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Dinitrogen pentoxide, \(\mathrm{N}_{2} \mathrm{O}_{5}\), decomposes by first- order kinetics with a rate constant of \(0.15 \mathrm{~s}^{-1}\) at \(353 \mathrm{~K}\). (a) What is the half-life (in seconds) for the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\) at \(353 \mathrm{~K}\) ? (b) If \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}=0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\), what will be the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) after \(2.0 \mathrm{~s}\) ? (c) How much time (in minutes) will elapse before the \(\mathrm{N}_{2} \mathrm{O}_{5}\) concentration decreases from \(0.0567 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to \(0.0135 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) ?

Short Answer

Expert verified
The half-life is \(4.62\) seconds, the concentration after 2 seconds is \(0.0351 \, \text{mol} \cdot \text{L}^{-1}\), and it takes about \(5.00\) minutes for the concentration to decrease from \(0.0567\) to \(0.0135 \, \text{mol} \cdot \text{L}^{-1}\).

Step by step solution

01

Determine the Half-life

To find the half-life, use the first-order half-life formula: \( t_{1/2} = \frac{\ln(2)}{k} \), where \( k \) is the rate constant. Substitute \( k = 0.15 \, \text{s}^{-1} \) to get \( t_{1/2} = \frac{\ln(2)}{0.15} \).
02

Calculate the Concentration after 2 seconds

Use the first-order rate law: \( \left[\mathrm{N}_{2}\mathrm{O}_{5}\right] = \left[\mathrm{N}_{2}\mathrm{O}_{5}\right]_{0} \exp(-kt) \), where \( t = 2.0 \, \text{s} \), \( k = 0.15 \, \text{s}^{-1} \), and \( \left[\mathrm{N}_{2}\mathrm{O}_{5}\right]_{0} = 0.0567 \, \text{mol} \cdot \text{L}^{-1} \). Plug in the values to get \( \left[\mathrm{N}_{2}\mathrm{O}_{5}\right] = 0.0567 \exp(-0.15 \times 2) \).
03

Find the Time for the Concentration to Decrease

Starting with the same rate law: \( \left[\mathrm{N}_{2}\mathrm{O}_{5}\right] = \left[\mathrm{N}_{2}\mathrm{O}_{5}\right]_{0} \exp(-kt) \). Let the final concentration be \( 0.0135 \, \text{mol} \cdot \text{L}^{-1} \). Solve for \( t \) with initial concentration \( 0.0567 \, \text{mol} \cdot \text{L}^{-1} \) to obtain \( t = \frac{\ln\left(\frac{0.0567}{0.0135}\right)}{0.15} \). Convert the time from seconds to minutes by dividing by 60.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the area of chemistry devoted to studying the rates at which chemical reactions occur and the factors which affect these rates. Kinetics is vital for understanding the world around us because it not only explains how fast reactions happen but also helps predict the outcome over time.

At the heart of kinetics is the reaction rate, the speed at which reactants transform into products. This rate can be influenced by conditions such as temperature, concentration of reactants, and catalysts. For instance, an increase in temperature generally increases the rate of a reaction. Similarly, higher concentrations of reactants can lead to more frequent collisions, thereby accelerating the reaction.

The first-order rate law relates the rate of reaction to the concentration of one of the reactants. For a reaction where substance A is converting to product B, the first-order rate may be expressed as Rate = k[A], where k is the rate constant and [A] is the concentration of A. This law allows for the calculation of how the concentration of a substance decreases over time.
Half-life Calculation
Half-life is a concept frequently associated with first-order reactions in chemical kinetics. It is defined as the time required for half the quantity of a reactant to be converted into the product. Mathematically, for a first-order reaction, the half-life does not depend on the initial concentration of the reactant and is given by the formula:
\( t_{1/2} = \frac{\ln(2)}{k} \).

In this formula, \( t_{1/2} \) represents the half-life, \( \ln(2) \) is the natural logarithm of 2, and k is the rate constant. This equation is derived from the integrated rate law for first-order kinetics, which illustrates that the time it takes for the concentration of a substance to drop to half its initial value is a constant.

