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Chapter 13: Problem 132

Use the initial rate data from the following table to determine the order of the decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}:\) $$\begin{array}{|c|c|c|}\text { Experiment } & \left[\mathrm{N}_{2} \mathrm{O}_{5}\right]_{0}(\mathrm{M}) & \text { Initial Rate }(\mathrm{M} / \mathrm{s}) \\\1 & 0.050 & 1.8 \times 10^{-5} \\\\\hline 2 & 0.100 & 3.6 \times 10^{-5} \\\\\hline\end{array}$$

Short Answer

Expert verified
Answer: The reaction order for the decomposition of N2O5 is approximately 1, making it a first-order reaction.

Step by step solution

01

Understand reaction order and rate law

Reaction order is the power to which the concentration of a reactant is raised in the rate law expression. A higher order indicates that the reaction rate is more sensitive to changes in the concentration of the reactant. The rate law for the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{5}\) can have the form: $$\text{Rate} = k[\mathrm{N}_{2}\mathrm{O}_{5}]^m$$ where \(k\) is the rate constant, \([\mathrm{N}_{2}\mathrm{O}_{5}]\) is the initial concentration of \(\mathrm{N}_{2}\mathrm{O}_{5}\), and \(m\) is the order of the reaction.
02

Use the rate law for Experiment 1

In Experiment 1, we have an initial concentration of 0.050 M and a rate of \(1.8 \times 10^{-5} \, \mathrm{M/s}\). We can plug these values into the rate law and solve for the rate constant \(k\): $$1.8 \times 10^{-5} \, \mathrm{M/s} = k(0.050 \, \mathrm{M})^m$$
03

Use the rate law for Experiment 2

In Experiment 2, we have an initial concentration of 0.100 M and a rate of \(3.6 \times 10^{-5} \, \mathrm{M/s}\). We can plug these values into the rate law and solve for the rate constant \(k\): $$3.6 \times 10^{-5} \, \mathrm{M/s} = k(0.100 \, \mathrm{M})^m$$
04

Divide rate laws for both experiments to find the reaction order

Divide the rate law for Experiment 2 by the rate law for Experiment 1. This will give us the following equation: $$\frac{3.6 \times 10^{-5}}{1.8 \times 10^{-5}} = \frac{k(0.100 \, \mathrm{M})^m}{k(0.050 \, \mathrm{M})^m}$$ Since the rate constants \(k\) will cancel each other, we have: $$2 = \frac{(0.100 \, \mathrm{M})^m}{(0.050 \, \mathrm{M})^m}$$ Next, rewrite the equation with the desired exponent: $$2 = \left(\frac{0.100 \, \mathrm{M}}{0.050 \, \mathrm{M}}\right)^m$$
05

Solve for reaction order

To find the reaction order \(m\), we can solve the equation by taking the logarithm of both sides: $$\log{2} = m\log{\left(\frac{0.100 \, \mathrm{M}}{0.050 \, \mathrm{M}}\right)}$$ Now, we can solve for \(m\): $$m = \frac{\log{2}}{\log{\left(\frac{0.100 \, \mathrm{M}}{0.050 \, \mathrm{M}}\right)}} \approx 1$$ The reaction order is approximately 1, which means the reaction is a first-order reaction with respect to the decomposition of \(\mathrm{N}_{2}\mathrm{O}_{5}\).

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Key Concepts

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

Rate Law
In chemistry, the rate law is a mathematical equation that links the rate of a chemical reaction to the concentration of its reactants. It is expressed as: \[ \text{Rate} = k[\text{Reactant}]^m \] where:
  • \(k\) is the rate constant, a factor that depends on the temperature and other specific reaction conditions.
  • \([\text{Reactant}]\) is the concentration of the reactant.
  • \(m\) is the reaction order, which indicates how the concentration of a reactant affects the rate.

The rate law is determined experimentally, as it describes how changes in concentration influence the speed of the reaction. This helps in predicting how a chemical reaction proceeds over time. It鈥檚 important to note that the reaction order (\(m\)) is not always equal to the stoichiometric coefficients from the balanced chemical equation.
Decomposition Reaction
A decomposition reaction involves a single reactant breaking down into two or more simpler products. These reactions are often characterized by their straightforward simplicity. For example: \[ \text{AB} \rightarrow \text{A} + \text{B} \]
  • These reactions can be observed in various forms, such as thermal decomposition, where heat causes the compound to split.
  • Photochemical decomposition, which is decomposition due to light, is another example.

In the context of \( \mathrm{N}_{2}\mathrm{O}_{5} \), the decomposition reaction follows a similar pattern where \(\mathrm{N}_{2}\mathrm{O}_{5}\) breaks down into \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\). Decomposition reactions are crucial in processes like recycling, environmental breakdown of substances, and even in biological systems where complex molecules are broken into simpler ones.
N2O5
Researching the properties of dinitrogen pentoxide (\(\mathrm{N}_{2}\mathrm{O}_{5}\)) can aid in understanding its decomposition. This compound is a powerful oxidizing agent. In the context of chemical reactions, it plays a key role as a reactant.
  • \(\mathrm{N}_{2}\mathrm{O}_{5}\) decomposes to release nitrogen dioxide (\(\mathrm{NO}_2\)) and oxygen (\(\mathrm{O}_2\)).
  • It is pertinent to note that \(\mathrm{N}_{2}\mathrm{O}_{5}\) is often studied in controlled environments to understand its reactivity.

