(10)

                For first order reactions, rate constant $k$ is given by:

                $\mathrm{k}=\frac{2.303}{\mathrm{t}}\log \frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{\mathrm{t}}}$

              When reaction is 99.9% complete, 0.1% reactants are left i.e.

              ${\mathrm{a}}_{\mathrm{t}}=\frac{0.1}{100}{\mathrm{a}}_{\mathrm{o}},$ $\frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{\mathrm{t}}}=1000$, thus

             $\mathrm{k}=\frac{2.303}{{\mathrm{t}}_{99.9\%}}\log 1000=\frac{2.303}{{\mathrm{t}}_{99.9\%}}\times 3$                        (I)

             When reaction is 50% complete, 50% reactants are left i.e.

             ${\mathrm{a}}_{\mathrm{t}}=\frac{50}{100}{\mathrm{a}}_{\mathrm{o}},$ $\frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{\mathrm{t}}}=2$, thus

             $\mathrm{k}=\frac{2.303}{{\mathrm{t}}_{50\%}}\log 2=\frac{2.303}{{\mathrm{t}}_{50\%}}\times 0.301$                     (II)

            Equate I and II

            $\frac{2.303}{{\mathrm{t}}_{99.9\%}}\times 3=\frac{2.303}{{\mathrm{t}}_{50\%}}\times 0.301$

            ${\mathrm{t}}_{99.9\%}\approx 10\times {\mathrm{t}}_{50\%}$

(63)

             For first order reaction,

             $\mathrm{k}=\frac{2.303}{\mathrm{t}}{\log }_{10}\frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{\mathrm{t}}}$

             At half life time i.e 36 hrs:

            $\mathrm{k}=\frac{2.303}{36}{\log }_{10}\frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{\mathrm{o}}/2}$

            $\mathrm{k}=\frac{2.303}{36}{\log }_{10}2$                                       (i)

           At after one day i.e 24 hrs:

           $\mathrm{k}=\frac{2.303}{24}{\log }_{10}\frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{(\mathrm{t}=24)}}$                           (ii)

            Equating (i) and (ii)

           $\frac{2.303}{24}{\log }_{10}\frac{{\mathrm{a}}_{\mathrm{o}}}{{\mathrm{a}}_{(\mathrm{t}=24)}}=\frac{2.303}{36}{\log }_{10}2$

            ${\log }_{10}\frac{a_o}{a_{(t=24)}}=\frac{24}{36}{\log }_{10}2$

            ${\log }_{10}\frac{a_o}{a_{(t=24)}}=\frac{24}{36}\times 0.301=0.20066$

            $\frac{a_o}{a_{(t=24)}}=\text{antilog}(0.2006)=1.587$

            $\frac{a_{(t=24)}}{a_o}=\frac{1}{1.587}=0.63=63\times {10}^{-2}$

             Thus the fraction which remains after one day is $63\times {10}^{-2}$.

(24)

          For a first-order reaction $\mathrm{A}\to$ products, the rate of the reaction is given by: $\mathrm{r}=\mathrm{k}{\left[\mathrm{A}\right]}_{\mathrm{t}}=\mathrm{k}{\left[\mathrm{A}\right]}_0{\mathrm{e}}^{-\mathrm{kt}}$

          Substituting value of rate at t = 10 min and t = 20 min: 

          at, t = 10 min :

          $0.04 \text{mol} {\text{L}}^{-1} {\text{s}}^{-1}=\mathrm{k}{\left[\mathrm{A}\right]}_0{\mathrm{e}}^{-10\mathrm{k}}$                                     (I)

          at, t = 20 min : 

         $0.03 \text{mol} {\text{L}}^{-1} {\text{s}}^{-1}=\mathrm{k}{\left[\mathrm{A}\right]}_0{\mathrm{e}}^{-20\mathrm{k}}$                                     (II)

         $\mathrm{I}÷\mathrm{II}$

         $\frac{4}{3}=\frac{e^{-10k}}{e^{-20k}}$

          $e^{10k}=\frac{4}{3}$

          $\ln$ $e^{10k}=\ln \frac{4}{3}$

          $10\mathrm{k}$ $\ln$ $e=2.303{\log }_{10}\frac{4}{3}$

          $10\mathrm{k}=2.303{\log }_{10}\frac{4}{3}$    $(\ln$ $e=1)$

           $\mathrm{k}=\frac{2.303}{10}{\log }_{10}\frac{4}{3}$                                                 (III)

