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Chapter 13: Problem 119

On the basis of the frequency factors and activation energy values of the following two reactions, determine which one will have the larger rate constant at room temperature \((298 \mathrm{K})\). \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}_{2}(g)\) \(A=8.0 \times 10^{-12} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{\mathrm{a}}=17.1 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{O}_{3}(g)+\mathrm{Cl}(g) \rightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g)\) \(A=2.9 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \quad E_{\mathrm{a}}=2.16 \mathrm{kJ} / \mathrm{mol}\)

Short Answer

Expert verified
The reaction between ozone and chlorine has a larger rate constant at room temperature (1.25 x 10^{-11} cm鲁/(molecules路s)) compared to the reaction between ozone and oxygen (6.43 x 10^{-14} cm鲁/(molecules路s)), making it proceed faster.

Step by step solution

01

Convert the activation energy values to Joules per mole

The activation energy values are given in kJ/mol. We must convert them to J/mol in order to use the values in the Arrhenius equation. Recall that 1 kJ = 1000 J. \(Ea_{Ozone-Oxygen} = 17.1\: \mathrm{kJ/mol} \times 1000\: \mathrm{J\: (kJ)^{-1}} = 17,100\: \mathrm{J/mol}\) \(Ea_{Ozone-Chlorine} = 2.16\: \mathrm{kJ/mol} \times 1000\: \mathrm{J\: (kJ)^{-1}} = 2,160\: \mathrm{J/mol}\)
02

Calculate the rate constant for both reactions using the Arrhenius equation

To calculate the rate constant for both reactions, we'll plug the given values of A, Ea, R, and T into the Arrhenius equation for both reactions. \(k_{Ozone-Oxygen} = A_{Ozone-Oxygen} \cdot e^{\frac{-Ea_{Ozone-Oxygen}}{RT}}\) \(k_{Ozone-Chlorine} = A_{Ozone-Chlorine} \cdot e^{\frac{-Ea_{Ozone-Chlorine}}{RT}}\) \(k_{Ozone-Oxygen} = 8.0 \times 10^{-12} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \cdot e^{\frac{-17,100\: \mathrm{J/mol}}{(8.314\: \mathrm{J\: (mol\: K)^{-1})\times 298\: \mathrm{K}}} \) \(k_{Ozone-Oxygen} 鈮� 6.43 \times 10^{-14} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s})\) \(k_{Ozone-Chlorine} = 2.9 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s}) \cdot e^{\frac{-2,160\: \mathrm{J/mol}}{(8.314\: \mathrm{J\: (mol\: K)^{-1})\times 298\: \mathrm{K}}} \) \(k_{Ozone-Chlorine} 鈮� 1.25 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s})\)
03

Compare the rate constants

Analyze the values of the rate constants for both reactions: \(k_{Ozone-Oxygen} = 6.43 \times 10^{-14} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s})\) \(k_{Ozone-Chlorine} = 1.25 \times 10^{-11} \mathrm{cm}^{3} /(\text { molecules } \cdot \mathrm{s})\) We can conclude that the reaction between ozone and chlorine has a larger rate constant (1.25 x 10^{-11} cm鲁/(molecules路s)) at room temperature, making it proceed faster than the reaction between ozone and oxygen (6.43 x 10^{-14} cm鲁/(molecules路s)).

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Key Concepts

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

Arrhenius Equation
The Arrhenius Equation is a key concept in understanding how chemical reactions occur at different temperatures. It provides a mathematical relationship between the rate constant of a reaction and the temperature at which the reaction occurs. This equation can be represented as:\[k = A imes e^{-\frac{E_a}{RT}}\]where:
  • \(k\) is the rate constant.
  • \(A\) is the frequency factor, which indicates how often molecules collide in a way that results in a reaction.
  • \(E_a\) is the activation energy, the minimum energy needed for the reaction to occur.
  • \(R\) is the gas constant \((8.314 \, \text{J/(mol K)})\).
  • \(T\) is the temperature in Kelvin.
The equation demonstrates that an increase in temperature \((T)\) or a decrease in activation energy \((E_a)\) will increase the rate constant \((k)\). As seen in the exercise, reactions with a lower activation energy have higher rate constants at the same temperature, making them proceed faster.
Activation Energy
Activation Energy \((E_a)\) is crucial for understanding the speed of chemical reactions. It is the minimum energy that reactant molecules must possess for a reaction to take place. Imagine rolling a ball over a hill; the ball needs enough energy to reach the top of the hill before it can roll down the other side. Similarly, molecules need a certain amount of energy to break existing bonds and form new ones during a chemical reaction.