Understanding \(\mathrm{N}_{2}\mathrm{O}_{5}\)'s behavior in reactions lets chemists predict and manipulate reaction pathways and rates. It also helps demonstrate overall concepts in chemistry, such as reaction kinetics and mechanisms.
Rate Constant
The rate constant (\(k\)) is a unique value in a rate law equation that indicates how fast a reaction occurs under specific conditions. It is dependent on factors such as temperature and the nature of the reactants involved.
  • The rate constant provides critical insight into the reaction鈥檚 dynamics. A higher \(k\) means a faster reaction.
  • In temperature-dependent studies, the Arrhenius equation often comes into play: \[ k = Ae^{-\frac{E_a}{RT}} \] where \(A\) is the pre-exponential factor, \(E_a\) is activation energy, \(R\) is the gas constant, and \(T\) is temperature in Kelvin.

By calculating the rate constant, chemists can evaluate the conditions under which a reaction is most efficient. In decomposition reactions like that of \(\mathrm{N}_{2}\mathrm{O}_{5}\), understanding \(k\) helps elucidate the relationship between reactants and reaction speed. It鈥檚 an essential component for anyone studying reaction kinetics to master.

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Most popular questions from this chapter

In the reaction between nitrogen dioxide and ozone, $$2 \mathrm{NO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{O}_{2}(g)$$ how are the rates of change in the concentrations of the reactants and products related?

During a smog event, trace amounts of many highly reactive substances are present in the atmosphere. One of these is the hydroperoxyl radical, \(\mathrm{HO}_{2},\) which reacts with sulfur trioxide, \(\mathrm{SO}_{3}\). The rate constant for the reaction $$2 \mathrm{HO}_{2}(g)+\mathrm{SO}_{3}(g) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{3}(g)+2 \mathrm{O}_{2}(g)$$ is \(2.6 \times 10^{11} M^{-1} s^{-1}\) at \(298 \mathrm{K} .\) The initial rate of the reaction doubles when the concentration of \(\mathrm{SO}_{3}\) or \(\mathrm{HO}_{2}\) is doubled. What is the rate law for the reaction?

Nitric oxide (NO) is a gaseous free radical that plays many biological roles, including regulating neurotransmission and the human immune system. One of its many reactions involves the peroxynitrite ion (ONOO'): $$\mathrm{NO}(g)+\mathrm{ONOO}^{-(a q) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{2}^{-}(a q)}$$ a. Use the following data to determine the rate law and rate constant of the reaction at the experimental temperature at which these data were generated. $$\begin{array}{cccc}\text { Experiment } & \text { [NO]o (M) } & \text { [ONOO }\left.^{-}\right]_{0}(M) & \text { Rate }(M / \mathrm{s}) \\\\\hline 1 & 1.25 \times 10^{-4} & 1.25 \times 10^{-4} & 2.03 \times 10^{-11} \\\\\hline 2 & 1.25 \times 10^{-4} & 0.625 \times 10^{-4} & 1.02 \times 10^{-11} \\\\\hline 3 & 0.625 \times 10^{-4} & 2.50 \times 10^{-4} & 2.03 \times 10^{-11} \\\\\hline 4 & 0.625 \times 10^{-4} & 3.75 \times 10^{-4} & 3.05 \times 10^{-11} \\\\\hline\end{array}$$ b. Draw the Lewis structure of peroxynitrite ion (including all resonance forms) and assign formal charges. Note which form is preferred. c. Use the average bond energies in Appendix Table A4.1 to estimate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) using the preferred structure from part (b).

Is NO a catalyst for the decomposition of \(\mathrm{N}_{2} \mathrm{O}\) in the following two-step reaction mechanism, or is \(\mathrm{N}_{2} \mathrm{O}\) a catalyst for the conversion of \(\mathrm{NO}\) to \(\mathrm{NO}_{2} ?\) (1) \(\quad \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \rightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g)\) (2) \(\quad 2 \mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\)

The chemistry of smog formation includes \(\mathrm{NO}_{3}\) as an intermediate in several reactions. a. If \(\Delta\left[\mathrm{NO}_{3}\right] / \Delta t=-2.2 \times 10^{5} \mathrm{m} M / \mathrm{min}\) in the following reaction, what is the rate of formation of \(\mathrm{NO}_{2} ?\) $$\mathrm{NO}_{3}(g)+\mathrm{NO}(g) \rightarrow 2 \mathrm{NO}_{2}(g)$$ b. What is the rate of change of \(\left[\mathrm{NO}_{2}\right]\) in the following$$ \begin{array}{l} \text { reaction if } \Delta\left[\mathrm{NO}_{3}\right] / \Delta t=-2.3 \mathrm{mM} / \mathrm{min} \text { ? } \\ \qquad 2 \mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \end{array}$$
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