Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry
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241.
Which of the following will react at the highest rate?
A
$$1{\text{ }}mole$$ of $$A$$ and $$1{\text{ }}mole$$ of $$B$$ in a $$1{\text{ - }}L$$ vessel
B
$$2{\text{ }}mole$$ of $$A$$ and $$2{\text{ }}mole$$ of $$B$$ in a $$2{\text{ - }}L$$ vessel
C
$$3{\text{ }}mole$$ of $$A$$ and $$3{\text{ }}mole$$ of $$B$$ in a $$3{\text{ - }}L$$ vessel
D
All would react at the same rate
Answer :
All would react at the same rate
Since all have same concentration of reactants, all would react at same time.
242.
For an exothermic chemical reaction occurring two steps as
$$\eqalign{
& \left( {\text{i}} \right)A + B \to X\left( {{\text{slow}}} \right) \cr
& \left( {{\text{ii}}} \right)X \to AB\left( {{\text{fast}}} \right) \cr} $$
The progress of the reaction can be best described by
A
B
C
D
Answer :
As conversion of $$X$$ to $$AB$$ is fast, it means the process has a very low activation energy.
243.
The hypothetical reaction $${A_2} + {B_2} \to 2AB;$$ follows the following mechanism \[{{A}_{2}}\xrightarrow{\text{Fast}}A+A,\] \[A+{{B}_{2}}\xrightarrow{\text{Slow}}AB+B,A+B\xrightarrow{\text{Fast}}AB\cdot \]
The order of the overall reaction is
A
$$0$$
B
$$1$$
C
$$2$$
D
$$\frac{3}{2}$$
Answer :
$$\frac{3}{2}$$
$$\eqalign{
& {A_2} + {B_2} \to 2AB; \cr
& {A_2} \to A + A\left( {{\text{Fast}}} \right) \cr
& A + {B_2} \to AB + B\left( {{\text{Slow}}} \right) \cr} $$
Rate law $$ = k\left[ A \right]\left[ {{B_2}} \right]$$ put value of $$\left[ A \right]$$ from $${{\text{I}}^{{\text{st}}}}$$ reaction since $$A$$ is intermediate $$\sqrt {k\left[ {{A_2}} \right]} = A$$
∴ Rate law equation $$ = K\sqrt {k\left[ {{A_2}} \right]} \left[ {{B_2}} \right]$$
∴ Order $$ = \frac{1}{2} + 1 = \frac{3}{2}$$
244.
The activation energy of the reaction, $$A + B \to C + D + 38\,kcal$$ is $$20\,kcal.$$ What would be the activation energy of the following reaction.
$$C + D \to A + B$$
A
20$$\,kcal$$
B
- 20$$\,kcal$$
C
18$$\,kcal$$
D
58$$\,kcal$$
Answer :
58$$\,kcal$$
$${E_{{a_2}}} = 58$$
245.
Find the values of $$A, B$$ and $$C$$ in the following table for the reaction $$X + Y \to Z.$$ The reaction is of first order $$w.r.t$$ $$X$$ and zero order $$w.r.t.$$ $$Y.$$
Exp.
$$\left[ X \right]\left( {mol\,{L^{ - 1}}} \right)$$
$$\left[ Y \right]\left( {mol\,{L^{ - 1}}} \right)$$
$${\text{Rate}} = k\left[ X \right]{\left[ Y \right]^0}$$
Rate is independent of the cone. of $$Y$$ and it depends only on the conc. of $$X$$ and it is the first order reaction.
From exp. $$\left( 1 \right),2 \times {10^{ - 2}} = k\left( {0.1} \right)\,\,\,...\left( {\text{i}} \right)$$
From exp. $$\left( 2 \right),4 \times {10^{ - 2}} = k\left( A \right)\,\,\,...\left( {{\text{ii}}} \right)$$
Dividing (ii) and (i), $$\frac{{4 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 2}}}} = \frac{{k\left( A \right)}}{{k\left( {0.1} \right)}} = \frac{A}{{0.1}}$$
$$ \Rightarrow 2 \times 0.1 = A \Rightarrow A = 0.2\,mol\,{L^{ - 1}}$$
From exp. $$\left( 3 \right),B = k\left( {0.4} \right)\,\,\,...\left( {{\text{iii}}} \right)$$
Dividing (iii) and (i), $$\frac{B}{{2 \times {{10}^{ - 2}}}} = \frac{{k\left( {0.4} \right)}}{{k\left( {0.1} \right)}} = 4$$
$$ \Rightarrow B = 4 \times 2 \times 10 - 2 = $$ $$8 \times {10^{ - 2}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}$$
From exp. $$\left( 4 \right),2 \times {10^{ - 2}} = k\left( C \right)\,\,\,...\left( {{\text{iv}}} \right)$$
Dividing (iv) and (i), $$\frac{{2 \times {{10}^{ - 2}}}}{{2 \times {{10}^{ - 2}}}} = \frac{{k\left( C \right)}}{{k\left( {0.1} \right)}} = \frac{C}{{0.1}}$$
$$ \Rightarrow C = 0.1\,mol\,{L^{ - 1}}$$
246.
