If and represent the time taken to complete and of reaction 1 and reaction 2, respectively, then the value of the ratio is

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Solution

Q.
Consider the two different first order reactions given below

$A + B \to C (Reaction 1)$

$P \to Q (Reaction 2)$

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 $2 / 3^{r d}$ and $4 / 5^{t h}$ of reaction 1 and reaction 2, respectively, then the value of the ratio $t_{1} : t_{2}$ is _______ $\times 10^{- 1}$ (nearest integer).

[Given: $\log_{10}$ (3) = 0.477 and $\log_{10}$ (5) = 0.699] [2024]

Ans.
(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 $2 / 3^{r d}$ complete, then $1 / 3^{r d}$ of reactants remain, $r_{t} = \frac{1}{3} r_{o}$

$k_{1} = \frac{2.303}{t_{1}} \log_{10} \frac{r_{o}}{(\frac{1}{3}) r_{o}}$

$t_{1} = \frac{2.303}{k_{1}} \log_{10} 3$ ...(i)

  Reaction 2: When reaction is $4 / 5^{t h}$ complete, then $1 / 5^{t h}$ of reactants remain, $r_{t} = \frac{1}{5} r_{o}$

   $k_{2} = \frac{2.303}{t_{2}} \log_{10} \frac{r_{o}}{(\frac{1}{5}) r_{o}}$

   $t_{2} = \frac{2.303}{k_{2}} \log_{10} 5$ ...(ii)

   $(i) \div (ii)$

   $\frac{t_{1}}{t_{2}} = \frac{k_{2}}{k_{1}} \frac{\log_{10} 3}{\log_{10} 5}$

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

So, $\frac{t_{1}}{t_{2}} = \frac{0.693 / {(t_{1 / 2})}_{2}}{0.693 / {(t_{1 / 2})}_{1}} \frac{\log_{10} 3}{\log_{10} 5}$

   $\frac{t_{1}}{t_{2}} = \frac{{(t_{1 / 2})}_{1}}{{(t_{1 / 2})}_{2}} \frac{\log_{10} 3}{\log_{10} 5} = \frac{5 \times 0.477}{2 \times 0.699} = 1.7 = 17 \times 10^{- 1}$

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