Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

In the absence of $$A,3\,dps$$   is zero error, hence
Initial count $$ = 23 - 3 = 20\,dps$$
After $$10\,\min = 13 - 3 = 10\,dps$$
After $$20\,\min = 5\,dps$$
( recorded $$ = 5 + 3 = 8\,dps$$    )
( $$50\% $$  fall in $$10\,\min ,{T_{50}} = 10\,\min $$     )

Instantaneous rate can be determined graphically by drawing a tangent on the curve.
Thus, volume change during 50 - 30 $$sec .$$  interval $$ = {V_5} - {V_2}$$
and not $${V_4} - {V_2}.$$   Hence, $$\frac{{{V_4} - {V_2}}}{{50 - 30}}$$   does not show instantaneous rate of reaction at 40^th second.

$$\eqalign{ & M \to N \cr & r = k{\left[ M \right]^x} \cr & {\text{when}}\,M = 2M;\,r = 8r,\,{\text{thus}} \cr & 8r = k{\left[ {2M} \right]^x} \cr & 8 = {\left( 2 \right)^x} \cr & x = 3 \cr} $$

$$2N{H_3} \to {N_2} + 3{H_2}$$
$${\text{Rate}} = - \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = \frac{{d\left[ {{N_2}} \right]}}{{dt}}$$     $$ = \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$
$$\frac{1}{2}{k_1}\left[ {N{H_3}} \right] = {k_2}\left[ {N{H_3}} \right] = \frac{1}{3}{k_3}\left[ {N{H_3}} \right];$$       $$1.5{k_1} = 3{k_2} = {k_3}$$

$$\eqalign{ & {\text{Initially rate }}\left( {{r_1}} \right) = k{\left[ X \right]^n}...\left( {\text{i}} \right) \cr & {\text{(let }}n{\text{ be the order of reaction)}} \cr & {\text{then,}}\,{r_1} \times 1.837 = k{\left( {1.5X} \right)^n}...\left( {{\text{ii}}} \right) \cr & {\text{dividing (ii) by (i)}} \cr & 1.837 = {\left( {1.5} \right)^n};\,\,\,\,\therefore \,\,n = 1.5 \cr & {\text{Thus, order of reaction is 1}}{\text{.5}}{\text{.}} \cr} $$

Order of a reaction can be fractional. Rest of all are true.
[ NOTE : Order of a reaction can be determined experimentally ]

From experiment I and II, it is observed that order of reaction $$w.r.t.$$  $$(c)$$ is zero.
From experiment II and III, $$\alpha $$ can be calculated as :
$$\eqalign{ & \frac{{1.386 \times {{10}^{ - 2}}}}{{6.93 \times {{10}^{ - 3}}}} = {\left( {\frac{{0.2}}{{0.1}}} \right)^\alpha }\,\,\therefore \,\,\alpha = 1 \cr & {\text{Now,}}\,\,{\text{Rate}} = \,K{\left[ A \right]^1} \cr & {\text{or}},\,\,6.93 \times {10^{ - 3}} = K\left( {0.1} \right) \cr & K = 6.93 \times {10^{ - 2}} \cr & {\text{For the reaction,}}\,{\text{2A + B}} \to {\text{Products}} \cr & {\text{2}}Kt = \ln \frac{{{{\left[ A \right]}_0}}}{{\left[ A \right]}} \cr & \therefore \,\,{t_{\frac{1}{2}}} = \frac{{0.693}}{{2K}} = \frac{{0.693}}{{0.693 \times {{10}^{ - 2}} \times 2}} \cr & {t_{\frac{1}{2}}} = 5 \cr} $$

We know that the activation energy of chemical reaction is given by formula
$$\frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303R}}\left[ {\frac{{{T_2} - {T_1}}}{{{T_1}{T_2}}}} \right],$$       where $${{k_1}}$$  is the rate constant at temperature $${T_1}$$  and $${k_2}$$  is the rate constant at temperature $${T_2}$$  and $${E_a}$$ is the activation energy. Therefore activation energy of chemical reaction is determined by evaluating rate constant at two different temperatures.

Let the initial concentration $$\left( A \right) = 100$$
Final concentration at $$298\,K = 100 - 10 = 90$$
Final concentration at $$308\,K = 100 - 25 = 75$$
Substituting the values in the 1^st order rate reaction
$$\eqalign{ & t = \frac{{2.303}}{{{k_{298}}}}{\text{log}}\frac{{100}}{{90}}\,\,...\left( {\text{i}} \right) \cr & t = \frac{{2.303}}{{{k_{308}}}}{\text{log}}\frac{{100}}{{75}}\,\,...\left( {{\text{ii}}} \right) \cr & {\text{From (i) and (ii)}}\,\frac{{{k_{308}}}}{{{k_{208}}}} = 2.73 \cr} $$
Substituting the value in the following relation
$$\eqalign{ & {E_a} = \frac{{2.303R \times {T_1} \times {T_2}}}{{{T_2} - {T_1}}}{\text{log}}\frac{{{k_2}}}{{{k_1}}} \cr & = \frac{{2.303 \times 8.314 \times 298 \times 308}}{{308 - 298}}{\text{log}}\,2.73 \cr & {E_a} = 76.62\,kJ\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\, = 18.39\,kcal\,mo{l^{ - 1}} \cr} $$

In equation $$k = A{e^{\frac{{ - {E_a}}}{{RT}}}};A = $$     Frequency factor $$K =$$  velocity constant, $$R = $$  gas constant and $${E_a} = $$  energy of activation