Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry
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251.
Consider the reaction $$A \to 2\,B + C,\Delta H = - 15\,kcal.$$ The energy of activation of backward reaction is $$20\,kcal\,mo{l^{ - 1}}.$$ In presence of catalyst
the energy of activation of forward reaction is $$3\,kcal\,mo{l^{ - 1}}.$$ At $$400\,K$$ the catalyst causes the rate of the reaction to increase by the number of times equal to
252.
For a reaction $$X \to Y,$$ the rate of reaction becomes twenty seven times when the concentration of $$X$$ is increased three times. What is the order of the reaction?
253.
The activation energy for a simple chemical reaction, $$A \to B$$ is $${E_a}$$ in forward direction. The activation energy for reverse reaction
A
can be less than or more than $${E_a}$$
B
is always double of $${E_a}$$
C
is negative of $${E_a}$$
D
is always less than $${E_a}$$
Answer :
can be less than or more than $${E_a}$$
The energy of activation of reverse reaction is less than or more than energy of activation $$\left( {{E_a}} \right)$$ of forward reaction.
$$\therefore \,\,\Delta H = {\left( {{E_a}} \right)_F} - {\left( {{E_a}} \right)_R}$$
Because it depends upon the nature of reaction.
If $${\left( {{E_a}} \right)_F} > {\left( {{E_a}} \right)_R},$$ reaction is endothermic
or $${\left( {{E_a}} \right)_F} < {\left( {{E_a}} \right)_R},$$ reaction is exothermic
254.
The reaction, $$A \to B$$ follows first order kinetics. The time taken for $$0.8\,mole$$ of $$A$$ to produce $$0.6\,mole$$ of $$B$$ is $$1h.$$ What is the time taken for the conversion of $$0.9\,mole$$ of $$A$$ to $$0.675\,mole$$ of $$B?$$
A
$$0.25\,h$$
B
$$2\,h$$
C
$$1\,h$$
D
$$0.5\,h$$
Answer :
$$1\,h$$
$$\eqalign{
& {\text{Rate constant of first order reaction}} \cr
& k = \frac{{2.303}}{t}{\text{lo}}{{\text{g}}_{10}}\frac{{{{\left( A \right)}_0}}}{{{{\left( A \right)}_t}}} \cr
& {\text{or}}\,\,k = \frac{{2.303}}{1} \times {\text{lo}}{{\text{g}}_{10}}\frac{{0.8}}{{0.2}}\,\,\,...{\text{(i)}} \cr
& \left( {{\text{because }}0.6{\text{ }}mole{\text{ of }}B{\text{ is formed}}} \right) \cr} $$
Suppose $${t_1}$$ hour are required for changing the concentration of $$A$$ from $$0.9\,mole$$ to $$0.675\,mole$$ of $$B.$$
Remaining mole of $$A = 0.9 - 0.675 = 0.225$$
$$\eqalign{
& \therefore \,\,k = \frac{{2.303}}{{{t_1}}}{\text{lo}}{{\text{g}}_{10}}\frac{{0.9}}{{0.225}}\,\,\,...{\text{(ii)}} \cr
& {\text{From Eqs}}{\text{. (i) and (ii)}} \cr
& \frac{{2.303}}{1}{\text{lo}}{{\text{g}}_{10}}\frac{{0.8}}{{0.2}} = \frac{{2.303}}{{{t_1}}}{\text{lo}}{{\text{g}}_{10}}\frac{{0.9}}{{0.225}} \cr
& 2.303\,{\text{lo}}{{\text{g}}_{10}}\,4 = \frac{{2.303}}{{{t_1}}}{\text{lo}}{{\text{g}}_{10}}4 \cr
& {t_1} = 1\,h \cr} $$
255.
The rate of formation of a dimer in a second order dimerisation reaction is $$9.1 \times {10^{ - 6}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$ at $$0.01\,mol\,{L^{ - 1}}$$ monomer concentration. What will be the rate constant for the reaction?
A
$$9.1 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
B
$$9.1 \times {10^{ - 6}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
C
$$3 \times {10^{ - 4}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
D
$$27.3 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
$$\eqalign{
& 2A \to {A_2} \cr
& {\text{Rate of formation of dimer}} = k{\left[ A \right]^2} \cr
& k = \frac{{{\text{Rate of}}\,{\text{formation of dimer}}}}{{{{\left[ A \right]}^2}}} \cr
& k = \frac{{9.1 \times {{10}^{ - 6}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}}}{{{{\left( {0.01\,mol\,{L^{ - 1}}} \right)}^2}}} \cr
& \,\,\,\, = 9.1 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}} \cr} $$
256.
