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 rate constant, $\mathrm{k}$ for the reaction $\displaystyle \mathrm{N} _{2}\mathrm{O} _{5}(\mathrm{g})\rightarrow 2\mathrm{N}\mathrm{O} _{2}(\mathrm{g})+\frac{1}{2}\mathrm{O} _{2}(\mathrm{g})$ ls $2.3\times 10^{-2}\mathrm{s}^{-1}$. Which equation given below describes the change of $[\mathrm{N} _{2}\mathrm{O} _{5}]$ with time?

$[\mathrm{N} _{2}\mathrm{O} _{5}] _{0}$ and $[\mathrm{N} _{2}\mathrm{O} _{5}] _{\mathrm{t}}$ correspond to concentration of $\mathrm{N} _{2}\mathrm{O} _{5}$ initially and at time $\mathrm{t}$.

  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. $log _{10}[N _{2}O _{5}] _{t}=log _{10}[N _{2}O _{5}] _{0}-kt$
  4. $ln\dfrac{[N _{2}O _{5}] _{0}}{[N _{2}O _{5}] _{t}}=kt$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The decomposition of  $\mathrm{N} _{2}\mathrm{O} _{5}$ follows first order kinetics.


The integrated rate law expression is $ln\dfrac{[N _{2}O _{5}] _{0}}{[N _{2}O _{5}] _{t}}=kt$.

It can also be represented as $log _{10}[N _{2}O _{5}] _{t}=log _{10}[N _{2}O _{5}] _{0}-\dfrac {kt} {2.303}.$


It can also be represented as $[N _{2}O _{5}] _{t}=[N _{2}O _{5}] _{0}e^{-kt}.$

Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions
For the reaction ; $2H _2O _2(aq)\rightarrow 2H _2O(l)+O _2(g)$, rate of decomposition for $H _2O _2=k[H _2O _2]^2$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For $2H _2O _2(aq)\rightarrow 2H _2O(l)+O _2(g)$ rate of decomposition for $H _2O _2=k[H _2O _2]$. It is a first order reaction.  It proceeds through following mechanism.

$\displaystyle H _2O _2 \xrightarrow {slow} H _2O + O $

$\displaystyle  O + O \xrightarrow {fast} O _2$

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

For the reaction; $2N _2O _5\rightarrow 4NO _2+O _2$, rate and rate constant are $1.02\times 10^{-4} M sec^{-1}$ and $3.4\times 10^{-5}  sec^{-1}$ respectively, then concentration of $N _2O _5$, at that time will be:

  1. $1.732\ M$
  2. $3\ M$
  3. $1.02\times 10^{-4} M$
  4. $3.5\times 10^{5} M$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
From the unit of rate constant we can identify the reaction as first order.

As we know,
$r=K[N _2O _5]$

$\therefore [N _2O _5]=\frac {r}{K}=\frac {1.02\times 10^{-4}}{3.4\times 10^{-5}}=3M$.
Multiple choice chemistry chemical kinetics kinetic study of some first order reactions rate of chemical reaction endothermic and exothermic reactions

For the first order reaction:-
$2N _2O _5(g)\rightarrow 4NO _2(g)+O _2(g)$

  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

Option (A),(B),(D) are correct.
(C) : The half-life of the reaction is independent of the initial concentration of the reactant. Half-life for first order reaction is :$t _{1/2} = 0.693/k$
A first-order reaction has a rate proportional to the concentration of one reactant.
First-order rate constants have units of $sec^{-1}$. In other words, a first-order reaction has a rate law in which the sum of the exponents is equal to 1. 

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

The reaction; $N _2O _5(g) \longrightarrow 2NO _2(g)+\frac {1}{2}O _2(g)$ is of first order for $N _2O _5$ with rate constant $6.2\times 10^{-4}s^{-1}$. What is the value of rate of reaction when $[N _2O _5]=1.25 \ mol L^{-1}$?

  1. $5.15\times 10^{-5}mol L^{-1}s^{-1}$
  2. $6.35\times 10^{-3}mol L^{-1}s^{-1}$
  3. $7.75\times 10^{-4}mol L^{-1}s^{-1}$
  4. $3.85\times 10^{-4}mol L^{-1}s^{-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As we know,
$r=K[N _2O _5]=6.2\times 10^{-4}\times 1.25=7.75\times 10^{-4} mol L^{-1} s^{-1}$.

