kt }\ \ln { \left( \cfrac { { r } _{ t }-{ r } _{ 0 } }{ { r } _{ 0 }-{ r } _{ \infty  } }  \right)  } =-kt\ k=\cfrac { 1 }{ t } \ln { \left( \cfrac { { r } _{ t }-{ r } _{ 0 } }{ { r } _{ 0 }-{ r } _{ \infty  } }  \right)  } $

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 hydrolysis reaction of an ester by dilute acid is $0.6931\times { 10 }^{ -3 }\ { s }^{ -1 }$. The time required to change the concentration of ester from $0.04$ $M$ to $0.01$ $M$ is:

  1. $6931$ sec
  2. $4000$ sec
  3. $2000$ sec
  4. $1000$ sec
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$k=0.06931{ s }^{ -1 }$

 So,
$ t=\cfrac { \ln { \left( \cfrac { 0.04 }{ 0.01 }  \right)  }  }{ 0.06931 } \ =2000{ s }^{ -1 }$

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

For the reaction of first order, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \rightleftharpoons 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, which of the following statements are correct?

  1. The concentration of the reactant decreases exponentially with time.

  2. The half-life of the reaction decreases with increasing temperature.

  3. The half-life of the reaction depends on the initial concentration of the reactant.
  4. The reaction proceeds to $99.6$% completion in eight half-life duration.
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation
For 1st order reaction,
$a _t=a _0e^{-kt}$
Also, $t _{1/2}=\frac{In2}{k}$ or $t _{1/2}\alpha \frac{1}{k}$
As the temperature increases, value of k also increases due to which $t _{1/2}$ decreases.
For 99.6% completion, $a _t=(\frac{100-99.6}{100})a _0=\frac{4a _0}{1000}$
$t=\frac{1}{k}In\frac{a _0}{4a-0/1000}=\frac{1}{k}In\frac{1000}{4}$
$=(\frac{t _{1/4}}{In2}).In250$
$t=8t _{1/2}$
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

For the reaction, ${\text{2}}{{\text{N}} _{\text{2}}}{{\text{O}} _{\text{5}}} \to {\text{4N}}{{\text{O}} _{\text{2}}} + {{\text{O}} _{\text{2}}}$, the value of rate and rate constant are $1.02\times 10^{-4} M/s$ and $3.4 \times {10^{ - 3}}{\sec ^{ - 1}}$ respectively. The concentration of ${{\text{N}} _{\text{2}}}{{\text{O}} _{\text{5}}}$ at that time will be: (in terms of molarity)

  1. $1.732$
  2. $3$
  3. ${\text{1}}{\text{.02}} \times {\text{1}}{{\text{0}}^{ - 4}}$
  4. $3.4 \times {10^4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

In the following reaction $2H _{2}O _{2}\rightarrow 2H _{2}O+O _{2}$ rate of formation of $O _{2}$ is 3.6 M $ min^{-1}.$ The rate of formation of $H _{2}O$ is:

  1. $7.2 \, mol litre^{-1}min^{-1}$
  2. $7.8 \, mol litre^{-1}min^{-1}$
  3. $7.9 \, mol litre^{-1}min^{-1}$
  4. $7.5 \, mol litre^{-1}min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The rate of formation of water is twice the rate of formation of oxygen.
$\frac {d[H _2O]} {dt}=2\frac {d[H _2O]} {dt}=2 \times 3.6  M  min^{-1} = 7.2 \, mol litre^{-1}min^{-1}$

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

At ${ 380 }^{ 0 }C$, the half life period for the first order decomposition of ${ H } _{ 2 }{ O } _{ 2 }$ is 360 minutes. The energy of activation of the reaction is 200 kJ ${ mol }^{ -1 }$. Calculate the time required for 75% decomposition at $450^{0}C$?

  1. 60 min

  2. 40 min

  3. 20.34 min

  4. 10 min
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$k _1 = \ \cfrac { 0.693 }{ 360 }= 1.92\times{ 10 }^{ -3 }{ min }^{ -1 }$

$\ log\cfrac { k _2 }{ k _1 } =\left( \cfrac { Ea }{ 2.303R }  \right) \left[ \left( \cfrac { { T } _{ 2 }-{ T } _{ 1 } }{ { T } _{ 1 }{ T } _{ 2 } }  \right)  \right]=  \cfrac { \left( 200\times { 10 }^{ 3 } \right)  }{ \left( 2.303\times 8.314 \right) \left[ \left( \cfrac { 723-653 }{ 653\times 723 }  \right)  \right]  } =  \cfrac { \left( 200\times { 10 }^{ 3 }\times 70 \right)  }{ \left( 2.303\times 8.314\times 653\times 723 \right)  } = 1.5487$

$\cfrac { k _2 }{ k _1 }  = Antilog (1.5487)= 35.38$, $k _2 = 35.38 \times 1.92 \times$ ${ 10 }^{ -3 } = 6.792 \times { 10 }^{ -2 }{ min }^{ -1 }$

Rate at $450^oC$, t = $\ \left( \cfrac { 2.303 }{ k _2 }  \right) log\left( \cfrac { 100 }{ 100-75 }  \right) $= $\ \left( \cfrac { 2.303 }{ 6.792\times { 10 }^{ -3 } }  \right) log\left( \cfrac { 100 }{ 25 }  \right) = \left( \cfrac { 2.303\times 0.6021 }{ 6.792\times { 10 }^{ 2 } }  \right)= 20.34$ minutes.

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

The reaction, $Sucrose\xrightarrow [  ]{ { H }^{ + } } Glucose+Fructose$, takes  place at certain temperature while the volume of solution is maintained at $1$ litre. At time zero the initial rotation of the mixture is ${ 34 }^{ o }C$.After $30$ minutes the total rotation of solution is ${ 19 }^{ o }C$ and after a very long time, the total rotation is ${ -11 }^{ o }C$. Find the time when solution was optically inactive?

  1. $135$ min
  2. $103.7$ min
  3. $38.7$ min
  4. $45$ min
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

rate constant $k = \dfrac{2.303}{t}log\dfrac{(r _0 - r _\infty) }{(r _t-r _\infty)} = 0.0135$
At the point of optical inactiveness, rotation is zero. 

So, time taken is $ t =\dfrac{ 2.303}{k}log(45/11) = 103.7$ min.