Half-life provides a useful measure to compare the kinetics of different reactions, and it is also widely used in fields such as pharmacology and nuclear chemistry to describe the decay of substances.
Rate Constant
The rate constant, symbolized by k, is an essential parameter in the rate equations of chemical kinetics. It quantifies the speed of a chemical reaction and is influenced by factors like temperature, the presence of catalysts, and the activation energy barrier of the reaction.

The value of the rate constant is specific to each reaction and its conditions, and it determines how concentration affects the rate. In first-order reactions, the rate constant enables the calculation of the reaction's half-life, as seen in the half-life formula. A larger rate constant indicates a faster reaction, as there would be a greater change in concentration of reactants or products per unit time.

Understanding the rate constant is vital for controlling chemical processes, developing synthetic routes in chemistry, and in the design of pharmaceutical products, where reaction rates can dictate the stability and effectiveness of drugs.
Exponential Decay
Exponential decay is a pattern of decrease that reduces a quantity by a constant proportion over fixed periods of time. This phenomenon is observable in various first-order processes, including certain chemical reactions and radioactive decay.

In the context of a first-order reaction like the decomposition of \( \mathrm{N}_{2}\mathrm{O}_{5} \), the concentration of the reactant diminishes exponentially over time, described by the equation:
\( [\mathrm{N}_{2}\mathrm{O}_{5}] = [\mathrm{N}_{2}\mathrm{O}_{5}]_{0} \exp(-kt) \),
where \( [\mathrm{N}_{2}\mathrm{O}_{5}]_{0} \) is the initial concentration, k is the rate constant, and t is the time elapsed.

This equation indicates that the concentration of a substance decreases at a rate proportional to its current value, leading to a smooth and continuous decline rather than a sudden drop off. The concept of exponential decay provides a predictive tool for understanding how quickly a reactant is used up in a reaction and plays a crucial role in numerous scientific and mathematical applications.

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Most popular questions from this chapter

Derive an expression for the half-life of the reactant A that decays by an \(n\) th-order reaction (with \(n>1\) ) with rate constant \(k\). Reaction Mechanisms

(a) Using a graphing calculator or standard graphing software, such as that on the Web site for this book, make an appropriate Arrhenius plot of the data shown here for the decomposition of iodoethane into ethene and hydrogen iodide, \(\mathrm{C}_{2} \mathrm{H} \mathrm{H}_{5} \mathrm{I}(\mathrm{g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})+\mathrm{HI}(\mathrm{g})\), and determine the activation energy for the reaction. (b) What is the value of the rate constant at \(400^{\circ} \mathrm{C}\) ? $$ \begin{array}{lcccc} T(\mathbf{K}) & 660 & 680 & 720 & 760 \\ k\left(\mathbf{s}^{-1}\right) & 7.2 \times 10^{-4} & 2.2 \times 10^{-3} & 1.7 \times 10^{-2} & 0.11 \end{array} $$

Write the overall reaction for the mechanism proposed below and identify any reaction intermediates. Step \(1 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Br} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{Br}^{-}\) Step \(2 \mathrm{C}_{4} \mathrm{H}_{9}{ }^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}{ }^{+}\) Step \(3 \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}_{2}^{+}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{H}_{3} \mathrm{O}^{+}\)

The second-order rate constant for the decomposition of \(\mathrm{NO}_{2}\) (to \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) ) at \(573 \mathrm{~K}\) is \(0.54 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1}\). Calculate the time for an initial \(\mathrm{NO}_{2}\) concentration of \(0.20 \mathrm{~mol} \cdot \mathrm{L}^{-1}\) to decrease to (a) one-half; (b) one-sixteenth; (c) one-ninth of its initial concentration.

The Michaelis constant \(\left(K_{M}\right)\) is an index of the stability of an enzyme-substrate complex. Does a high Michaelis constant indicate a stable or an unstable enzyme-substrate complex? Explain your reasoning.

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