Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry
Learn Chemical Kinetics MCQ questions & answers in Physical Chemistry are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
171.
Rate of a general reaction $$A + B \to $$ products can be expressed as follows on the basis of collision theory. $${\text{Rate}} = {Z_{AB}}{e^{ - \frac{{{E_a}}}{{RT}}}}$$
Which of the following statements is not correct for the above expression?
A
$$Z$$ is collision frequency and is equal to number of collisions per second per unit volume of the reaction mixture.
B
$${e^{ - \frac{{{E_a}}}{{RT}}}}$$ is the fraction of molecules with kinetic energy equal to or greater than $${E_a}.$$
C
$${E_a}$$ is activation energy of the reaction.
D
All the molecules which collide with one other are effective collisions.
Answer :
All the molecules which collide with one other are effective collisions.
Only those collisions in which molecules collide with sufficient energy, called threshold energy and proper orientation are effective collisions. Rest of the molecules collide and bounce back.
172.
The time taken for $$90\% $$ of a first order reaction to complete is approximately
173.
For a chemical reaction $${t_{\frac{1}{2}}}$$ is 2.5 hours at room temperature. How much of the reactant will be left after 7.5 hours if initial weight of reactant was $$160\,g?$$
A
10$$\,g$$
B
40$$\,g$$
C
80$$\,g$$
D
20$$\,g$$
Answer :
20$$\,g$$
$$\eqalign{
& {\text{Using the relation}} \cr
& \left[ A \right] = {\left[ A \right]_0}{\left( {\frac{1}{2}} \right)^n}\left[ {n = {\text{number of half - lives}}} \right] \cr
& T = n \times {t_{\frac{1}{2}}} \cr
& {\text{Here,}}\,n = \frac{{7.5}}{{2.5}} = 3 \cr
& \therefore \,\,\left[ A \right] = 160 \times {\left( {\frac{1}{2}} \right)^3} \cr
& = 160 \times \frac{1}{8} \cr
& = 20\,g \cr} $$
174.
Activation energy of a chemical reaction can be determined by _________.
A
determining the rate constant at standard temperature
B
determining the rate constants at two temperatures
C
determining probability of collision
D
using catalyst
Answer :
determining the rate constants at two temperatures
175.
Which of the following statements is incorrect about the collison theory of chemical reaction?
A
It considers reacting molecules or atoms to be hard spheres and ignores their structural features.
B
Number of effective collisions determines the rate of reaction.
C
Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
D
Molecules should collide with sufficient threshold energy and proper orientation for the collision to be effective.
Answer :
Collision of atoms or molecules possessing sufficient threshold energy results into the product formation.
Not only sufficient threshold energy of colliding atoms or molecules but also the proper orientation for the collision is required for the formation of products.
176.
The reaction $$2{N_2}{O_5} \rightleftharpoons 2{N_2}{O_4} + {O_2}$$ is
A
bimolecular and of second order
B
unimolecular and of first order
C
bimolecular and of first order
D
bimolecular and of zero order
Answer :
bimolecular and of first order
It is bimolecular first order reaction since $${\text{Rate}} \propto \left[ {{N_2}{O_5}} \right]$$
177.
For a reaction $$\frac{1}{2}A \to 2B,$$ rate of disappearance of $$‘A’$$ is related to the rate of appearance of $$'B’$$ by the expression
A
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}}$$
B
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ B \right]}}{{dt}}$$
C
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{{d\left[ B \right]}}{{dt}}$$
D
$$ - \frac{{d\left[ A \right]}}{{dt}} = 4\frac{{d\left[ B \right]}}{{dt}}$$
Answer :
$$ - \frac{{d\left[ A \right]}}{{dt}} = \frac{1}{4}\frac{{d\left[ B \right]}}{{dt}}$$
$$\eqalign{
& {\text{The rates of reactions for the reaction}} \cr
& \frac{1}{2}A \to 2B \cr
& {\text{can be written either as}} \cr
& - 2\frac{d}{{dt}}\left[ A \right]\,\,\,{\text{with respect to }}'A' \cr
& or\,\,\frac{1}{2}\frac{d}{{dt}}\left[ B \right]\,\,\,{\text{with respect to }}'B' \cr
& {\text{From the above, we have}} \cr
& - 2\frac{d}{{dt}}\left[ A \right] = \frac{1}{2}\frac{d}{{dt}}\left[ B \right] \cr
& or\,\, - \frac{d}{{dt}}\left[ A \right] = \frac{1}{4}\frac{d}{{dt}}\left[ B \right] \cr
& {\text{i}}{\text{.e}}{\text{., correct answer is (B)}} \cr} $$
178.
A photon of hard gamma radiation knocks a proton out of $$_{12}^{24}Mg$$ nucleus to form
179.
Under the same reaction conditions, initial concentration of $${\text{1}}{\text{.386 }}mol{\text{ }}d{m^{ - 3}}$$ of a substance becomes half in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio $$\left( {{k_1}/{k_0}} \right)$$ of the rate constant for first order $$\left( {{k_1}} \right)$$ and zero order $$\left( {{k_2}} \right)$$ of the reaction is -
A
$$0.5\,mo{l^{ - 1}}d{m^3}$$
B
$$1.0\,mol\,d{m^{ - 3}}$$
C
$$1.5\,mol\,d{m^{ - 3}}$$
D
$$2.0\,mo{l^{ - 1}}d{m^3}$$
Answer :
$$0.5\,mo{l^{ - 1}}d{m^3}$$
The values of rate constants $${k_0},{k_1}$$ for zero order and first order reaction, respectively, are given by the following equation :
$${k_0} = \frac{{{A_0}}}{{2 \times {t_{\frac{1}{2}}}}}$$ [ where $${{A_0} = }$$ initial concentration, and $${{t_{\frac{1}{2}}} = }$$ half - life period ]
and $${k_1} = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}$$
substituting various given values, we get
$$\eqalign{
& {k_0} = \frac{{{\text{1}}{\text{.386}}\,{\text{mol litr}}{{\text{e}}^{ - 1}}}}{{2 \times 20\,\sec }}\,\,...{\text{(i)}} \cr
& {\text{and}}\,{k_1} = \frac{{0.639}}{{40\,\sec }}\,\,...({\text{ii)}} \cr
& {\text{Dividing (ii) by (i), we get}} \cr
& \frac{{{k_1}}}{{{k_0}}} = \frac{{0.639}}{{40}} \times \frac{{2 \times 20}}{{1.386}}mo{l^{ - 1}}{\text{litre}} \cr
& = \frac{{0.639}}{{1.386}}mo{l^{ - 1}}{\text{litre}} \cr
& = 0.5\,mo{l^{ - 1}}{\text{litre}} \cr
& = 0.5\,mo{l^{ - 1}}d{m^3}\,\,\,\,\,\,\,\,\,\left[ {1\,{\text{litre}} = 1d{m^3}} \right] \cr
& {\text{Thus the correct answer is (A)}}{\text{.}} \cr} $$
180.
The rate law for the reaction below is given by the expression $$k\left[ A \right]\left[ B \right]$$
$$A + B \to {\text{Product}}$$
If the concentration of $$B$$ is increased from $$0.1$$ to $$0.3\,mole,$$ keeping the value of $$A$$ at $$0.1\,mole,$$ the rate constant will be :