          Also for a first-order reaction,

          ${\mathrm{t}}_{1/2}=\frac{2.303{\log }_{10}2}{\mathrm{k}}$ or $\mathrm{k}=\frac{2.303{\log }_{10}2}{{\mathrm{t}}_{1/2}}$          (IV)

         Equating III and IV:

         $\frac{2.303{\log }_{10}2}{{\mathrm{t}}_{1/2}}=\frac{2.303}{10}{\log }_{10}\frac{4}{3}$

          $\frac{{\log }_{10}2}{{\mathrm{t}}_{1/2}}=\frac{1}{10}({\log }_{10}4-{\log }_{10}3)$

           $\frac{{\log }_{10}2}{{\mathrm{t}}_{1/2}}=\frac{1}{10}(2{\log }_{10}2-{\log }_{10}3)$

            $\frac{0.3010}{t_{1/2}}=\frac{1}{10}(2\times 0.3010-0.4771)$

             ${\mathrm{t}}_{1/2}=\frac{0.3010}{0.01249} \text{min}=24.1 \text{min}$

(399)

             From rate expression it is clear that the reaction is a first order reaction. As 50% of the reactants decompose in 120 min, $t_{1/2}=120$ min. For first order reactions, the rate constant $k$ is given by:  $k=\frac{2.303}{t}\log \frac{a_o}{a_t}$

              When reaction is 90% complete, 10% reactants are left i.e. $a_t=\frac{10}{100}{a}_o,\frac{a_o}{a_t}=10$

              $\Rightarrow k=\frac{2.303}{t_{90\%}}\log 10$                                   (I)

               When reaction is 50% complete, 50% reactants are left i.e.  $a_t=\frac{50}{100}{a}_o,\frac{a_o}{a_t}=2$

              $\Rightarrow k=\frac{2.303}{t_{50\%}}\log 2$                                     (II)

              Equating I and II

              $\frac{2.303}{t_{90\%}}\log 10=\frac{2.303}{t_{50\%}}\log 2$

              $\frac{1}{t_{90\%}}\times 1=\frac{1}{120}\times 0.301$

               $t_{90\%}=398.67 \text{min}\approx 399 \text{min}$

(17190)

                ${}^{14}\mathrm{C}$ is radioactive and undergoes first order decay. In a piece of wood amount of ${}^{14}\mathrm{C}$ decreases because of radioactive decay but amount of ${}^{12}\mathrm{C}$ remains same. So if ratio of $\frac{{}^{14}\mathrm{C}}{{}^{12}\mathrm{C}}$ in a piece of wood is $\frac{1}{8}$, this implies ratio of amount ${}^{14}\mathrm{C}$ remained in the wood to the initial amount is $\frac{1}{8}$.

                In first order reaction equal percentage of reaction is completed in equal time interval. Thus concentration gets halved after every $t_{1/2}$.

                $1\stackrel{t_{1/2}}{\to }\frac{1}{2}\stackrel{t_{1/2}}{\to }\frac{1}{4}\stackrel{t_{1/2}}{\to }\frac{1}{8}$

                To get to ${\left(\frac{1}{8}\right)}^{\mathrm{th}}$ value of initial amount, it takes $3t_{1/2}$ time i.e.

                $3\times 5730 \text{years}=17190 \text{years}$

(50)

                           $\mathrm{A}$          $+$          $\mathrm{B}$        $\to$         $\mathrm{C}$

              $t=0$    $a_o$                     $a_o$                      0

              $t=t$     $a_o-x$              $a_o-x$                $x$

              $r=k{\left[A\right]}^{1/2}{\left[B\right]}^{1/2}=k{(a_o-x)}^{1/2}{(a_o-x)}^{1/2}$

