Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

The rate of disappearance of $$S{O_2}$$  and the rate of formation of $$S{O_3}$$  are same.

$$\eqalign{ & {e^{ - \,\frac{{{E_a}}}{{RT}}}} = {10^{ - 3}}\% = {10^{ - 5}}; \cr & {E_a} = 2.303 \times 2 \times 300 \times 5\,cal \cr & = 6.91\,k\,cal\,mo{l^{ - 1}} \cr} $$

TIPS/Formulae :
$$\eqalign{ & k = \frac{{2.303}}{t}\log \frac{{{{\left[ A \right]}_0}}}{{{{\left[ A \right]}_t}}} \cr & \therefore \,\,k = \frac{{2.303}}{t}\log \frac{{800}}{{50}} \cr & = 1.386 \times {10^{ - 4}}{s^{ - 1}} \cr} $$

NOTE : Individual rates of reactants and products become equal when each of these is divided by their respective stoichiometric coefficient. With time concentration of reactants decreases and is represented by negative sign whereas concentration of products increases and is represented. by positive sign.
The given reaction is
$${N_2}\left( g \right) + 3{H_2}\left( g \right) \to 2N{H_3}\left( g \right)$$
∴ Correct relationship amongst the rate expression is shown in (A)

The order of reaction is $$\frac{3}{2}$$  and molecularity is 2.

The definition of threshold energy.

Reactions of higher order (>3) are very rare due to very less chances of many molecules to undergo effective collisions.

The species $$_{13}A{l^{29}}$$  ( No. of neutrons = 16 ) contains more neutrons than the stable isotope $$_{13}A{l^{27}}$$  ( No. of neutrons= 14 ) .
Neutron on decomposition shows $$\beta $$ -emission.
$$\eqalign{ & _0{n^1}{ \to _{ + 1}}{p^1}{ + _{\, - 1}}{e^0} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\beta - {\text{particle}} \cr} $$

$${\text{Rate}} = k{\left[ A \right]^x}{\left[ B \right]^y}$$
$${\text{From exp}}{\text{.}}\left( 1 \right),5 \times {10^{ - 4}}$$     $$ = k{\left( {2.5 \times {{10}^{ - 4}}} \right)^x}{\left( {3 \times {{10}^{ - 5}}} \right)^y}...\left( {\text{i}} \right)$$
$${\text{From exp}}.\left( 2 \right),4 \times {10^{ - 3}}$$     $$ = k{\left( {5 \times {{10}^{ - 4}}} \right)^x}{\left( {6 \times {{10}^{ - 5}}} \right)^y}...\left( {{\text{ii}}} \right)$$
$${\text{Dividing }}\left( {{\text{ii}}} \right){\text{ by }}\left( {\text{i}} \right){\text{,}}\,\frac{{4 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 4}}}}$$       $$ = {2^x} \cdot {2^y} = 8$$
$${\text{From exp}}{\text{.}}\left( 3 \right),\,1.6 \times {10^{ - 2}}$$     $$ = k{\left( {1 \times {{10}^{ - 3}}} \right)^x}{\left( {6 \times {{10}^{ - 5}}} \right)^y}...\left( {{\text{iii}}} \right)$$
$${\text{Dividing }}\left( {{\text{iii}}} \right){\text{ by }}\left( {{\text{ii}}} \right){\text{,}}\frac{{1.6 \times {{10}^{ - 2}}}}{{4 \times {{10}^{ - 3}}}}$$       $$ = {2^x} = 4$$
$${\text{or}}\,\,x = 2,y = 1$$
Hence, order with respect to $$A$$  is 2 and with respect to $$B$$ is 1.

$$\Delta H = {E_a}\left( f \right) - {E_a}\left( b \right)$$
Thus energy of activation for reverse reaction depend upon whether reaction is exothermic or endothermic.
$$\eqalign{ & {\text{If reaction is exothermic,}} \cr & \Delta H = - ve,{E_a}\left( b \right) > {E_a}\left( f \right) \cr & {\text{If reaction is endothermic,}} \cr & \Delta H = + ve\,\,{E_a}\left( b \right) < {E_a}\left( f \right) \cr} $$