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

The hydrolysis of an ester was carried out with 0.1 M $H _2SO _4$ and 0.1 M HCl separately. Which of the following expressions between the rate consists is expected? The rate expression being rate = $k[H^{\oplus}][ester]$ 

  1. $k _{HCl}\, =\, k _{H _2SO _4}$
  2. $k _{HCl}\, >\, k _{H _2SO _4}$
  3. $k _{HCl}\, <\, k _{H _2SO _4}$
  4. $k _{ H _2SO _4}\, =\, k _{HCl}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$[H _2SO _4]\, =\, 0.1\, M\, =\, 0.1\, \times\, 2\, =\, 0.2 N$

$[HCl]$ = $0.1 N$

In case of $[H _2SO _4]$ 

$r _1\, =\, k[H^{\oplus}][Ester]$ 

$\displaystyle k _{H _2SO _4}\, = \frac{r _1}{2\, N\, \times\, [Ester]}$ 

In case of HCl, $r _1\, =\, k[H^{\oplus}]\, [Ester]$ 

$\displaystyle k _{HCl}\, =\, \frac{r _2}{1\, N\, [Ester]}$ 

Hence $K _{HCl}\, >\, K _{H _2SO _4}$

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

$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$

If $\displaystyle -\, \frac{d[N _2O _5]}{dt}\, =\, k _1[N _2O _5]$

$\displaystyle \frac{d[NO _2]}{dt}\, =\, k _2[N _2O _5]$

$\displaystyle \frac{d[O _2]}{dt}\, =\, k _3[N _2O _5]$
What is the relation between $k _1, k _2\, and\, k _3$ ?

  1. $k _1\, =\, k _2\, =\, k _3$
  2. $2k _1\, =\, k _2\, =\, 4k _3$
  3. $2k _1\ =\, 4k _2\, =\, k _3$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As we know,
for a reaction:
$2N _2O _5\, \rightarrow\, 4NO _2\, +\, O _2$
$\displaystyle -\, \frac{1}{2}\, \frac{d[N _2O _5]}{dt}\, =\, \frac{1}{4}\, \frac{d[NO _2]}{dt}\, =\, \frac{d[O _2]}{dt}$
So
$2k _1\, =\, k _2\, =\, 4k _3$

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

The inversion of cane sugar proceeds with the half-life of 500 min at pH 5 for any concentration of sugar. However, if pH = 6, the half life changes to 50 min. The rate law expression for the sugar inversion can be written as:

  1. $r\, =\, k[sugar]^2[H]^6$
  2. $r\, =\, k[sugar]^1[H]^0$
  3. $r\, =\, k[sugar]^0[H^{\oplus}]^6$
  4. $r\, =\, k[sugar]^0[H^{\oplus}]^1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given,
Since $t _{1/2}$ does not depends upon the sugar concentration means it is first order w.r.t [sugar]
$\therefore t _{1/2}\, \propto\, [sugar]^{1}$
$t _{1/2}\, \times\, a^{n\, -\, 1}\, =\, k$
$\displaystyle \frac{(t _1/2) _1}{(t _{1/2) _2}}\, =\, \frac{[H^{\oplus}] _1^{1-n}}{[H^{\oplus}] _2^{1-n}}$

$\displaystyle \frac{500}{50}\, =\, \left ( \frac{10^{-5}}{10^{-6}}\right )^{1-n}$

10 = $(10)^{1-n}\, \Rightarrow\, n\, =\, 0$

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

The reaction $2N _2O _5(g)\, \rightarrow\, 4NO _2(g)\, +\, O _2(g)$ is first order w.r.t. $N _2O _5$. Which of the following graphs would yield a straight line ?

  1. $log\, p _{N _2O _5}$ vs time with -ve slope
  2. $(p _{N _2O _5})^{-1}$ vs time
  3. $p _{N _2O _5}$ vs time
  4. $log\, p _{N _2O _5}$ vs time with +ve slope
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a first order reaction, the graph of logarithm of the partial pressure of reactant to the time is a straight line with negative slope. 


Hence, $\displaystyle log\, p _{N _2O _5}$ vs time t will give a straight line.

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

When ethyl acetate was hydrolyzed in the presence of $0.1 M$ $HCl$, the constant was found to be $5.40\, \times\, 10^{-5}\, s^{-1}$. But when $0.1$ $M\, H _2SO _4$ was used for hydrolysis, the rate constant was found to be $6.20\, \times\,10^{-5}\, s^{-1}$. From these we can say that:

  1. $H _2SO _4$ is stronger than $HCl.$
  2. $H _2SO _4$ and $HCl$ are both of the same strength.
  3. $H _2SO _4$ is weaker than $HCl.$
  4. The data is insufficient to compare the strength of $HCl$ ad $H _2SO _4$.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $k _{H _2SO _4}\, >\, k _{HCl},$ hence $H _2SO _4$ is stronger acid than HCl.

