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Chapter 16: Problem 47

For the simple decomposition reaction \(\mathrm{AB}(g) \longrightarrow \mathrm{A}(g)+\mathrm{B}(g)\) rate \(=k[\mathrm{AB}]^{2}\) and \(k=0.2 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s} .\) How long will it take for \([\mathrm{AB}]\) to reach one-third of its initial concentration of \(1.50 \mathrm{M}\) ?

Short Answer

Expert verified
The time is \ \frac{20}{3} \ \text{s} or approximately 6.67 seconds.

Step by step solution

01

- Identify the Reaction Order

The given rate law is \[ \text{rate} = k[\text{AB}]^2 \]. Since the rate depends on the concentration to the second power, it is a second-order reaction.
02

- Write the Integrated Rate Law for Second-Order Reactions

For a second-order reaction, the integrated rate law is \[ \frac{1}{[\text{AB}]} = kt + \frac{1}{[\text{AB}_0]} \], where \([\text{AB}_0]\) is the initial concentration and \(k\) is the rate constant.
03

- Plug in Initial and Final Concentrations

In this problem, \([\text{AB}_0] = 1.50 \text{ M}\) and \([\text{AB}] = \frac{1.50}{3} = 0.50 \text{ M}\).
04

- Substitute Known Values into the Integrated Rate Law

Substitute \(k = 0.2 \text{ L/mol路s}\), \([\text{AB}_0] = 1.50 \text{ M}\), and \([\text{AB}] = 0.50 \text{ M}\) into the integrated rate law \[ \frac{1}{[\text{AB}]} = kt + \frac{1}{[\text{AB}_0]} \].
05

- Solve for Time \(t\)

Plug in the values: \[ \frac{1}{0.50 \text{ M}} = 0.2 \text{ L/mol路s} \times t + \frac{1}{1.50 \text{ M}} \]. This simplifies to \[ 2 = 0.2t + \frac{2}{3} \]. Subtract \(\frac{2}{3}\) from both sides to get \[ \frac{4}{3} = 0.2t \]. Solving for \(t\), we get \[ t = \frac{4}{3 \times 0.2} = \frac{4}{0.6} = \frac{40}{6} = \frac{20}{3} \text{ s} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integrated Rate Law
The integrated rate law is a mathematical expression that connects the concentration of reactants to time.
For second-order reactions, the integrated rate law is given as:
\( \frac{1}{[AB]} = kt + \frac{1}{[AB_0]} \).
Here, \( [AB] \) is the concentration at time \( t \); \( k \) is the rate constant, and \( [AB_0] \) is the initial concentration.
This law helps us understand how the concentration of the reactant changes over time and is a powerful tool in reaction kinetics.
Reaction Order
Knowing the reaction order is crucial in determining how a reactant's concentration affects the rate of reaction.
In our example, the given rate law is \( \text{rate} = k[AB]^2 \). This shows a second-order reaction because the exponent of \( [AB] \) is 2.

It's essential to identify the reaction order to use the correct integrated rate law.
Reaction orders can be zero, first, or any higher integer, and they significantly impact the kinetics analysis.
Rate Constant Calculation
The rate constant \( k \) is a crucial parameter in reaction kinetics.
For our reaction, \( k = 0.2 \text{ L/mol路s} \). To calculate the time it takes for \( [AB] \) to reach a certain concentration, we substitute the values in the integrated rate law.
This shows the direct relationship between reaction rate and concentration.
The rate constant also helps predict how fast a reaction progresses under various conditions.
Concentration Change Over Time
To understand how concentration changes over time, we use the integrated rate law with specific initial and final concentrations.
In our scenario, we want to find the time it takes for \( [AB] \) to drop to one-third of its initial concentration (1.50 M to 0.50 M).
Substituting these values into the rate law: \( \frac{1}{0.50} = 0.2t + \frac{1}{1.50} \).
We solve this equation to find \( t \). Details like these demonstrate the impact of time and conditions on concentration, helping us make predictions about the reaction's future behavior.

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Most popular questions from this chapter

The rate constant of a reaction is \(4.50 \times 10^{-5} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(195^{\circ} \mathrm{C}\) and \(3.20 \times 10^{-3} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(258^{\circ} \mathrm{C}\). What is the activation energy of the reaction?

Nitrification is a biological process in which \(\mathrm{NH}_{3}\) in wastewater is converted to \(\mathrm{NH}_{4}^{+}\) and then removed according to the following reaction: \(\mathrm{NH}_{4}^{+}+2 \mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}\) The first-order rate constant is given as \(k_{1}=0.47 e^{0.095\left(T-15^{\circ} \mathrm{C}\right)}\) where \(k_{1}\) is in day \(^{-1}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). (a) If the initial concentration of \(\mathrm{NH}_{3}\) is \(3.0 \mathrm{~mol} / \mathrm{m}^{3},\) how long will it take to reduce the concentration to \(0.35 \mathrm{~mol} / \mathrm{m}^{3}\) in the spring \(\left(T=20^{\circ} \mathrm{C}\right) ?\) (b) In the winter \(\left(T=10^{\circ} \mathrm{C}\right) ?\) (c) Using your answer to part (a), what is the rate of \(\mathrm{O}_{2}\) consumption?

16.103 Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: rate \(=k\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]\). The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \longrightarrow 2 \mathrm{HI}(g)\) In the 1960 s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism: (1) \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) [fast] (2) \(\mathrm{H}_{2}(g)+\mathrm{I}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{I}(g)\) [fast] (3) \(\mathrm{H}_{2} \mathrm{I}(g)+\mathrm{I}(g) \longrightarrow 2 \mathrm{HI}(g) \quad[\) slow \(]\) Show that this mechanism is consistent with the rate law.

In a kinctics experiment, a chemist places crystals of iodine in a closed reaction vessel, introduces a given quantity of \(\mathrm{H}_{2}\) gas, and obtains data to calculate the rate of HI formation. In a second experiment, she uses the same amounts of iodine and hydrogen but first warms the flask to \(130^{\circ} \mathrm{C}\), a temperature above the sublimation point of iodine. In which of these experiments does the reaction proceed at a higher rate? Explain.

Explain why the coefficients of an elementary step equal the reaction orders of its rate law but those of an overall reaction do not.
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