Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

The activation energy of a reverse reaction decide whether the given reaction is exothermic or endothermic, so, the energy of activation of reverse reaction is either greater or less than 50 $$kcal.$$  In case of exothermic reaction, the activation energy for reverse reaction is more than activation energy of the forward reaction and in case of endothermic reaction, the activation energy for reverse reaction is less than activation energy of the forward reaction.

$$\eqalign{ & {\text{For a zero order reaction,}} \cr & {\text{rate}} = k = \frac{{dx}}{{dt}} \cr & {\text{Units of}}\,\,k = mol\,{L^{ - 1}}\,{s^{ - 1}} \cr} $$

\[P{{H}_{3}}\xrightarrow{W}P+\frac{3}{2}{{H}_{2}}\]
This is an example of surface catalysed unimolecular decomposition.
For the above reaction, rate is given as
$${\text{Rate}} = \frac{{k\alpha p}}{{1 + \alpha p}}$$
where, $$p =$$  partial pressure of absorbing substrate.
At low pressure, $$\alpha p < < 1\,\,{\text{or}}\,\,{\text{Rate = }}k\alpha p$$
So, $$\left( {\alpha p + 1} \right)$$  can be neglected.
Thus, the decomposition is predicted to be first order.

Activation energy can be calculated from the equation
$$\eqalign{ & \frac{{{\text{log}}\,{K_2}}}{{{\text{log}}\,{K_1}}} = \frac{{ - {E_a}}}{{2.303\,R}}\left( {\frac{1}{{{T_2}}} - \frac{1}{{{T_1}}}} \right) \cr & {\text{Given}}\,\frac{{{\text{log}}\,{K_2}}}{{{\text{log}}\,{K_1}}} = 2\,{T_2} = 308\,K;\,{T_1} = 298\,K \cr & \therefore \,\,{\text{log}}\,2 = \frac{{ - {E_a}}}{{2.303 \times 8.314}}\left( {\frac{1}{{308}} - \frac{1}{{298}}} \right) \cr & {E_a} = 52.9\,kJ\,mo{l^{ - 1}} \cr} $$

$${t_{\frac{1}{2}}}\,\,{\text{of}}\,\,{n^{th}}\,\,{\text{order reaction}} \propto \frac{1}{{{a^{n - 1}}}}$$
where, $$a=$$  initial concentration of reactant
             $$n=$$  order of reaction
$$\therefore \,\,{t_{\frac{1}{2}}}$$   for first order reaction $$\left( {n = 1} \right)$$
$$\eqalign{ & {t_{\frac{1}{2}}} \propto \frac{1}{{{a^{1 - 1}}}} \cr & {\text{or}}\,\,\,{t_{\frac{1}{2}}} \propto \frac{1}{{{a^0}}}\,\,\,\,\left( {{a^0} = 1} \right) \cr} $$
So, for a first order reaction half-life is independent on initial concentration of reactants.

$$\eqalign{ & {\text{For the reaction,}} \cr & {N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right) \cr & {\text{The rate of reaction }}wrt \cr} $$
$${N_2} = - \frac{{d\left[ {{N_2}} \right]}}{{dt}}$$     [ Rate of disappearance ]
The rate of reaction with respect to
$${H_2} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}}$$     [ Rate of disappearance ]
The rate of reaction with respect to
$$N{H_3} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}}$$     [ Rate of appearance ]
$$\eqalign{ & {\text{Hence, at a fixed time}} \cr & - \frac{{d\left[ {{N_2}} \right]}}{{dt}} = - \frac{1}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} = + \frac{1}{2}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} \cr & {\text{or}}\,\,\,{\text{ + }}\frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{2}{3}\frac{{d\left[ {{H_2}} \right]}}{{dt}} \cr & {\text{or}}\,\, + \frac{{d\left[ {N{H_3}} \right]}}{{dt}} = - \frac{{2d\left[ {{N_2}} \right]}}{{dt}} \cr} $$

For slowest step : rate $$ = k\left[ {NOB{r_2}} \right]\left[ {NO} \right]...\left( {\text{i}} \right)$$
For equilibrium also $${K_c} = \frac{{\left[ {NOB{r_2}} \right]}}{{\left[ {NO} \right]\left[ {B{r_2}} \right]}}...\left( {{\text{ii}}} \right)$$
By eqs. (i) and (ii), $$r = k \cdot {K_c}{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$
Rate $$ = k{\left[ {NO} \right]^2}\left[ {B{r_2}} \right]$$

$${N_t} = {N_0}{\left( {\frac{1}{2}} \right)^n}$$ where n is number of halflife periods.
$$\eqalign{ & n = \frac{{{\text{Total}}\,{\text{time}}}}{{{\text{half}}\,{\text{life}}}} \cr & = \frac{{24}}{4} \cr & = 6 \cr & \therefore \,{N_t} = 200{\left( {\frac{1}{2}} \right)^6} \cr & = 3.125g \cr} $$

There are $$5\,tens$$  hence $${\left( 2 \right)^5} = 32.$$

$$\eqalign{ & {\text{For first order reaction,}} \cr & k = \frac{{2.303}}{t}\log \frac{a}{{\left( {a - x} \right)}} \cr & \,\,\,\, = \frac{{2.303}}{{200}}\log \frac{{400}}{{25}} \cr & \,\,\,\, = \frac{{2.303}}{{200}} \times 1.204 \cr & \,\,\,\, = 0.01386\,{s^{ - 1}} \cr & \,\,\,\, = 1.386 \times {10^{ - 2}}{s^{ - 1}} \cr} $$