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

The rate law for the reaction : $:Ester+H^+\rightarrow Acid+Alcohol\,$ is
$V\,=\,k\;\left[ester \right]\;\left[H _3O^+ \right]^0$
What would be the new rate if
(a)$\;$conc. of ester is doubled
(b)$\;$conc. of $:H^{+}$ is doubled

  1. (a)$\;v\;$ (b)$\;2v$
  2. (a)$\;2v\;$ (b)$\;v$
  3. (a)$\;2v\;$ (b)$\;2v$

  4. None of the above

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

$\upsilon=k[ester][H _3O^+]$


(a) Conc. of ester is doubled rate also double that is $2\upsilon$ because rate of the reaction depends upon ester concentration.

(b) Conc. of $H^+$ is doubled rate does not change that is $\upsilon$ because rate of the reaction does not depends on $H _3O^+$ concentration.

So answer is B.

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

In the presence of acid, the initial concentration of cane-sugar was reduced from 0.2 M to 0.1 in 5 hr and to 0.05 M in 10 hr. The reaction must be of :

  1. Zero order

  2. First order

  3. Second order

  4. Fractional order
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle 0.2\, M \underset{t _{1/2}\, =\, 5hr}{\rightarrow} 0.1\, M \underset{t _{1/2}\, =\, 5hr}{\rightarrow} 0.05\, M$

$From\, 0.2\, M \underset{t\, =\, 10hr}{\rightarrow}\, 0.05\, M$
So $t _{1/2}$ is constant which is characteristic of first order reaction.
Hence, $t _{1/2}\, \propto\, (a)^0$.

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

The decomposition of $H _2O _2$ can be followed by titration with $KMnO _4$ and is found to be a first order reaction. The rate constant is $4.5\, \times\, 10^{-2}$. In an experiment, the initial titrate value was 25 mL. The titrate value will be 5 mL after a lapse of :

  1. $4.5\, \times\, 10^{-2}\, \times\, 5\, min$
  2. $\displaystyle \frac{log _{e}5}{4.5\, \times\, 10^{-2}}\, min$
  3. $\displaystyle \frac{log _{e}5/4}{4.5\, \times\, 10^{-2}}\, min$

  4. None of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

As we know,
$\displaystyle t\, =\, \frac{2.303}{k}\, log\, \frac{V _0}{V _1}$

$\displaystyle =\, \frac{1}{k}\, ln\, \frac{V _0)}{V _1}$

$\displaystyle =\, \frac{1}{4.5\, \times\, 10^{-2}\, min^{-1}}\, In\, \frac{25mL}{5mL}$

$\displaystyle =\, \frac{log _{e}5}{4.5\, \times\, 10^{-2}}min$ 

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 minutes, the volume of $O _2$ produced is 9 $mL$ and at the end of the reaction is 35 $mL$. The rate constant is equal to:

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

  4. None of the above

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

For a first order reaction,

$KT= ln (a/a-x)$

So, for the following reaction:

$N _2O _5\rightarrow N _2O _4+1/2O _2$

$K\times15\,=\,ln\begin{pmatrix}\displaystyle\frac{35-0}{35-9}\end{pmatrix}$


$N _2O _5\rightarrow N _2O _4+1/2O _2$

$K= \dfrac{1}{15} \,ln\begin{pmatrix}\displaystyle\frac{35}{26}\end{pmatrix}$

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

The reaction $N _{2}O _{5}$ (in $CCl _{4}$) $\rightarrow 2NO _{2}+1/2O _{2}(g)$ is the first order in $N _{2}O _{5}$ with rate constant $6.2\times 10^{-4}S^{-1}$. 


What is the value of the rate of reaction when $N _2O _5=1.25:mole:L^{-1}$ ?

  1. $7.75\times 10^{-4}mol\:L^{-1}S^{-1}$
  2. $6.35\times 10^{-3}mol\:L^{-1}S^{-1}$
  3. $5.15\times 10^{-5}mol\:L^{-1}S^{-1}$
  4. $3.85\times 10^{-4}mol\:L^{-1}S^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For the first-order reaction, the rate of the reaction is given by the expression

Rate $\displaystyle  = k [N _2O _5]$ where k is the rate constant.

Substitute values in the above expression

Rate $\displaystyle  = 6.2\times 10^{-4}S^{-1} \times 1.25\:mole\:L^{-1} = 7.75\times 10^{-4}mol\:L^{-1}S^{-1}$

So, the correct option is $A$