Chemical Kinetics — DPP 1

20 NEET questions from Chemical Kinetics, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

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1
According to collision theory, most molecular collisions do not lead to reaction. Which of the following are necessary for collisions to successfully lead to reaction?
I. The total kinetic energy of the colliding molecules must be greater than some minimum value.
II. A catalyst must be present at the collision.
III. The colliding particles must be properly oriented in space when they collide.
  1. A I, II and III
  2. B I only
  3. C I and III
  4. D II and III
Show answer & explanation

Correct answer: C

Colliding molecules must have energy greater than activation energy with proper orientation of collision.
Therefore, option (3) is correct answer
2
The rate constant k, for the reaction:
$\mathrm { N } _ { 2 } \mathrm { O } _ { 5 } ( \mathrm {~g} ) \longrightarrow 2 \mathrm { NO } _ { 2 } ( \mathrm {~g} ) + \frac { 1 } { 2 } \mathrm { O } _ { 2 } ( \mathrm {~g} )$, is $2.3 \times 10 ^ { - 2 } \mathrm {~s} ^ { - 1 }$.
Which equation given below describes the change of $\left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right]$ with time? $\left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { 0 }$ and $\left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { t }$ correspond to the concentration of $\mathrm { N } _ { 2 } \mathrm { O } _ { 5 }$ initially and at time t .
  1. A $\left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { \mathrm { t } } = \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { 0 } + \mathrm { kt }$
  2. B $\left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { 0 } = \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { 1 } \mathrm { e } ^ { - \mathrm { kt } }$
  3. C $\log \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { \mathrm { t } } = \log \left[ \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \right] _ { 0 } + \mathrm { kt }$
  4. D $\ln \frac{\left[\mathrm{N}_2 \mathrm{O}_5\right]_0}{\left[\mathrm{~N}_2 \mathrm{O}_5\right]_{\mathrm{t}}}=\mathrm{kt}$
Show answer & explanation

Correct answer: D

$\mathrm{k}=\frac{1}{\mathrm{t}} \ln \frac{\mathrm{C}_{0}}{\mathrm{C}_{\mathrm{t}}}=\frac{1}{\mathrm{t}} \ln \frac{\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}}{\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]_{\mathrm{t}}}$
Therefore, option (4) is correct answer.
3
For the hypothetical reaction, 2A → 3C, the reaction rate ' $r$ ' in terms of the rate of change of the concentration is given by:
  1. A $\mathrm { r } = - \frac { \mathrm { d } [ \mathrm { A } ] } { \mathrm { dt } }$
  2. B $\mathrm { r } = - \frac { 1 } { 2 } \frac { \mathrm {~d} [ \mathrm {~A} ] } { \mathrm { dt } }$
  3. C $\mathrm { r } = \frac { 1 } { 3 } \frac { \mathrm {~d} [ \mathrm {~A} ] } { \mathrm { dt } }$
  4. D $\mathrm { r } = \frac { \mathrm { d } [ \mathrm { A } ] } { \mathrm { dt } }$
Show answer

Correct answer: B

4
If the volume of closed vessel in which the following simple reaction is carried out is reduced to one-third of its original volume, the rate of reaction becomes:
$2 \mathrm { NO } ( \mathrm {g} ) + \mathrm { O } _ { 2 } ( \mathrm {g} ) \longrightarrow 2 \mathrm { NO } _ { 2 } ( \mathrm {g} )$
  1. A one third.
  2. B three times.
  3. C nine times.
  4. D twenty seven times.
Show answer & explanation

Correct answer: D

$\mathrm{V}_{2}=\frac{\mathrm{V}_{1}}{3} \Rightarrow \mathrm{C}_{2}=3 \mathrm{C}_{1} \left(\right.$Conc. $\left.\propto \frac{1}{\text { volume }}\right)$
$\begin{aligned} \mathrm{r}_{1} & =\mathrm{k}[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right] \\ \mathrm{r}_{2} & \Rightarrow \mathrm{k}[3 \mathrm{NO}]^{2}\left[3 \mathrm{O}_{2}\right] \\ & \Rightarrow 27 \mathrm{k}[\mathrm{NO}]^{2}\left[\mathrm{O}_{2}\right]=27 \mathrm{r}_{1}\end{aligned}$
Therefore, option (4) is correct answer.
5
Which order reaction obeys the expression, $\mathrm { t } _ { 1 / 2 } \propto \mathrm { l } / [ \mathrm { A } ] _ { 0 }$ ?
  1. A First
  2. B Second
  3. C Third
  4. D Zero
Show answer

