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 introduction to chemical kinetics understanding chemical kinetics rate of chemical reaction

In a reaction mechanism consisting of elementary reaction steps where the relative rate of each is given, which of the following is most likely to be the rate-determining step?

  1. A step labeled fast

  2. A step labeled moderate

  3. A step labeled slow

  4. It is not possible to tell which step is rate determining from this information.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a multi-step reaction, the elementary step having the slowest-rate is always the rate determining step of the overall reaction.

Therefore, a step labeled slow will be the rate determining step.

Multiple choice chemistry chemical kinetics introduction to chemical kinetics understanding chemical kinetics rate of chemical reaction

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

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

$2 SO _2 + O _2 \rightleftharpoons  2 SO _3$


$\dfrac{1}{2}\dfrac{d[SO _3]}{dt} = 1.6 \times 10^{-3} kg/min = \dfrac{1.6 \times 10^{-3}}{80} mol/min$                             $[\because M _{SO _3}= 80 \times 10^{-3} Kg/mol]$


$\dfrac{-1}{2}\dfrac{d[SO _2]}{dt} = \dfrac{+1}{2}\dfrac{d[SO _3]}{dt}$

$= -\dfrac{1.6 \times 10^{-3}}{80 \times 10} $ mol/min

$= -\dfrac{1}{50} \times 64 \times 10^{-3} = -1.28 \times 10^{-3} $ kg/min

Multiple choice chemistry chemical kinetics introduction to chemical kinetics understanding chemical kinetics rate of chemical reaction

For a reaction
$2A+B\rightarrow C+D$, the active mass of $B$ is kept constant but that of $A$ is tripled. The rate of reaction will -

  1. Decrease by $3$ times
  2. Increased by $9$ times
  3. Increase by $3$ times

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

According to the given reaction, $ Rate = k[A]^{2}[B]^{1} $

When A becomes 3A i.e. triples, the rate will become: $ Rate^{'} = k[3A]^{2}[B]^{1} = 9k[A]^{2}[B] $
Thus, rate of reaction is increased by 9 times. 

Multiple choice chemistry chemical kinetics introduction to chemical kinetics understanding chemical kinetics rate of chemical reaction

The container of $2$ litrer contains $4$ mole of $N _{2}O _{5}$. On heating to $100^{\circ}C, N _{2}O _{5}$ undergoes complete dissociation to $NO _{2}$ and $O _{2}$. Select the correct answers if rate constant for decomposition of $N _{2}O _{5}$ is $6.2\times 10^{-4}sec^{-1}$.
1. The mole ratio before and after dissociation is $4 : 2$
2. Half life of $N _{2}O _{5}$ is $1117\ sec$ and it is independent of concentration.
3. Time required to complete $40$% of reaction is $824\ sec$.
4. If volume of container is doubled, the initial rate of decomposition becomes half of the initial rate.

  1. $1, 3, 4$
  2. $1, 2, 3, 4$
  3. $3, 4$
  4. $2, 3, 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Because reaction is first order

$\vartheta =k[A]$
$A=A _{0}e^{-kt}$

for which

$t _{40}=824$sec
if$v _{f}=2V _{i}\Rightarrow C _{f}=\frac{C _{i}}{2}\Rightarrow \vartheta _{f}=\frac{\vartheta _{i}}{2}$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

In a reaction $2HI \rightarrow H _{2} + I _{2}$, the concentration of $HI$ decreases from $0.5\ mol\ L^{-1}$ to $0.4\ mol\ L^{-1}$ in $10$ minutes. What is the rate of reaction during this interval?

  1. $5\times 10^{-3} M\ min^{-1}$
  2. $2.5\times 10^{-3} M\ min^{-1}$
  3. $5\times 10^{-2} M\ min^{-1}$
  4. $2.5\times 10^{-2} M\ min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Average rate $= -\dfrac {1}{2} \dfrac {\triangle [R]}{\triangle t} = -\dfrac {1}{2}\times \dfrac {0.4 - 0.5}{10}$


$= \dfrac {1}{2}\times \dfrac {0.1}{10} = 5\times 10^{-3}M\ min^{-1}$.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Consider the chemical reaction:
$N _2(g)+3H _2(g)\rightarrow 2NH _3(g)$


The rate of this reaction can be expressed; in terms of time and of concentration of $N _2(g), H _2(g)$ $NH _3(g)$. Identify the correct relationship amongst the rate expressions.

  1. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=+\dfrac{1}{2}\dfrac{d[NH _3]}{dt}$
  2. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{3d[H _2]}{dt}=\dfrac{2d[NH _3]}{dt}$
  3. Rate $=-\dfrac{d[N _2]}{dt}=-\dfrac{1}{3}\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$
  4. Rate $=-\dfrac{d[N _2]}{dt}=\dfrac{d[H _2]}{dt}=\dfrac{d[NH _3]}{dt}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As the reaction proceed, the concentration of the reactant decrease and that of the product increases.
Thus, the rate of the reaction of the reactant and product can be given as follow:
$Rate =-\cfrac{d}{dt}[N _2]=-\cfrac{1}{3}\cfrac{d}{dt}[H _2]=+\cfrac{1}{2}\cfrac{d}{dt}[NH _3]$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Rate of formation of $SO _3$ in the following reaction $2SO _2+O _2\rightarrow 2SO _3$ is $100g$ $min^{-1}$. 


