Chemistry · Chemical Kinetics

JEE Advanced 2024 — Paper 2 — Question 35

A sample initially contains only U-238 isotope of uranium. With time, some of the U-238 radioactively decays into $\mathrm{Pb}-206$ while the rest of it remains undisintegrated. When the age of the sample is $\mathbf{P} \times 10^8$ years, the ratio of mass of Pb-206 to that of U-238 in the sample is found to be 7. The value of $\mathbf{P}$ is $\qquad$ ………. [Given: Half-life of U-238 is $4.5 \times 10^9$ years; $\log _{\mathrm{e}} 2=0.693$ ]

Answer: 143

Numerical answer — enter this value.

Step-by-step solution

$\begin{aligned} & \text { Mass }=\text { Number of moles } \times \text { Molar mass } \\ & \frac{(\text { Mass }) \mathrm{P}_{\mathrm{b}}}{(\text { Mass }) \mathrm{U}}=\frac{\mathrm{n}_{\mathrm{Pb}} \times 206}{\mathrm{n}_{\mathrm{u}} \times 238}=\frac{7}{1} \\ & \frac{\mathrm{n}_{\mathrm{Pb}}}{\mathrm{n}_{\mathrm{u}}}=\frac{238 \times 7}{206} \\ & \ell \mathrm{n}\left(1+\frac{\mathrm{n}_{\mathrm{Pb}}}{\mathrm{n}_{\mathrm{u}}}\right)=\lambda \mathrm{t} \\ & \ell \mathrm{n}\left(1+\frac{238 \times 7}{206}\right)=\frac{\ell \mathrm{n} 2}{\mathrm{t}_{1 / 2}} \times \mathrm{t} \\ & \ell \mathrm{n}\left(\frac{206+1666}{206}\right)=\frac{\ell \mathrm{n} 2}{\mathrm{t}_{1 / 2}} \times \mathrm{t} \\ & \ell \mathrm{n}(9)=\frac{\ell \mathrm{n} 2}{\mathrm{t}_{1 / 2}} \times \mathrm{t}\end{aligned}$

$\begin{aligned} & \mathrm{t}=\frac{\ln 9}{\ln 2} \times \mathrm{t}_{1 / 2} \\ & \mathrm{t}=\frac{2 \times 0.4771}{0.3010} \times 4.5 \times 10^9 \\ & \mathrm{t}=14.265 \times 10^9=142.65 \times 10^8 \text { years } \\ & \mathrm{P}=142.65\end{aligned}$

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Chemistry
Chapter: Chemical Kinetics
Topic: Integrated Rate Laws