Correct answer: B

6
Select the correct statement regarding activation energy.
  1. A Activation energy may be greater than heat of reaction.
  2. B Activation energy is less than threshold energy.
  3. C Rate of reaction is inversely proportional to the activation energy.
  4. D All of these.
Show answer & explanation

Correct answer: D

According to Arrhenius equation
$\mathrm{k}=\mathrm{A} \cdot \mathrm{e}^{\frac{-\mathrm{E}_{\mathrm{a}}}{\mathrm{RT}}}$
The rate constant is inversely related to activation energy.
It is energy required to make it equal to threshold level and
it can be greater than ΔH for the reaction.
Therefore, option (4) is correct answer.
7
For the reaction, $\mathrm { N } _ { 2 } + 3 \mathrm { H } _ { 2 } \rightleftharpoons 2 \mathrm { NH } _ { 3 }$

$\left( \frac { \mathrm { dx } } { \mathrm { dt } } \right) = 1 \times 10 ^ { 2 } \left[ \mathrm {~N} _ { 2 } \right] \left[ \mathrm { H } _ { 2 } \right] ^ { 3 } - 1 \times 10 ^ { - 3 } \left[ \mathrm { NH } _ { 3 } \right] ^ { 2 }$ and at some instant if $\frac { \left[ \mathrm { N } _ { 2 } \right] \left[ \mathrm { H } _ { 2 } \right] ^ { 3 } } { \left[ \mathrm { NH } _ { 3 } \right] ^ { 2 } } = 10 ^ { - 5 } \mathrm { M } ^ { 2 }$ then at this instant value of $\left( \frac { \mathrm { dx } } { \mathrm { dt } } \right)$ is:
  1. A 0
  2. B $1 \times 10 ^ { 5 }$
  3. C $1 \times 10 ^ { - 5 }$
  4. D $1 \times 10 ^ { - 3 }$
Show answer

Correct answer: A

8
In the reaction $2 \mathrm {~A} + \mathrm { B } \longrightarrow \mathrm { A } _ { 2 } \mathrm {~B}$, if the concentration of A is doubled and that of $B$ is halved, then the rate of reaction will:
  1. A increase by 4 times.
  2. B decrease by 2 times.
  3. C increase by 2 times.
  4. D remain the same.
Show answer & explanation

Correct answer: C

Initially rate =k$\mathrm{[A]^{2}[B]}$
New rate =k$\mathrm{[2 A]^{2}}\left[\frac{B}{2}\right]$=2$k\mathrm{[A]^{2}[B]}$ i.e., increased 2 times.
Therefore, option (3) is correct answer.
9
A catalyst lowers the activation energy of a reaction from $20 \mathrm {~kJ} \mathrm {~mol} ^ { - 1 }$ to $10 \mathrm {~kJ} \mathrm {~mol} ^ { - 1 }$. The temperature at which the uncatalysed reaction will have the same rate as that of the catalysed at $27 ^ { \circ } \mathrm { C }$ is:
  1. A $- 123 ^ { \circ } \mathrm { C }$
  2. B $327 ^ { \circ } \mathrm { C }$
  3. C $372 ^ { \circ } \mathrm { C }$
  4. D $23 ^ { \circ } \mathrm { C }$
Show answer

Correct answer: B

10
The half life a first order reaction is 24 hours. If we start with 10 M initial concentration of the reactant then concentration after 96 hours will be:
  1. A 6.25 M
  2. B 1.25 M
  3. C 0.125 M
  4. D 0.625 M
Show answer & explanation

