Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry
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1.
The rate of a first order reaction is \[1.5\times {{10}^{-2}}mol\,{{L}^{-1}}{{\min }^{-1}}\] at $$0.5\,M$$ concentration of the reactant. The half life of the reaction is
A
$$0.383\,\min $$
B
$$23.1\,\min $$
C
$$8.73\,\min $$
D
$$7.53\,\min $$
Answer :
$$23.1\,\min $$
$$\eqalign{
& {\text{For a first order reaction,}}\,A \to {\text{Products}} \cr
& r = k\left[ A \right]\,{\text{or}}\,k = \frac{1}{{\left[ A \right]}} \cr
& \Rightarrow k = \frac{{1.5 \times {{10}^{ - 2}}}}{{0.5}} = 3 \times {10^{ - 2}} \cr
& {\text{Further}},\,{t_{\frac{1}{2}}} = \frac{{0.693}}{k} = \frac{{0.693}}{{3 \times {{10}^{ - 2}}}} = 23.1 \cr} $$
2.
Consider the following statements :
(i) Increase in concentration of reactant increases the rate of a zero order reaction.
(ii) Rate constant $$k$$ is equal to collision frequency $$A$$ if $${E_a} = 0.$$
(iii) Rate constant $$k$$ is equal to collision frequency $$A$$ if $${E_a} = \infty .$$
(iv) $${\text{ln}}\,k\,\,{\text{vs}}\,\,T$$ is a straight line.
(v) $${\text{In}}\,k\,\,{\text{vs}}\,\,\frac{1}{T}$$ is a straight line.
Correct statements areCorrect statements are
A
(i) and (iv)
B
(ii) and (v)
C
(iii) and (iv)
D
(ii) and (iii)
Answer :
(ii) and (v)
According to Arrhenius equation,
$$\eqalign{
& k = A{e^{ - \,\frac{{Ea}}{{RT}}}} \cr
& \therefore \,\,{\text{when}}\,{E_a} = 0,k = A \cr} $$
Also $${\text{ln}}\,k\,\,{\text{vs}}\,\,\frac{1}{T}$$ is a straight line with slope $$ = - \frac{{{E_a}}}{R}.$$
∴ Statements (ii) and (v) are correct.
3.
Which of the following factors are responsible for the increase in the rate of a surface catalysed reaction?
(i) A catalyst provides proper orientation for the reactant molecules to react.
(ii) Heat of adsorption of reactants on a catalyst helps reactant molecules to overcome activation energy.
(iii) The catalyst increases the activation energy of the reaction.
A
(i) and (iii)
B
(i) and (ii)
C
(ii) and (iii)
D
(i), (ii) and (iii)
Answer :
(i) and (ii)
The catalyst decreases the activation energy of the reaction and thus increases the rate of reaction.
4.
For a first order reaction, the ratio of the time take for $${\frac{7}{8}^{th}}$$ of the reaction to complete to that of half of the reaction to complete is
A
3 : 1
B
1: 3
C
2 : 3
D
3 : 2
Answer :
3 : 1
$$k = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$$
For $$\frac{7}{8}$$ of the reaction to complete, $$t = \frac{7}{{8t}}$$
$$\eqalign{
& a - x = a - \frac{{7a}}{8} = \frac{a}{8} \cr
& \therefore {t_{\frac{7}{8}}} = \frac{{2.303}}{k}\log \frac{a}{{\frac{a}{8}}} = \frac{{2.303}}{k}\log 8 \cr} $$
For half of the reaction to complete, $$t = {t_{\frac{1}{2}}},$$
$$\eqalign{
& x = a - \frac{a}{2} = \frac{a}{2} \cr
& \therefore {t_{\frac{1}{2}}} = \frac{{2.303}}{k}\log \frac{a}{{\frac{a}{2}}} = \frac{{2.303}}{k}\log 2 \cr
& \therefore \frac{{{t_{\frac{7}{8}}}}}{{{t_{\frac{1}{2}}}}} = \frac{{\log 8}}{{\log 2}} = \frac{{\log {2^3}}}{{\log 2}} = \frac{{3\log 2}}{{\log 2}} = 3 \cr} $$
5.
A positron is emitted from $$_{11}^{23}Na.$$ The ratio of the atomic mass and atomic number of the resulting nuclide is
A
$$\frac{{22}}{{10}}$$
B
$$\frac{{22}}{{11}}$$
C
$$\frac{{23}}{{10}}$$
D
$$\frac{{23}}{{12}}$$
Answer :
$$\frac{{23}}{{10}}$$
$$_{11}^{23}Na \to _{10}^{23}X + _1^0\beta $$
6.
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 1 hour. What is the time taken for conversion of $$0.9\,mole$$ of $$A$$ to produce $$0.675\,mole$$ of $$B?$$
7.
A reaction takes place in various steps. The rate constant for first, second, third and fifth steps are $${k_1},{k_2},{k_3}$$ and $${k_5}$$ respectively. The overall rate constant is given by $$k = \frac{{{k_2}}}{{{k_3}}}{\left( {\frac{{{k_1}}}{{{k_5}}}} \right)^{\frac{1}{2}}}$$ If activation energy are 40, 60, 50 and $$10\,kJ/mol$$ respectively, the overall energy of activation $$\left( {kJ/mol} \right)$$ is :