Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & \left[ {{N_2}{O_5}} \right] = \frac{{{\text{Rate}}}}{k} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{1.02 \times {{10}^{ - 4}}\,mol\,{L^{ - 1}}{s^{ - 1}}}}{{3.4 \times {{10}^{ - 5}}\,{s^{ - 1}}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\,mol\,{L^{ - 1}} \cr} $$

$$\eqalign{ & {\text{For}}\,{\text{zero}}\,{\text{order}}\,{\text{reaction,}} \cr & = k{\left[ A \right]^{0\,}}\,i.e.\,rate = k \cr & {\text{hence}}\,{\text{unit}}\,{\text{of}}\,k = M.{\sec ^{ - 1}} \cr & {\text{For}}\,{\text{the}}\,{\text{fast}}\,{\text{order}}\,{\text{reaction,}} \cr & {\text{rate}} = k\left[ A \right] \cr & k = \frac{{M.{{\sec }^{ - 1}}}}{M} \cr & = {\sec ^{ - 1}} \cr} $$

As per Arrhenius equation $$\left( {k = A{e^{ - {E_a}/RT}}} \right),$$    the rate constant increases exponentially with temperature.

$$\eqalign{ & {\text{For}}\,\,{\text{the}}\,\,{\text{fast}}\,\,{\text{order}}\,\,{\text{reaction,}} \cr & k = \frac{{2.303}}{t}\log \frac{{100}}{{100 - 99}} \cr & \frac{{0.693}}{{6.93}} = \frac{{2.303}}{t}\log \frac{{100}}{1} \cr & \frac{{0.693}}{{6.93}} = \frac{{2.303 \times 2}}{t} \Rightarrow t = 46.06\,\min \cr} $$

For $$P,$$ if $${t_{50\% }} = X$$  then $${t_{75\% }} = 2x$$
This is true only for first order reaction.
So, order with respect to $$P$$ is 1.
Further the graph shows that concentration of $$Q$$ decreases with time. So rate, with respect to $$Q,$$ remains constant. Hence, it is zero order wrt $$Q.$$
So, overall order is $$1+0=1$$

It is a constant of a particular reaction at a given temperature. It does not depend upon initial concentration of the reactants, time of reaction and extent of reaction.

$$\eqalign{ & \frac{{d\left[ {I{O^ - }} \right]}}{{dt}} = k\frac{{{{\left[ {{I^ - }} \right]}^1}{{\left[ {OC{l^ - }} \right]}^1}}}{{{{\left[ {O{H^ - }} \right]}^1}}} \cr & {\text{Order of reaction}} = 1 + 1 - 1 = 1 \cr} $$

The equilibrium is attained faster in case of reaction in which catalyst is used as it lowers the activation energy.

For exothermic reaction, $$\Delta H = - ve$$
$$\Delta H = {E_{a{\text{(forward}}\,\,{\text{reaction)}}}} - {E_{a{\text{(backward}}\,\,{\text{reaction)}}}}$$
For $$\Delta H$$  to be $$ - ve,{E_{a,b}} > {E_{a,f}}$$     which is true in graph (I) only.

$$\eqalign{ & {\text{For a zero order reaction,}} \cr & {\text{rate}} = k{\left[ A \right]^0} = k \cr & {\text{Units}} = mol\,{L^{ - 1}}\,{\text{tim}}{{\text{e}}^{ - 1}} \cr} $$