Correct answer: D

Number of half-lives $=\frac{96}{24}=4$
Concentration remaining after $96 \mathrm{hrs}=\frac{10}{(2)^{4}}=0.625 \mathrm{~M}$
Therefore, option (4) is correct answer.
11
Which is true for a first order reaction?
  1. A It can have rate constant $1 \times 10 ^ { - 2 } \mathrm {~L} \mathrm {~mol} ^ { - 1 } \mathrm {~s} ^ { - 1 }$.
  2. B Its half-life is inversely proportional to its initial concentration.
  3. C Time to complete $75 \%$ reaction is twice of half life.
  4. D $t _ { 50 \% } = \frac { 2.303 } { k }$
Show answer

Correct answer: C

12
Half life period for a first order reaction is 20 minutes. How much time is required to change the concentration of the reactants from 0.08 M to 0.01 M ?
  1. A 20 minutes
  2. B 60 minutes
  3. C 40 minutes
  4. D 50 minutes
Show answer

Correct answer: B

13
The rate constant k$_ { 1 }$ and k$_ { 2 }$ for two different reactions are $10 ^ { 16 } \mathrm { e } ^ { - 2000 / \mathrm { T } }$ and $10 ^ { 15 } \mathrm { e } ^ { - 1000 / \mathrm { T } }$ respectively. The temperature at which k$_ { 1 }$ = k$_ { 2 }$ is:
  1. A 2000 K
  2. B $\frac { 1000 } { 2.303 } \mathrm {~K}$
  3. C 1000 K
  4. D $\frac { 2000 } { 2.303 } \mathrm {~K}$
Show answer & explanation

Correct answer: B

k$_{1}$ = 10$^{16}$.e$^{–2000/T}$ and k$_{2}$ = 10$^{15}$.e$^{–1000/T}$
Now, k$_{1}$ = k$_{2}$
10$^{16}$.e$^{–2000/T}$ =10$^{15}$.e$^{-1000/T}$
$\begin{array}{l}\frac{\mathrm{e}^{-2000 / \mathrm{T}}}{\mathrm{e}^{-1000 / \mathrm{T}}}=\frac{10^{15}}{10^{16}} \\ \mathrm{e}^{-1000 / \mathrm{T}}=10^{-1}\end{array}$
Taking ln both sides, we get,
ln e$^{–1000/T}$ = ln 10$^{–1}$
$\begin{aligned} \frac{-1000}{\mathrm{~T}} & =2.303 \log 10^{-1} \\ \frac{-1000}{\mathrm{~T}} & =-2.303 ;(\log 10=1) \\ \mathrm{T} & =\frac{1000}{2.303} \mathrm{~K} .\end{aligned}$
Therefore, option (2) is correct answer.
14
For a reaction $\mathrm { A } + \mathrm { B } \rightarrow$ Product, it was found that rate of reaction increases four times if concentration of ' A ' is doubled but the rate of reaction remains unaffected, If concentration of ' B ' is doubled. Hence, the rate law for the reaction is:
  1. A Rate $= \mathrm { k } [ \mathrm { A } ] [ \mathrm { B } ]$
  2. B Rate $= \mathrm { k } [ \mathrm { A } ] ^ { 2 }$
  3. C Rate = k [A] [B]$^ { 1 }$
  4. D Rate $= \mathrm { k } [ \mathrm { A } ] [ \mathrm { B } ] ^ { 2 }$
Show answer & explanation

