Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry
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161.
Bombardment of aluminium by $$o-$$ particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products $$X, Y$$ and $$Z$$ respectively are,
162.
In the presence of a catalyst, the heat evolved or absorbed during the reaction _________.
A
increases
B
decreases
C
remains unchanged
D
may increase or decrease
Answer :
remains unchanged
Catalyst alters the activation energy of both forward and backward reactions equally hence heat of reaction remains unchanged.
163.
Two reactions $${R_1}$$ and $${R_2}$$ have identical pre-exponential factors. Activation energy of $${R_1}$$ exceeds that of $${R_2}$$ by $$10\,kJ\,mo{l^{ - 1}}.$$ If $${k_1}$$ and $${k_2}$$ are rate constants for reactions $${R_1}$$ and $${R_2}$$ respectively at $$300 K$$ , then In $$\left( {{k_2}/{k_1}} \right)$$ is equal to :
$$\left( {R = 8.314\,J\,mo{l^{ - 1}}{K^{ - 1}}} \right)$$
164.
Rate of a reaction can be expressed by Arrhenius equation as : $$k = A{e^{ - \,\frac{E}{{RT}}}}$$
In this equation, $$E$$ represents
A
the total energy of the reacting molecules at a temperature, $$T$$
B
the fraction of molecules with energy greater than the activation energy of the reaction
C
the energy above which all the colliding molecules will react
D
the energy below which all the colliding molecules will react
Answer :
the energy above which all the colliding molecules will react
In Arrhenius equation $$k = A{e^{ - \,\frac{E}{{RT}}}},E$$ is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products.
Since the slow step is the rate determining step hence if we consider option (A) we find
$${\text{Rate}} = k\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]$$
Now if we consider option (B) we find
$$\eqalign{
& {\text{Rate}} = k\left[ {C{l_2}} \right]\left[ {H{S^ - }} \right]\,\,\,\,\,...\left( {\text{i}} \right) \cr
& {\text{From equation (i)}} \cr
& k = \frac{{\left[ {{H^ + }} \right]\left[ {H{S^ - }} \right]}}{{{H_2}S}}\,\,{\text{or}}\,\left[ {H{S^ - }} \right] = \frac{{k\left[ {{H_2}S} \right]}}{{{H^ + }}} \cr} $$
Substituting this value in equation (i) we find
$${\text{Rate}} = k\left[ {C{l_2}} \right]K\frac{{\left[ {{H_2}S} \right]}}{{{H^ + }}} = k'\frac{{\left[ {C{l_2}} \right]\left[ {{H_2}S} \right]}}{{\left[ {{H^ + }} \right]}}$$
hence only , mechanism (A) is consistent with the given
rate equation.
166.
For a reaction, activation energy $$\left( {{E_a}} \right) = 0$$ and rate constant $$\left( k \right) = 3.2 \times {10^6}{s^{ - 1}}$$ at $$300\,K.$$ What is the value of the rate constant at $$310\,K$$
A
$$3.2 \times {10^{ - 12}}{s^{ - 1}}$$
B
$$3.2 \times {10^6}{s^{ - 1}}$$
C
$$6.4 \times {10^{12}}{s^{ - 1}}$$
D
$$6.4 \times {10^6}{s^{ - 1}}$$
Answer :
$$3.2 \times {10^6}{s^{ - 1}}$$
When $${E_a} = 0,$$ rate constant is independent of temperature.
167.
In a first order reaction, $$A \to B,$$ if $$k$$ is rate constant and initial concentration of the reactant $$A$$ is $$0.5$$ $$M,$$ then the half-life is
168.
The activation energies of the forward and backward reactions in the case of a chemical reaction are $$30.5$$ and $$45.4\,kJ/mol$$ respectively. The reaction is :
A
exothermic
B
endothermic
C
neither exothermic nor endothermic
D
independent of temperature
Answer :
exothermic
Exothermic because of activation energy $${E_b} > {E_f}$$
169.
The decomposition of dinitrogen pentoxide $$\left( {{N_2}{O_5}} \right)$$ follows first order rate law. What will be the rate constant from the given data?
$$\eqalign{
& {\text{At}}\,t = 800\,s{\text{,}}\left[ {{N_2}{O_5}} \right] = 1.45\,mol\,{L^{ - 1}} \cr
& {\text{At}}\,t = 1600\,s,\left[ {{N_2}{O_5}} \right] = 0.88\,mol\,{L^{ - 1}} \cr} $$