Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & C{H_3}CHO \to C{H_4} + CO \cr & {\text{Rate}} = k{\left[ {C{H_3}CHO} \right]^{\frac{3}{2}}} \cr} $$

$${\text{Rate of appearance/disappearance}}$$
\[=\pm \frac{1}{\begin{align} & \text{stoichiometric} \\ & \text{coefficient} \\ \end{align}}\times \]    $$\frac{{{\text{[reactant or product]}}}}{{{\text{time taken}}}}$$
$${\text{For the reaction,}}$$
$$BrO_3^ - \left( {aq} \right) + 5B{r^ - }\left( {aq} \right) + 6{H^ + }$$      $$ \to 3B{r_2}\left( l \right) + 3{H_2}O\left( l \right)$$
$${\text{Rate of appearance of bromine}}$$       $$\left( {B{r_2}} \right)$$
$$ = + \frac{1}{3}\frac{{d\left[ {B{r_2}} \right]}}{{dt}}$$
$${\text{Rate of disappearance of bromide}}$$       $$ion\,\left( {B{r^ - }} \right)$$
$$\eqalign{ & = - \frac{1}{5}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}} \cr & {\text{or}}\,\,\,\frac{{d\left[ {B{r_2}} \right]}}{{dt}} = - \frac{3}{5}\frac{{d\left[ {B{r^ - }} \right]}}{{dt}} \cr} $$

$$\eqalign{ & \log k = \log A - \frac{{{E_a}}}{{2.303\,RT}} \cr & \log k = \log \left( {1.45 \times {{10}^{11}}} \right) - \frac{{35 \times {{10}^3}}}{{2.303 \times 2 \times 573}} \cr & \log \,k = 11.16 - 13.26 = - 2.1 \cr & {\text{Taking antilog,}}\,k = 7.94 \times {10^{ - 3}}\,{s^{ - 1}} \cr} $$

If we write rate of reaction in terms of concentration of $${N{H_3}}$$  and $${H_2},$$  then
Rate of reaction $$ = \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
So, $$\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{2}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$

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

$$\eqalign{ & {t_{\frac{1}{4}}} = \frac{{2.303}}{k}{\text{log}}\frac{1}{{\frac{3}{4}}} \cr & = \frac{{2.303}}{k}{\text{log}}\frac{4}{3} \cr & = \frac{{2.303}}{k}\left( {{\text{log}}\,4 - {\text{log}}\,3} \right) \cr & = \frac{{2.303}}{k}\left( {2\,{\text{log}}\,2 - {\text{log}}3} \right) \cr & = \frac{{2.303}}{k}\left( {2 \times 0.301 - 0.4771} \right) \cr & = \frac{{0.29}}{k} \cr} $$

Such reaction in which two lighter nucleus are fused together to form a heavier nuclei is called nuclear fusion.
NOTE : In hydrogen bomb, a mixture of deuterium oxide $$\left( {{H_2}O} \right)$$  and tritium Oxide $$\left( {{T_2}O} \right)$$  is enclosed in a space surrounding an ordinary atomic bomb. The temperature produced by the atomic bomb initiates the fusion reaction between $$_1{H^3}$$ and $$_1{H^2}$$ releasing a large amount of energy.

Let the order of reaction with respect to $$A$$  and $$B$$  is $$x$$  and $$y$$  respectively. So, the rate law can be given as
$$R = k{\left[ A \right]^x}{\left[ B \right]^y}\,...\left( {\text{i}} \right)$$
When the concentration of only $$B$$  is doubled, the rate is doubled, so
$${R_1} = k{\left[ A \right]^x}{\left[ {2B} \right]^y} = 2R\,...\left( {{\text{ii}}} \right)$$
If concentrations of both the reactants $$A$$  and $$B$$  are doubled, the rate increases by a factor of 8, so
$$\eqalign{ & R'' = k{\left[ {2A} \right]^x}{\left[ {2B} \right]^y} = 8R\,\,\,...\left( {{\text{iii}}} \right) \cr & \Rightarrow k{2^x}{2^y}{\left[ A \right]^x}{\left[ B \right]^y} = 8R\,\,\,...\left( {{\text{iv}}} \right) \cr & {\text{From Eqs}}{\text{. (i) and (ii), we get}} \cr & \Rightarrow \frac{{2R}}{R} = \frac{{{{\left[ A \right]}^x}{{\left[ {2B} \right]}^y}}}{{{{\left[ A \right]}^x}{{\left[ B \right]}^y}}} \cr & 2 = {2^y} \cr & \therefore \,\,y = 1 \cr & {\text{From Eqs}}{\text{. (i) and (iv), we get}} \cr & \Rightarrow \,\frac{{8R}}{R} = \frac{{{2^x}{2^y}{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}{{{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}\,\,{\text{or}}\,\,8 = {2^x}{2^y} \cr & {\text{Substitution of the value of }}y{\text{ gives,}} \cr & {\text{8 = }}{{\text{2}}^x}{2^1} \cr & 4 = {2^x} \cr & {\left( 2 \right)^2} = {\left( 2 \right)^x} \cr & \therefore x = 2 \cr} $$
Substitution of the value of $$x$$  and $$y$$  in Eq. (i) gives,
$$R = k{\left[ A \right]^2}\left[ B \right]$$

$$\eqalign{ & A \to B,\Delta H = - 10\,kJ\,mo{l^{ - 1}} \cr & {\text{It is an exothermic reaction}}{\text{.}} \cr & {E_{a\left( b \right)}} = {E_{a\left( f \right)}} - \left( {\Delta H} \right) \cr & = 50 - \left( { - 10} \right) \cr & = 60\,kJ \cr} $$

$$\eqalign{ & {e^{ - \frac{{{E_a}}}{{RT}}}} = 3.8 \times \frac{{{{10}^{ - 16}}}}{{100}} \cr & - \frac{{{E_a}}}{{RT}} = ln\,3.8 \times {10^{ - 18}} \cr & {E_a} = 100\,kJ/mol \cr} $$