The rate of the reaction : $$C{H_3}COO{C_2}{H_5} + NaOH \to $$ $$C{H_3}COONa + {C_2}{H_5}OH$$ is given by the equation, $${\text{rate}} = k\left[ {C{H_3}COO{C_2}{H_5}} \right]\left[ {NaOH} \right]$$
If concentration is expressed in $$mol/L,$$ the units of $$k$$ are
A
$$mo{l^{ - 2}}\,{L^2}\,{s^{ - 1}}$$
B
$$mol\,{L^{ - 1}}\,{s^{ - 1}}$$
C
$$L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
D
$${s^{ - 1}}$$
Answer :
$$L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
For a second order reaction, $$\frac{{dx}}{{dt}} = k{\left[ A \right]^2}$$
\[\frac{\text{conc}\text{.}}{\text{time}}=k{{\left[ \text{conc}\text{.} \right]}^{2}}\]
$$\frac{{mol\,{L^{ - 1}}}}{2} = k\,mol\,{L^{ - 1}} \times mol\,{L^{ - 1}};$$ $$k = L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
247.
A reaction having equal energies of activation for forward and reverse reaction has :
A
$$\Delta G = 0$$
B
$$\Delta H = 0$$
C
$$\Delta H = \Delta G = \Delta S = 0$$
D
$$\Delta S = 0$$
Answer :
$$\Delta H = 0$$
$$\Delta H = {E_a}\left( f \right) - {E_a}\left( b \right) = 0$$
248.
If the rate of a reaction is equal to the rate constant, the order of the reaction is
A
2
B
3
C
0
D
1
Answer :
0
Rate of reaction is equal to the rate constant for zero order reaction.
Let us consider the following hypothetical change.
$$A \to B + C$$
Suppose this reaction is zero order, then
$$\eqalign{
& {\text{rate}} \propto {\left[ A \right]^0} \cr
& \therefore \,\,{\text{rate}} = k{\left[ A \right]^0} = k \cr} $$
249.
For the reaction, $${N_2}{O_5}\left( g \right) \to 2N{O_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right)$$ The value of rate of disappearance of $${N_2}{O_5}$$ is given as $$6.25 \times {10^{ - 3}}mol\,{L^{ - 1}}{s^{ - 1}}.$$ The rate of formation of $$N{O_2}$$ and $${O_2}$$ is given respectively as
Key Idea Rate of disappearance of reactant = rate of appearance of product
$${\text{or}}$$ $$ - \frac{1}{{{\text{Stoichiometric coefficient of reactant}}}}$$ $$\frac{{{\text{d [reactant]}}}}{{dt}}$$
$$ = + \frac{1}{{{\text{Stoichiometric coefficient}}\,{\text{of product}}}}$$ $$\frac{{{\text{d[product]}}}}{{dt}}$$
250.
For the reaction system : $$2NO\left( g \right) + {O_2}\left( g \right) \to 2\,N{O_2}\left( g \right)$$ volume is
suddenly reduced to half its value by increasing the pressure on it. If the reaction is of first order
with respect to $${O_2}$$ and second order with respect to $$NO,$$ the rate of reaction will
A
diminish to one-eighth of its initial value
B
increase to eight times of its initial value
C
increase to four times of its initial value
D
diminish to one-fourth of its initial value
Answer :
increase to eight times of its initial value
$$r = k\left[ {{O_2}} \right]{\left[ {NO} \right]^2}.$$ When the volume is reduced to $$\frac{1}{2},$$ the $$conc.$$ will double
∴ New rate $$ = k\left[ {2{O_2}} \right]{\left[ {2NO} \right]^2} = 8\,k\left[ {{O_2}} \right]{\left[ {NO} \right]^2}$$
The new rate increases to eight times of its initial.