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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2024 · 6 April · Evening Shift · Chemistry · Thermodynamics (C)

Consider the two different first order reactions given below (A+B...

Consider the two different first order reactions given below (\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}) (Reaction 1)) (P \rightarrow Q) (Reaction 2) The ratio of the half life of Reaction 1 : Reaction 2 is 5:25: 2. If t1t_1 and t2t_2 represent the time taken to complete (2 / 3^{\text{rd }}) and (45^{\text{th }}) of Reaction 1 and Reaction 2, respectively, then the value of the ratio t1:t2t_1: t_2 is ______ (\times 10^{-1}) (nearest integer). [Given : (\log _{10}(3)=0.477) and (\log _{10}(5)=0.699)]

Correct answer

0

Step-by-step solution

(t1/2)1(t1/2)II=K2 K1=52∴K1t1=ln⁡11−23=ln⁡3 K2t2=ln⁡11−45=ℓn5⇒K1 K2×t1t2=0.4770.699⇒t1t2=0.4770.699×52=1.7=17×10−1\begin{aligned} & \frac{\left(\mathrm{t}_{1/2}\right)_1}{\left(\mathrm{t}_{1/2}\right)_{\mathrm{II}}}=\frac{\mathrm{K}_2}{\mathrm{~K}_1}=\frac{5}{2} \\ & \therefore \mathrm{K}_1 \mathrm{t}_1=\ln \frac{1}{1-\frac{2}{3}}=\ln 3 \\ & \mathrm{~K}_2 \mathrm{t}_2=\ln \frac{1}{1-\frac{4}{5}}=\ell \mathrm{n} 5 \\ & \Rightarrow \frac{\mathrm{K}_1}{\mathrm{~K}_2} \times \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{0.477}{0.699} \\ & \Rightarrow \frac{\mathrm{t}_1}{\mathrm{t}_2}=\frac{0.477}{0.699} \times \frac{5}{2}=1.7=17 \times 10^{-1}\end{aligned}
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