Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & k = A{e^{ - \,\frac{{{E_a}}}{{RT}}}} \cr & \therefore \,\,\frac{{ - {E_a}}}{R} = - 40000 \cr & \therefore \,\,{E_a} = 40000 \times 2 = 80000\,cal \cr} $$

In Arrhenius equation $$k = A{e^{ - \,\frac{{{E_a}}}{{RT}}}},{E_a}$$    is the energy of activation, which is required by the colliding molecules to react resulting in the formation of products.

$$\eqalign{ & 2{N_2}{O_5} \to 4N{O_2} + {O_2} \cr & \frac{{ - d\left[ {{N_2}{O_5}} \right]}}{{dt}} = k.\left[ {{N_2}{O_5}} \right] \cr & 1.02 \times {10^{ - 4}} = 3.4 \times {10^{ - 5}}{s^{ - 1}} \times \left[ {{N_2}{O_5}} \right] \cr & \therefore \,\,\left[ {{N_2}{O_5}} \right] = \frac{{1.02 \times {{10}^{ - 4}}}}{{3.4 \times {{10}^{ - 5}}}} = 3 \cr} $$

$$Rt = \log {C_o} - \log {C_t}$$
It is clear from the equation that if we plot a graph between $$\log {C_t}$$  and time, a straight line with a slope equal to $$ - \frac{k}{{2.303}}$$  and intercept equal to log $$\left[ {{A_0}} \right]$$  will be obtained.

If $$x$$ $$atm$$  be the decrease in pressure of $$A$$  at time $$t$$  and one mole each of $$B$$  and $$C$$  is being formed, the increase in pressure of $$B$$  and $$C$$  will also be $$x$$ $$atm$$  each.
\[\begin{matrix} {} & {{A}_{\left( g \right)}} & \to \\ \text{At}\,\,t=0 & {{p}_{i}}\,\,atm & {} \\ \text{At time}\,\,t & \left( {{p}_{i}}-x \right)atm & {} \\ \end{matrix}\,\,\,\,\,\,\,\,\,\,\begin{matrix} {{B}_{\left( g \right)}} & + & {{C}_{\left( g \right)}} \\ 0\,atm & {} & 0\,atm \\ x\,atm & {} & x\,atm \\ \end{matrix}\]

where, $${p_t}$$  is the initial pressure at time $$t = 0$$
$$\eqalign{ & {p_t} = \left( {{p_i} - x} \right) + x + x = {p_i} + x \cr & x = \left( {{p_t} - {p_i}} \right) \cr} $$
where, $${p_A} = {p_i} - x = {p_i} - \left( {{p_t} - {p_i}} \right)$$       $$ = 2{p_i} - {p_t}$$
$$\eqalign{ & k = \left( {\frac{{2.303}}{t}} \right)\left( {\log \frac{{{p_i}}}{{{p_A}}}} \right) \cr & \,\,\,\, = \frac{{2.303}}{t}\log \frac{{{p_i}}}{{\left( {2{p_i} - {p_t}} \right)}} \cr} $$

For $${10^ \circ }$$ rise in temperature, $$n = 1$$
$${\text{so rate}} = {2^n} = {2^1} = 2$$
When temperature is increased from $${10^ \circ }C$$  to $${100^ \circ }C,$$  change in temperature
$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = 100 - 10 = {90^ \circ }C \cr & {\text{i}}{\text{.e}}{\text{.}}\,\,n = 9 \cr & {\text{so,}}\,{\text{rate}} = {2^9} = 512\,{\text{times}} \cr} $$
Alternate method with every $${10^ \circ }$$ rise in temperature, rate becomes double,
So $$\frac{{r'}}{r} = {2^{\left( {\frac{{100 - 10}}{{10}}} \right)}} = {2^9} = 512\,{\text{times}}{\text{.}}$$

It is a second order reaction, first order both $$w.r.t\,{S_2}O_8^{2 - }$$   and $${I^ - }\,.$$
$$\therefore \,\,r = k\left[ {{S_2}{O_8}^{2 - }} \right]\left[ {{I^ - }} \right]$$
All other options are of first order reaction.

For the reaction
$$3A + 2B \to C + D$$
Rate of disappearance of $$A=$$  Rate of appearance of $$C$$ reaction
$$\eqalign{ & = - \frac{1}{3}\frac{{d\left[ A \right]}}{{dt}} \cr & = \frac{{d\left[ C \right]}}{{dt}} \cr & = k{\left[ A \right]^n}{\left[ B \right]^m} \cr} $$

$$\eqalign{ & {\text{For reaction,}} \cr & 3A \to 2B \cr} $$
$${\text{Rate}} = - \frac{1}{3}\frac{{d\left[ A \right]}}{{dt}}$$     $${\text{[Rate of disappearance]}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\, = + \frac{1}{2}\frac{{d\left[ B \right]}}{{dt}}$$     $$[{\text{Rate of appearance]}}$$
$$\therefore \,\,{\text{ + }}\frac{{d\left[ B \right]}}{{dt}} = - \frac{2}{3}\frac{{d\left[ A \right]}}{{dt}}$$

For the first order reaction for small finite change
$${k_1} = \frac{1}{{\left[ A \right]}}\frac{{\Delta \left[ A \right]}}{{\Delta t}} \Rightarrow \frac{{\frac{{\Delta \left[ A \right]}}{{\left[ A \right]}}}}{{\Delta t}} = 1.5\% $$       \[{{\min }^{-1}}\]
\[=0.015\,{{\min }^{-1}}\]
\[{{t}_{\frac{1}{2}}}=\frac{0.693}{0.015\,{{\min }^{-1}}}=46.2\,\min \approx 46\,\min \]