Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & Rat{e_1} = k{\left[ A \right]^n}{\left[ B \right]^m};\,\,\,\,\,Rat{e_2} = k{\left[ {2A} \right]^n}{\left[ {\frac{1}{2}B} \right]^m} \cr & \therefore \,\,\frac{{Rat{e_2}}}{{Rat{e_1}}} = \frac{{k{{\left[ {2A} \right]}^n}{{\left[ {\frac{1}{2}B} \right]}^m}}}{{k{{\left[ A \right]}^n}{{\left[ B \right]}^m}}} \cr & = {\left[ 2 \right]^n}{\left[ {\frac{1}{2}} \right]^m} \cr & = {2^n}{.2^{ - m}} \cr & = {2^{n - m}} \cr} $$

$$k = \frac{{2.303}}{t}\log \frac{a}{{\left( {a - x} \right)}}$$
$$\left( {a - x} \right)$$  $${\text{is the concentration left after}}\,100\,\sec .$$
$$2.7 \times {10^{ - 3}} = \frac{{2.303}}{{100}}\log \frac{{0.29}}{{\left( {a - x} \right)}}$$
$$ \Rightarrow \frac{{0.27}}{{2.303}} = \log \frac{{0.29}}{{\left( {a - x} \right)}} \Rightarrow $$      $$0.117 = \log \frac{{0.29}}{{\left( {a - x} \right)}}$$
$$ \Rightarrow \left( {a - x} \right) = 0.22\,M$$

$$\eqalign{ & {\text{Order }}w.r.t.{\text{ }}A = 1;{\text{ order }}w.r.t.{\text{ }}B = 1 \cr & {\text{Rate}} = \frac{1}{2}\frac{{d\left( {AB} \right)}}{{dt}} = {k_r}\left[ A \right]\left[ B \right]; \cr & \frac{1}{2} \times \left( {2.5 \times {{10}^4}} \right) \cr & = {k_r}\left( {0.1} \right)\left( {0.1} \right) \cr & {k_r} = 1.25 \times {10^{ - 2}} \cr} $$

When the temperature is increased, energy in form of heat is supplied which increases the kinetic energy of the reacting molecules. this will increase the number of collisions and ultimately the rate of reaction will be enhanced.

Let the order of reaction wrt $$C{H_3}COC{H_3},B{r_2}$$    and $${{H^ + }}$$ are $$x, y$$  and $$z$$ respectively. Thus,
$$\eqalign{ & {\text{Rate}}\left( r \right) = {\left[ {C{H_3}COC{H_3}} \right]^x}{\left[ {B{r_2}} \right]^y}{\left[ {{H^ + }} \right]^z} \cr & 5.7 \times {10^{ - 5}} = {\left( {0.30} \right)^x}{\left( {0.05} \right)^y}{\left( {0.05} \right)^z}...{\text{(i)}} \cr & 5.7 \times {10^{ - 5}} = {\left( {0.30} \right)^x}{\left( {0.10} \right)^y}{\left( {0.05} \right)^z}...{\text{(ii)}} \cr & 1.2 \times {10^{ - 4}} = {\left( {0.30} \right)^x}{\left( {0.10} \right)^y}{\left( {0.10} \right)^z}...{\text{(iii)}} \cr & 3.1 \times {10^{ - 4}} = {\left( {0.40} \right)^x}{\left( {0.05} \right)^y}{\left( {0.20} \right)^z}...{\text{(iv)}} \cr & {\text{From Eqs}}{\text{. (i) and (ii)}} \cr & 1 = {\left( {\frac{1}{2}} \right)^y}\,\,{\text{or}}\,\,\,{1^ \circ } = {\left( {\frac{1}{2}} \right)^y} \cr & y = 0 \cr & {\text{From Eqs}}{\text{. (ii) and (iii)}} \cr & z = 1 \cr & {\text{From Eggs}}{\text{. (i) and (iv)}} \cr & x = 1 \cr} $$
$${\text{Thus, rate law}} \propto $$    $${\left[ {C{H_3}COC{H_3}} \right]^1}{\left[ {B{r_2}} \right]^0}{\left[ {{H^ + }} \right]^1}$$
$$ = k\left[ {C{H_3}COC{H_3}} \right]\left[ {{H^ + }} \right]$$

Let the order of reaction with respect to $$A$$  is $$x$$  and with respect to $$B$$  is $$y.$$ Thus,
$$\eqalign{ & {\text{rate}} = \,k{\left[ A \right]^x}{\left[ B \right]^y} \cr & \left( {x{\text{ and }}y{\text{ are stoichiometric coefficient}}} \right) \cr & {\text{For the given cases,}} \cr & \left( {\text{i}} \right){\text{rate}} = k{\left( {0.1} \right)^x}{\left( {0.1} \right)^y} = 6.0 \times {10^{ - 3}} \cr & \left( {{\text{ii}}} \right){\text{rate}} = k{\left( {0.3} \right)^x}{\left( {0.2} \right)^y} = 7.2 \times {10^{ - 2}} \cr & \left( {{\text{iii}}} \right){\text{rate}} = k{\left( {0.3} \right)^x}{\left( {0.40} \right)^y} = 2.88 \times {10^{ - 1}} \cr & \left( {{\text{iv}}} \right){\text{rate}} = k{\left( {0.4} \right)^x}{\left( {0.1} \right)^y} = 2.40 \times {10^{ - 2}} \cr & {\text{Dividing Eq}}{\text{.}}\left( {\text{i}} \right){\text{by Eq}}{\text{.}}\left( {{\text{iv}}} \right){\text{,}}\,{\text{we get}} \cr & {\left( {\frac{{0.1}}{{0.4}}} \right)^x}{\left( {\frac{{0.1}}{{0.1}}} \right)^y} = \frac{{6.0 \times {{10}^{ - 3}}}}{{2.4 \times {{10}^{ - 2}}}} \cr & {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {\frac{1}{4}} \right)^x} = {\left( {\frac{1}{4}} \right)^1} \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1 \cr & {\text{On dividing Eq}}{\text{.}}\left( {{\text{ii}}} \right){\text{by Eq}}{\text{.}}\left( {{\text{iii}}} \right){\text{,}}\,{\text{we get}} \cr & {\left( {\frac{{0.3}}{{0.3}}} \right)^x}{\left( {\frac{{0.2}}{{0.4}}} \right)^y} = \frac{{7.2 \times {{10}^{ - 2}}}}{{2.88 \times {{10}^{ - 1}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{or}}{\left( {\frac{1}{2}} \right)^y} = \frac{1}{4} \cr & {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {\frac{1}{2}} \right)^y} = {\left( {\frac{1}{2}} \right)^2} \cr & \therefore \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y = 2 \cr & {\text{Thus, rate law is,}} \cr & {\text{rate}} = k{\left[ A \right]^1}{\left[ B \right]^2} \cr & \,\,\,\,\,\,\,\,\,\, = k\left[ A \right]{\left[ B \right]^2} \cr} $$

Linear plots are obtained in the graph of $$\left( {a - x} \right)$$  vs $$t$$  for zero order reaction and $$\frac{1}{{a - x}}$$  vs $$t$$  for second order reaction.

The molecularity of a reaction is the number of reactant molecules taking part in a single step of the reaction.
NOTE: The reaction involving two different reactant can never be unimolecular.

The slowest step determines the rate.

A catalyst affects equally both forward and backward reactions, therefore it does not affect equilibrium constant of reaction.