Factors Influencing Activation Energy

Activation energy can be influenced by different factors, such as:
  • Nature of reactants: Some reactants naturally have lower activation energies due to their atomic or molecular structure.
  • Catalysts: These substances lower the activation energy, allowing reactions to proceed more quickly.
  • Temperature: While it doesn't change \(E_a\) directly, higher temperatures can make it easier for molecules to reach the necessary energy level.
Lower activation energy means that more molecules can participate in the reaction at a given temperature, increasing the reaction rate as observed in the ozone and chlorine reaction.
Chemical Reaction Kinetics
Chemical Reaction Kinetics is the study of speed or rate at which a chemical reaction occurs. It explores the factors that influence reaction rates, including the concentrations of reactants, presence of catalysts, and the temperature and pressure at which the reaction is carried out.

Understanding Reaction Rates

The rate of a chemical reaction is determined by how quickly the reactants are transformed into products. By examining reaction kinetics, chemists aim to understand and predict these rates.
  • Reactions can be either fast (like explosions) or slow (like rusting).
  • Several methods are used to measure reaction rates, such as monitoring changes in concentration over time.
  • The study also helps in the development of new processes and technologies, by optimizing reaction rates for industrial and laboratory applications.
In our exercise example, comparing reaction rates at room temperature helps us understand which molecular transformations occur more readily, illustrating the principles of chemical kinetics.

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Most popular questions from this chapter

The rate at which drugs are metabolized depends upon age: children metabolize some drugs more rapidly than adults, while the elderly metabolize drugs more slowly. Diazepam is used to treat anxiety disorders and seizures in patients in all age groups. Its half-life in hours is estimated to be equal to the patient's age in years; in a 50 -year-old, for example, diazepam would have a 50 -hour half-life. How long will it take for \(95 \%\) of a dose of diazepam to be metabolized in a 5-year-old child compared to a 50 -year-old adult assuming a first order process?

Radioactive isotopes such as \(^{32} \mathrm{P}\) are used to follow biological processes. The following radioactivity data (in relative radioactivity values) were collected for a sample containing \(^{32} \mathrm{P}\) : $$\begin{array}{cc} \text { Time (days) } & \text { Relative Radioactivity } \\\\\hline 0 & 10.00 \\\\\hline 1 & 9.53 \\\\\hline 2 & 9.08 \\\\\hline 5 & 7.85 \\\\\hline 10 & 6.16 \\\\\hline 20 & 3.79 \\\\\hline\end{array}$$ a. Write the rate law for the decay of \(^{32} \mathrm{P}\). b. Determine the value of the rate constant. c. Determine the half-life of \(^{32} \mathrm{P}\). d. How many days does it take for \(99 \%\) of a sample of \(^{32} \mathrm{P}\) to decay?

Predict the order with respect to NO for the reaction \(\mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightarrow \mathrm{NOBr}_{2}(g)\) under each of the following conditions: a. The rate doubles when \([\mathrm{NO}]\) is doubled and \(\left[\mathrm{Br}_{2}\right]\) remains constant. b. The rate increases by 1.56 times when \([\mathrm{NO}]\) is increased 1.25 times and \(\left[\mathrm{Br}_{2}\right]\) remains constant. c. The rate is halved when \([\mathrm{NO}]\) is doubled and \(\left[\mathrm{Br}_{2}\right]\) remains constant.

The kinetics of the reaction between chlorine dioxide and ozone are relevant to the study of atmospheric ozone destruction. The value of the rate constant for the reaction between chlorine dioxide and ozone was measured at four temperatures between 193 and \(208 \mathrm{K}\). The results were as follows: $$\begin{array}{cc}T(\mathrm{K}) & k\left(M^{-1} \mathrm{s}^{-1}\right) \\\193 & 34.0 \\\\\hline 198 & 62.8 \\\\\hline 203 & 112.8 \\\\\hline 208 & 196.7 \\\\\hline\end{array}$$ a. Calculate the values of the activation energy and the frequency factor for the reaction. b. What is the value of the rate constant higher in the stratosphere where \(T=245 \mathrm{K} ?\)

Nitrogen and oxygen can combine to form different nitrogen oxides that play a minor role in the chemistry of smog. Write balanced chemical equations for the reactions of \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) that produce (a) \(\mathrm{N}_{2} \mathrm{O}\) and (b) \(\mathrm{N}_{2} \mathrm{O}_{5}\).
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