The p-Block Elements — Chemistry STD 12 Science — Question

$\ln {k_t} = \ln {k_0} + \left( {\frac{{\ln \left( {\frac{5}{2}} \right)}}{{10}}} \right) \times t\left( {t \geqslant 0{}\,^0C} \right)$

$K_0$ =rate constant at $0\,^o C$ ;  $k_t$ = rate constant at $t\,^o C$

Temperature coefficient of reaction, assuming that the rate constant increases same number of times on each $10\,^oC$ rise in temperature, is

$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ (Reaction 1$)$

$\mathrm{P} \rightarrow \mathrm{Q}$ (Reaction $2$)

The ratio of the half life of Reaction $1$ : Reaction $2$ is $5: 2$. If $t_1$ and $t_2$ represent the time taken to complete $2 / 3^{\text {dd }}$ and $4 / 5^{\text {dd }}$ of Reaction $1$ and

Reaction $2$, respectively, then the value of the ratio $\mathrm{t}_1: \mathrm{t}_2$ is . . . .$\times 10^{-1}$ (nearest integer).

[Given: $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$ ]

Reason: Due to formation of a complex

$\mathop {\begin{array}{*{20}{c}}
  O \\ 
  {||} \\ 
  {Ph - C - Ph} 
\end{array}}\limits_{(I)} $        $\mathop {\begin{array}{*{20}{c}}
  {\,\,\,\,\,\,O} \\ 
  {\,\,\,\,\,||} \\ 
  {C{H_3} - C - H} 
\end{array}}\limits_{(II)} $          $\mathop {\begin{array}{*{20}{c}}
  {\,O} \\ 
  {||} \\ 
  {C{H_3} - C - C{H_3}} 
\end{array}}\limits_{(III)} $