Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$2{N_2}{O_4} \rightleftharpoons 4N{O_2}$$
$${\text{Rate}} = - \frac{1}{2}\frac{{d\left[ {{N_2}{O_4}} \right]}}{{dt}} = + \frac{1}{4}\frac{{d\left[ {N{O_2}} \right]}}{{dt}}$$       $$ = \frac{1}{2}k = \frac{1}{4}k'\,\,{\text{or}}\,\,k' = 2k$$

Rate of a zero order reaction is independent of the concentration of reactants.

$$\eqalign{ & {\text{For the reaction,}} \cr & A + B \to {\text{Products}} \cr} $$
On doubling the initial concentration of $$A$$  only, the rate of the reaction is also doubled, therefore
$$\eqalign{ & {\text{Rate}} \propto {\left[ A \right]^1}\,\,\,...{\text{(i)}} \cr & {\text{Let initial rate law is}} \cr & {\text{Rate}} = k\left[ A \right]{\left[ B \right]^y}\,\,\,...{\text{(ii)}} \cr} $$
If concentration of $$A$$  and $$B$$  both are doubled, the rate gets changed by a factor of 8.
$$\eqalign{ & 8 \times {\text{rate}} = k\left[ {2A} \right]{\left[ {2B} \right]^y}\,\,\,...{\text{(iii)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\because \,{\text{Rate}} \propto {{\left[ A \right]}^1}} \right] \cr & {\text{Dividing Eq}}{\text{. (iii) by Eq}}{\text{. (ii), we get}} \cr & \,\,\,\,\,\,\,8 = 2 \times {2^y} \cr & \,\,\,\,\,\,\,4 = {2^y} \cr & {\left( 2 \right)^2} = {\left( 2 \right)^y} \cr & \therefore \,\,y = 2 \cr & {\text{Hence, rate law is, rate}} = k\left[ A \right]{\left[ B \right]^2} \cr} $$

TIPS/Formulae :
Find the order of reaction and then use appropriate equation.
As unit of $$k$$ is $${\sec ^{ - 1}}.$$  reaction is of first order,
$$\eqalign{ & r = k\left[ {{N_2}{O_5}} \right]; \cr & \therefore \,\left[ {{N_2}{O_5}} \right] = \frac{{2.4 \times {{10}^{ - 5}}}}{{3 \times {{10}^{ - 5}}}} \cr & = 0.8\,mol/L \cr} $$

Consider the following rate law equation,
$$\eqalign{ & \frac{{dx}}{{dt}} = k{\left[ A \right]^m}{\left[ {{B_2}} \right]^n} \cr & 1.6 \times {10^{ - 4}} = k{\left[ {0.50} \right]^m}{\left[ {0.50} \right]^n}\,\,\,...{\text{(i)}} \cr & 3.2 \times {10^{ - 4}} = k{\left[ {0.50} \right]^m}{\left[ {1.0} \right]^n}\,\,\,...{\text{(ii)}} \cr & 3.2 \times {10^{ - 4}} = k{\left[ {1.00} \right]^m}{\left[ {1.0} \right]^n}\,\,\,...{\text{(iii)}} \cr & {\text{By dividing Eq}}{\text{. (iii) by (ii) we get,}} \cr & \frac{{3.2 \times {{10}^{ - 4}}}}{{3.2 \times {{10}^{ - 4}}}} = \frac{{k{{\left[ {1.00} \right]}^m}{{\left[ {1.0} \right]}^n}}}{{k{{\left[ {0.50} \right]}^m}{{\left[ {1.0} \right]}^n}}} \cr & 1 = {2^m}\,\,\,\,\,{\text{or}}\,\,\,\,{{\text{2}}^0} = {2^m} \cr & \therefore \,\,\,m = 0 \cr & {\text{By dividing Eq}}{\text{. (ii) by (i)}} \cr & \frac{{3.2 \times {{10}^{ - 4}}}}{{1.6 \times {{10}^{ - 4}}}} = \frac{{{{\left[ {0.50} \right]}^m}{{\left[ {1.0} \right]}^n}}}{{{{\left[ {0.50} \right]}^m}{{\left[ {0.50} \right]}^n}}} \cr & 2 = {2^n} \cr & {\text{or}}\,\,\,{{\text{2}}^1} = {2^n} \cr & \therefore \,\,\,n = 1 \cr & {\text{Hence rate,}} \cr & \left( {\frac{{dx}}{{dt}}} \right) = k{\left[ A \right]^0}{\left[ {{B_2}} \right]^1} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = k\left[ {{B_2}} \right] \cr} $$

$$O_3\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}{O}_2+\left[O\right]\left(\text{fast}\right)$$
\[\left[ O \right]+{{O}_{3}}\xrightarrow{{{k}_{2}}}2{{O}_{2}}\left( \text{slow} \right)\]
Rate of reaction is determined by slow step hence, $${\text{Rate}} = {k_2}\left[ O \right]\left[ {{O_3}} \right]$$
$$\left[ O \right]$$  is unstable intermediate so substitute the value of $$\left[ O \right]$$  in above equation.
Rate of forward reaction $$ = {k_1}\left[ {{O_3}} \right]$$
Rate of backward reaction $$ = {k_{ - 1}}\left[ {{O_2}} \right]\left[ O \right]$$
At equilibrium,
Rate of forward reaction = Rate of backward reaction
$${k_1}\left[ {{O_3}} \right] = {k_{ - 1}}\left[ {{O_2}} \right]\left[ O \right];$$     $$\left[ O \right] = \frac{{{k_1}\left[ {{O_3}} \right]}}{{{k_{ - 1}}\left[ {{O_2}} \right]}}$$
$${\text{Rate}} = {k_2}\left( {\frac{{{k_1}\left[ {{O_3}} \right]}}{{{k_{ - 1}}\left[ {{O_2}} \right]}}} \right)\left[ {{O_3}} \right]$$       $$ = \frac{{k{{\left[ {{O_3}} \right]}^2}}}{{\left[ {{O_2}} \right]}}$$

For first order reaction, we know that
$${\text{log}}\left[ A \right] = - \frac{{kt}}{{2.303}} + {\text{log}}{\left[ A \right]_0}$$
On comparing it with the equation of straight line,
$${\text{i}}{\text{.e}}{\text{.}}\,\,\,y = mx + c$$
Plot of $$\log \left[ A \right]$$  versus time → straight line
$${\text{slope}} = \frac{{ - k}}{{2.303}}\left( {{\text{negative}}} \right)$$

The rate constant of a first order reaction has only time unit. It has no concentration unit. This means the numerical value of $$k$$  for a first order reactions is independent of the unit in which concentration is expressed.

The plot of $$\ln \,k\,\,vs\,\,\frac{1}{T}$$   gives a straight line according to the equation, $$\ln k = - \frac{{{E_a}}}{{RT}} + \ln A$$

Graph (C) represents a large gap between energy of products and reactants.