Consider the two different first order reactions given below \(\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}\) (Reaction 1) \(\mathrm{P} \rightarrow \mathrm{Q}\) (Reaction 2) The ratio of the half life of Reaction 1 : Reaction 2 is \(5: 2\). If \(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) represent the time taken to complete \(2/3\)rd and \(4/5\)th 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 \(\log_{10}(5)=0.699\)]

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Chemical Kinetics

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

rd and

4/5

th of Reaction 1 and Reaction 2, respectively, then the value of the ratio

t_{1} : t_{2}

is ______

\times 1 0^{- 1}

(nearest integer).
[Given:

lo g_{10} (3) = 0.477

and

lo g_{10} (5) = 0.699

]

Answer: 17

Since

t_{1/2} \propto 1/ k

, the given ratio

t_{1/2, 1} : t_{1/2, 2} = 5 : 2

gives

k_{1} : k_{2} = 2 : 5

. For first-order kinetics,

k t = ln \frac{1}{1 - x}

. Reaction 1 at

x = 2/3

k_{1} t_{1} = ln 3

. Reaction 2 at

x = 4/5

k_{2} t_{2} = ln 5

. Then

\frac{t _{1}}{t _{2}} = \frac{ln 3/ k _{1}}{ln 5/ k _{2}} = \frac{lo g 3}{lo g 5} \times \frac{k _{2}}{k _{1}} = \frac{0.477}{0.699} \times \frac{5}{2} \approx 1.706 = 17 \times 1 0^{- 1}

JEE Main · 2024 · 06 Apr 2024 · Shift 2 · Integer answer