              $\frac{dx}{dt}=k(a_o-x)$

              $\frac{dx}{(a_o-x)}=kdt$

               Integrate both sides

               ${\int }_0^{\mathrm{x}}\frac{\mathrm{dx}}{({\mathrm{a}}_{\mathrm{o}}-\mathrm{x})}={\int }_0^{\mathrm{t}}\mathrm{k} \mathrm{dt}$

                ${\left[\frac{\ln (a_o-x)}{-1}\right]}_0^x={\left[kt\right]}_0^t$

               $[\frac{\ln (a_o-x)}{-1}-\frac{\ln \left(a_o\right)}{-1}]=kt$

               $\ln \left(\frac{a_o}{a_o-x}\right)=kt$

               $2.303{\log }_{10}\left(\frac{a_o}{a_o-x}\right)=kt$

               $\text{In the question }{a}_0=1 \text{M and }(a_0-x)=0.1 \text{M}$

               $2.303{\log }_{10}\left(\frac{1}{0.1}\right)=4.6\times {10}^{-2} t$

              $2.303{\log }_{10}10=4.6\times {10}^{-2} t$

              $2.303\times 1=4.6\times {10}^{-2} t$

             $t=50 \text{s}$

(3)

                Rate constant of a first order reaction is given as:  $k=\frac{2.303}{t}\log \frac{a_o}{a_t}$

                When reaction is 90% complete, then only 10% reactants remain i.e. at $t_{90\%}$,          $a_t=0.1$ $a_0$

               $k=\frac{2.303}{t_{90\%}}\log \frac{a_0}{0.1a_0}=\frac{2.303}{t_{90\%}}\log 10=\frac{2.303}{t_{90\%}}\times 1=\frac{2.303}{t_{90\%}}$                                      ...(i)

               When reaction is 99.9% complete, then only 0.1% reactants remain i.e. $t_{99.9\%},$ $a_t=0.001$ $a_0$

                $k=\frac{2.303}{t_{99.9\%}}\log \frac{a_0}{0.001a_0}=\frac{2.303}{t_{99.9\%}}\log {10}^3=\frac{2.303}{t_{99.9\%}}\times 3\log 10=\frac{2.303}{t_{99.9\%}}\times 3$           ...(ii)

               Equate I and II

               $\frac{2.303}{t_{90\%}}=\frac{2.303}{t_{99.9\%}}\times 3$

                $t_{99.9\%}=t_{90\%}\times 3$

(17)

            For a first order reaction, rate constant $k$ is related to initial conc. of reactants ($r_o$) and conc. of reactants at time $t$($r_t$) as:

            $k=\frac{2.303}{t}{\log }_{10}\frac{r_o}{r_t}$

           Reaction 1: When reaction is $2/3^{rd}$ complete, then $1/3^{rd}$ of reactants remain, $r_t=\frac{1}{3}{r}_o$

           $k_1=\frac{2.303}{t_1}{\log }_{10}\frac{r_o}{\left(\frac{1}{3}\right)r_o}$

           $t_1=\frac{2.303}{k_1}{\log }_{10}3$                                          ...(i)

          Reaction 2: When reaction is $4/5^{th}$ complete, then $1/5^{th}$ of reactants remain, $r_t=\frac{1}{5}{r}_o$

          $k_2=\frac{2.303}{t_2}{\log }_{10}\frac{r_o}{\left(\frac{1}{5}\right)r_o}$

          $t_2=\frac{2.303}{k_2}{\log }_{10}5$                                           ...(ii)

          $\left(\mathrm{i}\right)÷\left(\mathrm{ii}\right)$

          $\frac{t_1}{t_2}=\frac{k_2}{k_1}\frac{{\log }_{10}3}{{\log }_{10}5}$

           For first order reaction, $t_{1/2}=\frac{0.693}{k}$

           So, $\frac{t_1}{t_2}=\frac{0.693/{\left(t_{1/2}\right)}_2}{0.693/{\left(t_{1/2}\right)}_1}\frac{{\log }_{10}3}{{\log }_{10}5}$

                  $\frac{t_1}{t_2}=\frac{{\left(t_{1/2}\right)}_1}{{\left(t_{1/2}\right)}_2}\frac{{\log }_{10}3}{{\log }_{10}5}=\frac{5\times 0.477}{2\times 0.699}=1.7=17\times {10}^{-1}$