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

Biotransformation of an organic compound having concentration (x) can be modelled using an ordinary differential equation $\frac{dx}{dt} + kx^2 = 0$, where k is the reaction rate constant. If x = a at t = 0, then the solution of the equation is

  1. x = ae-kt
  2. $\frac{1}{x} = \frac{1}{a} + kt$

  3. x = a (1 - e-kt)

  4. x = a + kt
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

dx/dt = -kx^2. Separating variables: dx/x^2 = -k*dt. Integrating both sides: -1/x = -kt + C. At t=0, x=a, so -1/a = C. Thus, -1/x = -kt - 1/a, which simplifies to 1/x = 1/a + kt.

Multiple choice

A portion of wastewater sample was subjected to standard BOD test (5 days, 20 oC), yielding a value of 180 mg/l. The reaction rate constant (to the base ‘e’) at 20 oC was taken as 0.18 per day. The reaction rate constant at other temperature may be estimated by kr = k20 (1.047)T-20. The temperature at which the other portion of the sample should be tested, to exert the same BOD in 2.5 days, is

  1. 4.9 oC

  2. 24.9 oC

  3. 31.7 oC

  4. 35.0 oC
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the BOD equation BOD_t = L(1 - exp(-kt)), we set the BOD equal for two conditions. Since the BOD is the same, k1*t1 = k2*t2. Solving for k2 at 2.5 days gives k2 = 0.36. Using the temperature correction formula 0.36 = 0.18 * (1.047)^(T-20) leads to T = 35 degrees Celsius.

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

Fill up the following with suitable terms.

(i) Activation energy $=$ Threshold energy $-$ ____.

(ii) Half-life period of zero order reaction $=$ __.

(iii) Average rate of reaction $=$ _.

(iv) Instantaneous rate of reaction $=$ ___.

  1. Potential energy, $\dfrac {0.693}{k}, \dfrac {dx}{dt}, \dfrac {\triangle [A]}{\triangle t}$
  2. Energy of reactants, $\dfrac {1}{k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$
  3. Energy of reaction, $\dfrac {\log k}{t}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$
  4. Average kinetic energy of reactants, $\dfrac {a}{2k}, \dfrac {\triangle [A]}{\triangle t}, \dfrac {dx}{dt}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
a) Activation energy = Threshold energy$-$ the Average kinetic energy of molecules(by definition)

b) half-life of a zero order reaction = ${ t } _{ \frac { 1 }{ 2 }  } =\frac { a }{ 2k } $

c) Average rate of reaction = $\dfrac {\displaystyle \text{ Total change concentration reactants }}{ \displaystyle \text{Total time taken} }$  = $\dfrac { \triangle \left[ A \right]  }{ \triangle t }$

d) Instantaneous rate of reaction =$ \dfrac { \displaystyle \text {Instantaneous change in concentration} }{\displaystyle \text { Instantaneous  time} } $=$\dfrac { dx }{ dt } $                    
Multiple choice chemistry chemical kinetics introduction to chemical kinetics understanding chemical kinetics rate of chemical reaction

If reaction A and B are given with Same temperature and same concentration but rate of $A$ is double than $B$. Pre exponential factor is same for both the reaction then difference in activation energy $E _{A} - E _{B}$ is?

  1. $-RT\ ln2$
  2. $RT\ ln2$
  3. $2RT$
  4. $\dfrac {RT}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac {r _{A}}{r _{B}} = \dfrac {A _{1}e^{\dfrac {-E _{A}}{RT}}}{A _{2}e^{\dfrac {-E _{B}}{RT}}}$
$\dfrac {2}{1} = \dfrac {e^{\dfrac {-E _{A}}{RT}}}{e^{\dfrac {-E _{B}}{RT}}}$
$ln2 = E _{B} - E _{A} / RT$
$E _{B} - E _{A} = RT\ ln2$
$E _{A} - E _{B} = -RT\ ln2$.

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

The rate of chemical reaction is directly proportional to the equilibrium constant.
In which of the following process reaction will be completed first?

  1. $K=10$
  2. $K=1$
  3. $K={ 10 }^{ 3 }$
  4. $K={ 10 }^{ -2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The rate of chemical reaction is directly proportional to the equilibrium constant.

$\therefore $ For $K={ 10 }^{ 3 }$, the reaction will complete fastest.

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.