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Chapter 12: Problem 55

A certain substance, initially at 0.10\(M\) in solution, decomposes by second- order kinetics. If the rate constant for this process is 0.40 $\mathrm{L} / \mathrm{mol} \cdot \min$ , how much time is required for the concentration to reach 0.020 \(\mathrm{M}\) ?

Short Answer

Expert verified
The time required for the concentration of the substance to reach 0.020 M, given that it decomposes by second-order kinetics with a rate constant of 0.40 L/mol·min, is 10 minutes.

Step by step solution

01

Write the integrated rate law for second-order kinetics

\(\frac{1}{[A]_{t}} - \frac{1}{[A]_{0}} = kt\)
02

Plug in the given values

Here, \([A]_{0} = 0.10 \: M\), \([A]_{t} = 0.020 \: M\), and \(k = 0.40 \: L/mol \cdot min\). Plug these values into the formula: \(\frac{1}{0.020} - \frac{1}{0.10} = 0.40t\)
03

Simplify the equation

Simplifying the equation gives: \(\frac{50 - 10}{10} = 0.40t \)
04

Solve for t

Solve for time t: \(t = \frac{40}{0.4 \times 10} = \frac{40}{4} = 10 \: min\). So, it would take 10 minutes for the concentration of the substance to reach 0.020 M.

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