Chemical Kinetics MCQ Questions & Answers in Physical Chemistry | Chemistry

For the exothermic reaction the energy of products is always less than the reactants. If $${E_a}$$ is the energy of activation for the forward reaction, the energy of activation for backward reaction is $${E_a} + \Delta H$$

$$\eqalign{ & {\text{For second order, rate constant}} \cr & = \frac{{mol\,{L^{ - 1}}}}{s} \times \frac{1}{{{{\left( {mol\,{L^{ - 1}}} \right)}^2}}} \cr & = mo{l^{ - 1}}\,L\,{s^{ - 1}} \cr & {\text{For first order, rate constant}} \cr & = \frac{{mol\,{L^{ - 1}}}}{s} \times \frac{1}{{mol\,{L^{ - 1}}}} = {s^{ - 1}} \cr} $$

When a biochemical reaction is carried out in laboratory from outside of human body in the absence of enzyme, then rate of reaction obtained is $${10^{ - 6}}$$  times than activation energy of reaction in the presence of enzyme. It is different from $${E_a}$$  obtained in laboratory because for a given chemical reaction.
$$k = A{e^{ - \frac{{{E_a}}}{{RT}}}}{\text{(arrhenius equation)}}$$
Also activation energy have different value in absence or presence of enzyme.

For a first order reaction
$$\eqalign{ & k = \frac{{2.0303}}{t}\log \frac{a}{{a - x}} \cr & = \frac{{2.303}}{{40}}\log \frac{{0.1}}{{0.025}} \cr & = \frac{{2.303}}{{40}}\log 4 \cr & = \frac{{2.303 \times 0.6020}}{{40}} \cr & = 3.47 \times {10^{ - 2}} \cr & R = K{\left( A \right)^1} \cr & = 3.47 \times {10^{ - 2}} \times 0.01 \cr & = 3.47 \times {10^{ - 4}} \cr} $$

We know that, slowest step is the rate determining step.
$$\eqalign{ & \therefore \,\,{\text{Rater}}\left( r \right) = {K_1}\left[ X \right]\left[ {{Y_2}} \right]\,\,\,...\left( {\text{i}} \right) \cr & {\text{Now, from equation}}{\text{. (i), i}}{\text{.e}}{\text{.}} \cr & {X_2} \to 2X\left[ {{\text{fast}}} \right] \cr & {K_{eq}} = \frac{{{{\left[ X \right]}^2}}}{{\left[ {{X_2}} \right]}} \cr & \left[ X \right] = {\left\{ {{K_{eq}}\left[ {{X_2}} \right]} \right\}^{\frac{1}{2}}}\,\,\,...\left( {{\text{ii}}} \right) \cr} $$
Now, substitute the value of $$\left[ X \right]$$  from equation. (ii) in equation. (i), we get
$$\eqalign{ & {\text{Rate}}\left( r \right) = {K_1}{\left( {{K_{eq}}} \right)^{\frac{1}{2}}}{\left[ {{X_2}} \right]^{\frac{1}{2}}}\left[ {{Y_2}} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = K{\left[ {{X_2}} \right]^{\frac{1}{2}}}\left[ {{Y_2}} \right] \cr & \therefore \,\,{\text{Order of reaction}} \cr & = \frac{1}{2} + 1 \cr & = \frac{3}{2} \cr & = 1.5 \cr} $$