Then the rate of disappearance of $O _2$ is:

  1. $50g$ $min^{-1}$
  2. $40g$ $min^{-1}$
  3. $200g$ $min^{-1}$
  4. $20g$ $min^{-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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


rate $=\dfrac { 1 }{ 2 } \dfrac { \left[ { \triangle SO } _{ 2 } \right]  }{ \triangle t } =\dfrac { \triangle \left[ { O } _{ 2 } \right]  }{ \triangle t } =\dfrac { 1 }{ 2 } \dfrac { \triangle \left[ { SO } _{ 3 } \right]  }{ \triangle t } $

rate of formation of ${ SO } _{ 3 }=$ rate of disappearance of ${ O } _{ 2 }$

$\dfrac { 1 }{ 2 } \dfrac { \triangle \left[ { SO } _{ 3 } \right]  }{ \triangle t } =-\dfrac { \triangle \left[ { O } _{ 2 } \right]  }{ \triangle t } $

$-\dfrac { \triangle w _{{ O } _{ 2 }}   }{ \triangle t }=\dfrac { 1 }{ 2 } \times \dfrac{100}{80}\times 32g{ min }^{ -1 }=20g{ min }^{ -1 }$

So, the correct option is $D$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

If concentration of reactants is increased by a factor x then the rate constant k becomes:

  1. $\ln{\frac{k}{x}}$
  2. $\frac{k}{x}$
  3. $k+x$
  4. $k$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Rate Constant doesn't depend on the concentration of the reactants.

So, If the conc of reactants is increased by a factor x, then the rate constant k becomes k.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For the non-equilibrium process, $A + B \rightarrow Products$, the rate is first order with respect to $A$ and second-order with respect to $B$. If $1.0$ mole each of $A$ and $B$ are introduced into a 1-litre vessel and the initial rate was $1.0 \times 10^{-2}$ mol/litre-sec. The rate (in mol $litre^{-1} sec^{-1}$) when half of the reactants have been used:

  1. $1.2 \times 10^{-3}$
  2. $1.2 \times 10^{-2}$
  3. $2.5 \times 10^{-4}$

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

$rate = k[A]{[B]}^{2}$


Initially 1 mol each of A and B are present and the rate was $1 \times {10}^{-2}\space mol/lit-sec$

When Half of the Reactants are used, the rate becomes $\dfrac{1}{2} \times {[\dfrac{1}{2}]}^{2} \times {10}^{-2}$ = $1.2 \times {10}^{-3}$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction
 Time  0  5min  10min  15min
 [A]  20mol  18mol  16mol  16 mol

For the reaction $A\longrightarrow Products$; $\frac { -d[A] }{ dt } =k$ and at different time interval, IAI values are given. At $20$ minute, rate will be :

  1. $12 mol /min$
  2. $10 mol/min$
  3. $8 mol/min$
  4. $0.4 mol/min$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The concentration decreases linearly with time

Hence moles of A at $20 min$
$=12$
Therefore, $k=\dfrac{-d[A]}{dt}$
$=-\dfrac{12-14}{20-15}$
$=0.4\ mol/min$

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

$H _2 + l _2 \rightarrow 2 Hl$ (An elementary reaction)
If the volume of the container containing the gaseous mixture is increased to two times, then final rate of the reaction

  1. Become four time
  2. Become $\dfrac{1}{4} th$ of the original rate
  3. Become $2$ times
  4. Become $\dfrac{1}{2}$ of the original rate
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For the elementary reaction H2 + I2 -> 2HI, the rate law is given by Rate = k [H2][I2]. When the volume of the container is increased to two times, the concentrations of both reactants are halved because concentration is moles divided by volume. Substituting the new concentrations into the rate law, the new rate becomes k([H2]/2)([I2]/2) = (1/4) k[H2][I2], which is 1/4th of the original rate.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Assuming an element reaction $H _2O _2+ 3I^-+ 2H^+\to 2H _2O+ I _3^-.$ The effect on the rate of this reaction brought about by doubling the concentration of $I^-$ without changing the order?

  1. The rate would increases by a factor of $3$
  2. The rate would increase by a factor of $8$
  3. The rate would decrease by a factor of $1/3$
  4. The rate would increase by a factor of $9$
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

Instanteneous rate of reaction can be found be :

  1. slope of a rate of reaction vs time

  2. slope of a concentration vs time graph

  3. taking any two points on the graph
  4. both $B$ and $C$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We determine an instantaneous rate at time t:

  1. by calculating the negative of the slope of the curve of concentration of a reactant versus time at time t.
  2. by calculating the slope of the curve of concentration of a product versus time at time t.

Multiple choice chemistry reaction kinetics determining rates graphically calculating rate of reaction graphically rate of a chemical reaction

For the reaction, $2{ N } _{ 2 }{ O } _{ 5 }\left( g \right) \longrightarrow 4N{ O } _{ 2 }\left( g \right) +{ O } _{ 2 }\left( g \right) $, if the concentration of $N{ O } _{ 2 }$ increases by $5.2\times { 10 }^{ -3 }M$ in $100$ sec, then the rate of reaction is:

  1. $1.3\times { 10 }^{ -5 }M{ s }^{ -1 }$
  2. $5\times { 10 }^{ -4 }M{ s }^{ -1 }$
  3. $7.6\times { 10 }^{ -4 }M{ s }^{ -1 }$
  4. $2\times { 10 }^{ -3 }M{ s }^{ -1 }$
  5. $2.5\times { 10 }^{ -5 }M{ s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vartheta =\cfrac { 1 }{ 4 } \cfrac { d }{ dt } [{ NO } _{ 2 }]=\cfrac { 1 }{ 4 } (5.2\times { 10 }^{ -5 }M{ s }^{ -1 })\ =1.3\times { 10 }^{ -5 }{ Ms }^{ -1 }$