Correct answer: B

A + B → Product
r$_{1}$ = k[A]$^{m}$ [B]$^{n}$ ...(i)
r$_{2}$ = 4 = k[2A]$^{m}$[B]$^{n}$ ...(ii)
r$_{3}$ = 4 = k[2A]$^{m}$[2B]$^{n}$ ...(iii)
By dividing eq. (i) and (ii) we get,
$\frac{\mathrm{r}_{1}}{\mathrm{r}_{2}}=\frac{1}{4}=\left(\frac{1}{2}\right)^{\mathrm{m}}$
m = 2
Rate does not depend upon [B].
Hence, rate $=\mathrm{k}[\mathrm{A}]^{2}$
Therefore, option (2) is correct answer.
15
In acidic medium, the rate of reaction between $\mathrm { BrO } _ { 3 } ^ { - }$and $\mathrm { Br } ^ { - }$ions is given by the expression:
$- \left\{ \frac { \mathrm { d } \left[ \mathrm { BrO } _ { 3 } ^ { - } \right] } { \mathrm { dt } } \right\} = \mathrm { k } \left[ \mathrm { BrO } _ { 3 } ^ { - } \right] \left[ \mathrm { Br } ^ { - } \right] \left[ \mathrm { H } ^ { + } \right] ^ { 2 } \text {. It means: }$
  1. A rate constant of the reaction depends upon the concentration of $\mathrm { H } ^ { + }$ions.
  2. B rate of reaction is independent of the concentration of acid.
  3. C the change in pH of the solution will not affect the rate.
  4. D doubling the concentration of $\mathrm { H } ^ { + }$ions will increase the reaction rate by 4 times.
Show answer

Correct answer: D

16
What is the molecularity of following reactions?
(i) $\mathrm { PCl } _ { 5 } \rightarrow \mathrm { PCl } _ { 3 } + \mathrm { Cl } _ { 2 }$
(ii) $2 \mathrm { CO } + \mathrm { O } _ { 2 } \rightarrow 2 \mathrm { CO } _ { 2 }$
  1. A 1 and 3
  2. B 1 and 2
  3. C 2 and 3
  4. D 3 and 1
Show answer & explanation

Correct answer: A

Molecularity = Number of reacting species
Therefore, option (1) is correct answer.
17
A reaction that is of the first order with respect to reactant $A$ has a rate constant $6 \mathrm {~min} ^ { - 1 }$. If we start with $[ \mathrm { A } ] = 0.3 \mathrm {~mol} \mathrm {~L} ^ { - 1 }$, when would $[ \mathrm { A } ]$ reach the value $0.03 \mathrm {~mol} \mathrm {~L} ^ { - 1 }$ ?
  1. A 0.384 min
  2. B 15 min
  3. C 10 min
  4. D 3.84 min
Show answer

Correct answer: A

18
Consider the reaction: $2 \mathrm {~N} _ { 2 } \mathrm { O } _ { 5 } \rightarrow 4 \mathrm { NO } _ { 2 } + \mathrm { O } _ { 2 }$.
In the reaction, $\mathrm { NO } _ { 2 }$ is being formed at the rate of $0.0125 \mathrm {~mol} \mathrm {~L} ^ { - 1 } \mathrm {~s} ^ { - 1 }$. What is the rate of reaction at this time?
  1. A $0.0018 \mathrm {~mol} \mathrm {~L} ^ { - 1 } \mathrm {~s} ^ { - 1 }$
  2. B $0.0031 \mathrm {~mol} \mathrm {~L} ^ { - 1 } \mathrm {~s} ^ { - 1 }$
  3. C $0.0041 \mathrm {~mol} \mathrm {~L} ^ { - 1 } \mathrm {~s} ^ { - 1 }$
  4. D $0.0050 \mathrm {~mol} \mathrm {~L} ^ { - 1 } \mathrm {~s} ^ { - 1 }$
Show answer

Correct answer: B

19
If $t _ { 1 / 2 }$ does not change with initial concentration, then order of the reaction is:
  1. A }$ zero.
  2. B second.
  3. C first.
  4. D third.
Show answer

Correct answer: C

20
What is the activation energy for the decomposition of $\mathrm { N } _ { 2 } \mathrm { O } _ { 5 }$ ?
$\mathrm { N } _ { 2 } \mathrm { O } _ { 5 } \rightleftharpoons 2 \mathrm { NO } _ { 2 } + \frac { 1 } { 2 } \mathrm { O } _ { 2 }$
If the values of the rate constants are $3.45 \times 10 ^ { - 5 }$ and $6.9 \times 10 ^ { - 3 }$ at $27 ^ { \circ } \mathrm { C }$ and $67 ^ { \circ } \mathrm { C }$ respectively.
  1. A $102 \times 10 ^ { 2 } \mathrm {~kJ}$
  2. B 488.5 kJ
  3. C 112 kJ
  4. D 14.7 kJ
Show answer

Correct answer: D

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