$$\eqalign{ & {\text{For a first order reaction,}} \cr & {\text{Rate constant }}\left( k \right) = \frac{{2.303}}{t}{\text{log}}\frac{a}{{a - x}} \cr & {\text{where, }}a = {\text{initial concentration}} \cr & a - x = {\text{concentration after time}}\,'t' \cr & t = {\text{time in sec}}{\text{.}} \cr & {\text{Given,}}\,a = 20\,g,\,a - x = 5\,g,\,k = {10^{ - 2}} \cr & \therefore \,\,t = \frac{{2.303}}{{{{10}^{ - 2}}}}{\text{log}}\,\frac{{20}}{5} \cr & \,\,\,\,\,\,\,\,\,\, = 138.6\,s \cr & {\text{Alternatively,}} \cr & {\text{Half - life for the first order reaction,}} \cr & \frac{{{t_{\frac{1}{2}}}}}{2} = \frac{{0.693}}{k} \cr & \,\,\,\,\,\,\,\, = \frac{{0.693}}{{{{10}^{ - 2}}}} \cr & \,\,\,\,\,\,\,\, = 69.3s \cr} $$
Two half-lives are required for the reduction of $$20$$ $$g$$ of reactant into $$5$$ $$g.$$
\[20\,g\xrightarrow{{{t}_{\frac{1}{2}}}}10\,g\xrightarrow{{{t}_{\frac{1}{2}}}}5\,g.\]
$$\therefore {\text{Total time}} = 2{t_{\frac{1}{2}}} = 2 \times 69.3 = 138.6\,s$$

Order of reaction may be zero, whole number or fraction number.

$${\text{Since}}$$  $${t_{\frac{1}{2}}} = \frac{{0.693}}{k},\,\,\therefore \,\,k = \frac{{0.693}}{{{t_{\frac{1}{2}}}}}$$
$${\text{Given}}\,{\text{:}}\,{t_{\frac{1}{2}}} = 30\,\min \,\,{\text{at}}\,\,{27^ \circ }C$$       $${\text{and}}\,\,{t_{\frac{1}{2}}} = 10\,\min \,\,{\text{at}}\,\,{47^ \circ }C$$
\[\begin{align} & \therefore \,{{k}_{{{27}^{\circ }}C}}=\frac{0.693}{30}{{\min }^{-1}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=0.0231\,{{\min }^{-1}} \\ \end{align}\]
$${\text{and}}$$  \[{{k}_{{{47}^{\circ }}C}}=\frac{0.693}{10}=0.0693\,{{\min }^{-1}}\]
$$\eqalign{ & {\text{We know that}} \cr & \log \frac{{{k_{{{47}^ \circ }C}}}}{{{k_{{{27}^ \circ }C}}}} = \frac{{{E_a}}}{{2.303R}} \times \frac{{\left( {{T_2} - {T_1}} \right)}}{{{T_2} \times {T_1}}} \cr & \therefore \,{E_a} = \frac{{2.303\,R \times {T_1} \times {T_2}}}{{\left( {{T_2} - {T_1}} \right)}}\log \frac{{{k_{{{47}^ \circ }C}}}}{{{k_{{{27}^ \circ }C}}}} \cr} $$
$${\text{or}}$$  $${E_a} = \frac{{2.303 \times 8.314 \times {{10}^{ - 3}} \times 300 \times 320}}{{\left( {320 - 300} \right)}}$$       $$\log \frac{{0.0693}}{{0.0231}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,= 43.848\,kJ\,mo{l^{ - 1}}$$

$${\text{Arrhenius equation}}\,\,k = A{e^{ - \,\frac{{{E^*}}}{{RT}}}}$$
$$\ln \,k = \ln \,A - \frac{{{E^*}}}{{RT}}$$       ( $${{E^*} = }$$  energy of activation )
$${\text{or}}\,\,{\text{log}}\,k = {\text{log}}\,A = \frac{{{E^*}}}{{2.303\,RT}}$$
Compare this equation with the straight line equation,
$${\text{i}}{\text{.e}}{\text{.}}\,\,y = mx + c$$
where $$'m’$$ is slope and $$'c’$$ is intercept
Hence, $${E^*}$$ can be calculated with the help of following slope

$$\begin{array}{l}\text{For first order reaction ,}\\ \text{Rate = }k\text{ [conc}\text{. of reactant]}\\ \text{Since 0}\text{.1 }M\text{ of }X\text{ changes to 0}\text{.025 }M\text{in 40 minutes,}\\ t_{\frac{1}{2}}\text{of reaction}=\frac{40}{2}=20\text{minutes}\\ \text{Rate of reaction of}\\ X=k\left[X\right]\\ =\frac{0.693}{t\frac{1}{2}}\times \left[X\right]\\ =\frac{0.693}{20}\times 0.01\\ =3.47\times {10}^{-4}M{\min }^{-1}\end{array}$$