Which of the following statements is not correct?
A
For a zero order reaction, $${t_{\frac{1}{2}}}$$ is proportional to initial concentration.
B
The relationship of variation of rate constant with temperature is given by $$\log \frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303R}}\left[ {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right]$$
C
The unit of rate constant for a reaction is $$mo{l^{1 - n}}\,{L^{n - 1}}\,{s^{ - 1}}$$ where $$n$$ is order of the reaction.
D
The unit of rate of reaction changes with order of reaction.
Answer :
The unit of rate of reaction changes with order of reaction.
The unit of rate of reaction is $$mol\,{L^{ - 1}}\,{s^{ - 1}}.$$ It does not change with order.
257.
The rate constant of the reaction $$A \to B$$ is $$0.6 \times {10^{ - 3}}\,mole$$ per second. If the concentration of $$A$$ is $$5$$ $$M$$ then concentration of $$B$$ after $$20\,\min $$ is
A
1.08$$\,M$$
B
3.60$$\,M$$
C
0.36$$\,M$$
D
0.72$$\,M$$
Answer :
0.72$$\,M$$
Key Concept For a zero order reaction unit of rate constant is mole per second. Hence, we can easily calculate concentration of $$B$$ after $$20$$ $$min$$ by the following formula,
$$\eqalign{
& x = Kt \cr
& x = Kt = 0.6 \times {10^{ - 3}} \times 20 \times 60 = 0.72\,M \cr} $$
258.
The order of a reaction, with respect to one of the reacting component $$Y,$$ is zero. It implies that :
A
the reaction is going on at a constant rate
B
the rate of reaction does not vary with temperature
C
the reaction rate is independent of the concentration of $$Y$$
D
the rate of formation of the activated complex is zero
Answer :
the reaction rate is independent of the concentration of $$Y$$
Let us consider a reaction,
$$\eqalign{
& xX + yY \to aA + bB \cr
& rate = {\left[ X \right]^x}{\left[ Y \right]^y} \cr} $$
It is given that order of reaction $$w.r.t$$ . component $$Y$$ is zero.
Hence, rate $$ = {\left[ X \right]^x}$$
i.e., rate becomes independent of the concentration of $$Y.$$
259.
The reactions rate $${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$$ was measured $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = 2 \times {10^{ - 4}}\,mol\,{\sec ^{ - 1}}.$$ The rates of reactions expressed in terms of $${N_2}$$ and $${H_2}$$ are:
Rate in terms of $${N_2}.$$ $$\left( {mol\,{L^{ - 1}}\,{{\sec }^{ - 1}}} \right)$$
Rate in terms of $${H_2}$$ $$\left( {mol\,{L^{ - 1}}\,{{\sec }^{ - 1}}} \right)$$
(a)
$$2 \times {10^{ - 4}}$$
$$2 \times {10^{ - 4}}$$
(b)
$$3 \times {10^{ - 4}}$$
$$1 \times {10^{ - 4}}$$
(c)
$$1 \times {10^{ - 4}}$$
$$3 \times {10^{ - 4}}$$
(d)
$$2 \times {10^{ - 1}}$$
$$2 \times {10^{ - 3}}$$
A
(a)
B
(b)
C
(c)
D
(d)
Answer :
(c)
$${N_2} + 3{H_2} \rightleftharpoons 2N{H_3};$$ Rate is given by any of the expressions
$$\frac{{ - d\left[ {{N_2}} \right]}}{{dt}} = - \frac{{1d\left[ {{H_2}} \right]}}{{3dt}} = \frac{1}{2}\frac{{d\left( {N\left. {{H_3}} \right]} \right.}}{{dt}}$$
Rate of disappearance of $${N_2} = \frac{1}{2}$$ the rate of formation of $$N{H_3} = 1 \times {10^{ - 4}}$$
Rate of disappearance of $${H_2} = \frac{3}{2}$$ the rate of formation of $$N{H_3} = 3 \times {10^{ - 4}}$$
260.
Which of the following expressions is correct for the rate of reaction given below?
$$5Br_{\left( {aq} \right)}^ - + BrO_{3\left( {aq} \right)}^ - + 6H_{\left( {aq} \right)}^ + \to $$ $$3B{r_{2\left( {aq} \right)}} + 3{H_2}{O_{\left( l \right)}}$$