Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\frac{n}{p}$$  ratio of $$^{24}Na$$  nuclide is $$\frac{{13}}{{11}}$$  i.e. greater than unity and hence $$^{24}Na$$  is radioactive. To achieve stability, it would tend to adjust its $$\frac{n}{p}$$  ratio to the proper value of unity. This can be done by breaking a neutron into proton and electron.
$$_0{n^1}{ \to _{ + 1}}{P^1}{ + _{ - e}}{e^0}\,\,or\,{\beta ^ - }$$
NOTE: The proton will stay inside the nucleus whereas electron which cannot exist in the nucleus, will be emitted out as $$\beta $$ -ray.

After every 30 minutes the amount is reduced to $$\frac{1}{2}$$  therefore $${t_{\frac{1}{2}}}$$  is 30 minutes. In 90 minutes the amount is reduced to $$\frac{1}{8}$$  i.e. $$\frac{1}{{{2^n}}}.$$  Here $$n = 3.$$  True for 1^st order reaction.

Lower the activation energy, faster is the reaction.

$$\eqalign{ & {\text{Given}} - \frac{{dMnO_4^ - }}{{dt}} = 4.56 \times {10^{ - 3}}M{s^{ - 1}} \cr & {\text{From the reaction given,}} \cr & - \frac{1}{2}\frac{{dMnO_4^ - }}{{dt}} = \frac{{4.56 \times {{10}^{ - 3}}}}{2}M{s^{ - 1}} \cr & - \frac{1}{2}\frac{{dMnO_4^ - }}{{dt}} = \frac{1}{5}\frac{{d{I_2}}}{{dt}} \cr & \therefore \,\, - \frac{5}{2}\frac{{dMnO_4^ - }}{{dt}} = \frac{{d{I_2}}}{{dt}} \cr & {\text{On substituting the given value}} \cr & \therefore \,\,\frac{{d{I_2}}}{{dt}} = \frac{{4.56 \times {{10}^{ - 3}} \times 5}}{2} \cr & = 1.14 \times {10^{ - 2}}M/s \cr} $$

The rate constant doubles for $${10^ \circ }C$$  rise in temperature.
For $${20^ \circ }C$$  rise, the rate constant will be 4 times
$$\therefore \,\,{k_1} = 4{k_2}\,\,{\text{or}}\,\,{k_2} = 0.25\,{k_1}$$

$$\eqalign{ & _{92}^{238}M \to _{88}^{230}N + 2_2^4He \cr & _{88}^{230}N \to _{86}^{230}L + 2{\beta ^ + } \cr} $$
The number of neutrons in element $$L = 230 - 86 = 144$$

$$\eqalign{ & {\text{For the reaction}} \cr & A \to {\text{ Product ;}}\,{\text{given }}{t_{\frac{1}{2}}} = 1\,{\text{hour}} \cr & {\text{for a zero order reaction}} \cr & {{\text{t}}_{{\text{completion}}}} = \frac{{\left[ {{A_0}} \right]}}{k} = \frac{{{\text{initial conc}}{\text{.}}}}{{{\text{rate constant}}}} \cr} $$
$$\therefore \,\,\,{t_{\frac{1}{2}}} = \frac{{\left[ {{A_0}} \right]}}{{2k}}\,{\text{or}}\,k = \frac{{\left[ {{A_0}} \right]}}{{2{t_{\frac{1}{2}}}}} = \frac{2}{{2 \times 1}}$$       $$ = 1\,mol\,li{t^{ - 1}}h{r^{ - 1}}$$
$$\eqalign{ & {\text{Further}}\,\,{\text{for}}\,\,{\text{a}}\,\,{\text{zero}}\,\,{\text{order}}\,\,{\text{reaction}} \cr & k = \frac{{dx}}{{dy}} = \frac{{{\text{change in concentration}}}}{{{\text{time}}}} \cr & 1 = \frac{{0.50 - 0.25}}{{time}}\,\,\,\therefore \,\,{\text{time}} = 0.25\,hr. \cr} $$

$$\eqalign{ & {\text{Rate}} = \frac{1}{4}\frac{{d\left[ {N{O_2}} \right]}}{{dt}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{4} \times 0.0125 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0031\,mol\,{L^{ - 1}}\,{s^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Specific rate constant,}} \cr & k = \frac{{0.693}}{{{t_{\frac{1}{2}}}}} \cr & \,\,\,\,\, = \frac{{0.693}}{{1386}} \cr & \,\,\,\,\, = 0.5 \times {10^{ - 3}}{s^{ - 1}} \cr} $$

As the concentration of reactant decreases from 0.8 to 0.4 in 15 minutes hence the $${t_{\frac{1}{2}}}$$ is 15 minutes. To fall the concentration from 0.1 to 0.025 we need two half lives i.e., 30 minutes.