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Chapter 12: Problem 51

For the reaction \(\mathrm{A} \rightarrow\) products, successive half-lives are observed to be \(10.0,20.0\), and \(40.0\) min for an experiment in which \([\mathrm{A}]_{0}=\) \(0.10 \mathrm{M} .\) Calculate the concentration of \(\mathrm{A}\) at the following times. a. \(80.0 \mathrm{~min}\) b. \(30.0 \mathrm{~min}\)

Short Answer

Expert verified
The concentrations of A at the given times are: a. \(80.0\, min: [A] = 0.0125\, M\) b. \(30.0\, min: [A] = 0.0250\, M \)

Step by step solution

01

Identify the order of the reaction

Given the successive half-lives (10 min, 20 min, and 40 min), it's clear that the half-life is doubling after each period. This indicates that this reaction is second-order with respect to A.
02

Calculate the rate constant (k)

For a second-order reaction, the relationship between the half-life, initial concentration, and rate constant (k) can be described by the equation : \[t_{1/2} = \frac{1}{k[A]_{0}}\] We can use the first given half-life (10 min) and the initial concentration of A (0.10 M) to calculate the rate constant (k). \[10.0\, min = \frac{1}{k(0.10\, M)}\] Solving for k, we find: \[k = \frac{1}{10.0\, min \times 0.10\,M} = 1.0\, M^{-1} min^{-1}\]
03

Calculate the concentration of A at 80.0 min

For a second-order reaction, the integrated rate law equation is: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\) We are given the time (80.0 min) and the initial concentration of A (0.10 M). We can plug these values and the determined rate constant (k = 1.0 M⁻¹ min⁻¹) into the equation to solve for the concentration of A at 80.0 minutes. \(\frac{1}{[A]_{80}} - \frac{1}{0.10\,M} = 1.0\, M^{-1} min^{-1} \times 80.0\,min\) Solving for [A]_{80}, we find: \([A]_{80} = 0.0125\, M\)
04

Calculate the concentration of A at 30.0 min

Using the same integrated rate law equation for a second-order reaction, we can calculate the concentration of A at 30.0 minutes: \(\frac{1}{[A]_{30}} - \frac{1}{0.10\,M} = 1.0\, M^{-1} min^{-1} \times 30.0\,min\) Solving for [A]_{30}, we find: \([A]_{30} = 0.0250\, M\) To summarize, the concentrations of A at the given times are: a. 80.0 min: [A] = 0.0125 M b. 30.0 min: [A] = 0.0250 M

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Most popular questions from this chapter

A certain reaction has the following general form: \(\mathrm{aA} \longrightarrow \mathrm{bB}\) At a particular temperature and \([\mathrm{A}]_{0}=2.00 \times 10^{-2} M\), concentration versus time data were collected for this reaction, and a plot of \(\ln [\mathrm{A}]\) versus time resulted in a straight line with a slope value of \(-2.97 \times 10^{-2} \mathrm{~min}^{-1}\). a. Determine the rate law, the integrated rate law, and the value of the rate constant for this reaction. b. Calculate the half-life for this reaction. c. How much time is required for the concentration of \(\mathrm{A}\) to decrease to \(2.50 \times 10^{-3} M ?\)

Hydrogen reacts explosively with oxygen. However, a mixture of \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) can exist indefinitely at room temperature. Explain why \(\mathrm{H}_{2}\) and \(\mathrm{O}_{2}\) do not react under these conditions.

The initial rate of a reaction doubles as the concentration of one of the reactants is quadrupled. What is the order of this reactant? If a reactant has a \(-1\) order, what happens to the initial rate when the concentration of that reactant increases by a factor of two?

In the Haber process for the production of ammonia, $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

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