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Chapter 14: Problem 74

The gas-phase reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ has an activation energy of \(103 \mathrm{kJ} / \mathrm{mol},\) and the rate constant is 0.0900 min \(^{-1}\) at 328.0 K. Find the rate constant at \(318.0 \mathrm{K}\).

Short Answer

Expert verified
The rate constant at 318.0 K is approximately 0.0274 min\(^{-1}\).

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation is used to find how the rate constant changes with temperature. It is given by \[ k = A e^{-\frac{E_a}{RT}} \] where \( k \) is the rate constant, \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant (8.314 J/mol路K), and \( T \) is the temperature in Kelvin.
02

Use Arrhenius Equation Rearranged

We can use a derived form of the Arrhenius equation for solving problems involving two temperatures: \[ \ln \left( \frac{k_2}{k_1} \right) = -\frac{E_a}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] where \( k_1 \) and \( k_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively.
03

List Known Variables

From the problem, know the following:- \( k_1 = 0.0900 \) min\(^{-1}\) at \( T_1 = 328.0 \) K- \( T_2 = 318.0 \) K- \( E_a = 103000 \) J/mol (converted from kJ/mol)- \( R = 8.314 \) J/mol路K.
04

Calculate the Factor \( \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \)

Compute \( \frac{1}{T_2} - \frac{1}{T_1} \):\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{318.0} - \frac{1}{328.0} = \frac{328.0 - 318.0}{318.0 \times 328.0} \]Calculate:\[ \frac{10}{318.0 \times 328.0} = \frac{10}{104304} \approx 9.59 \times 10^{-5} \text{ K}^{-1} \]
05

Compute \( \ln \left( \frac{k_2}{k_1} \right) \)

Using the Arrhenius equation, substitute the known values:\[ \ln \left( \frac{k_2}{0.0900} \right) = -\frac{103000}{8.314} \times 9.59 \times 10^{-5} \]Calculate the expression on the right:\[ \ln \left( \frac{k_2}{0.0900} \right) = -12390 \times 9.59 \times 10^{-5} \approx -1.1877 \]
06

Solve for \( k_2 \)

Exponentiate both sides to solve for \( \frac{k_2}{0.0900} \):\[ \frac{k_2}{0.0900} = e^{-1.1877} \approx 0.305 \]Thus, solve for \( k_2 \):\[ k_2 = 0.0900 \times 0.305 \approx 0.0274 \text{ min}^{-1} \]

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Key Concepts

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

Activation Energy
Activation energy is a crucial concept in chemistry that determines the speed at which a reaction can occur. It is the minimum energy required to start a chemical reaction. This energy barrier must be overcome for reactants to convert into products.
  • Think of it as the initial push needed to get a ball rolling down a hill.
  • Chemical reactions rely on this energy to help break and form chemical bonds.
In the context of the Arrhenius equation, activation energy (\(E_a\)) is an important factor that influences the rate constant and shows how energetically demanding a reaction is.
A higher activation energy implies a slower reaction at a given temperature, since more energy is needed to overcome the barrier, while a lower activation energy suggests a faster reaction.
Rate Constant
The rate constant, denoted as \(k\), is a fundamental component in the rate equation for a chemical reaction. It defines the reaction speed and how effectively the reactants are converted into products.
  • It is influenced by factors like temperature and activation energy.
  • The Arrhenius equation shows how \(k\) changes with temperature: \\[ k = A e^{-\frac{E_a}{RT}} \\]
Understanding \(k\) helps predict if a reaction will be fast or slow under certain conditions.
In the exercise, the rate constant changes with temperature, showcasing its dependence on both temperature and activation energy.
Gas Constant
The gas constant, represented by \(R\), is a universal constant connected to energy amounts per mole per degree Kelvin. In the context of the Arrhenius equation, it serves as the bridge between energy (in joules) and temperature (in Kelvin).
  • The value of \(R\) is \(8.314\) J/mol路K.
  • It is used to convert energy scales in the Arrhenius equation.
The gas constant appears frequently in thermodynamics and chemical kinetics to harmonize energy with temperature.
In calculations, \(R\) helps determine how factors like activation energy affect reaction rates.
Temperature Dependence of Reaction Rates
The rate of a chemical reaction is highly dependent on temperature, which is explained by the Arrhenius equation. As temperature increases, the reaction rate generally increases because particles have more energy, making it easier to overcome the activation energy barrier.
  • The relationship is captured by the equation:\( k = A e^{-\frac{E_a}{RT}} \)
  • An increase in temperature results in a higher rate constant \(k\), showing faster reactions.
At higher temperatures, molecules move more quickly and collide more often, leading to increased chances of successful reactions.
In the example exercise, you can see how a small change in temperature (from 328 K to 318 K) affects the rate constant, highlighting the temperature sensitivity of reaction rates.

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Most popular questions from this chapter

Give the relative rates of disappearance of reactants and formation of products for each of the following reactions. (a) \(2 \mathrm{O}_{3}(\mathrm{g}) \rightarrow 3 \mathrm{O}_{2}(\mathrm{g})\) (b) \(2 \mathrm{HOF}(\mathrm{g}) \rightarrow 2 \mathrm{HF}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g})\)

Using the rate equation Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}],\) define the order of the reaction with respect to A and B. What is the total order of the reaction?

Many biochemical reactions are catalyzed by acids. A typical mechanism consistent with the experimental results (in which HA is the acid and X is the reactant) is Step \(1: \quad\) Fast, reversible: \(\quad \mathrm{HA} \rightleftarrows \mathrm{H}^{+}+\mathrm{A}^{-}\) Step \(2: \quad\) Fast, reversible: \(\quad \mathrm{X}+\mathrm{H}^{+} \rightleftharpoons \mathrm{XH}^{+}\) Step 3: Slow \(\mathrm{XH}^{+} \rightarrow\) products What rate law is derived from this mechanism? What is the order of the reaction with respect to HA? How would doubling the concentration of HA affect the reaction?

The substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another molecule L Iwhere L is an electron-pair donor such as \(\mathrm{P}\left(\mathrm{CH}_{3}\right)_{3}\) ) was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. \(90,\) p. \(6927,1968 .\) A detailed study of the kinetics of the reaction led to the following mechanism: Slow: \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast: \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) increased the reaction rate by a factor of 2 Doubling the concentration of L had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C}\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M},\) what is the concentration of the product after 5.0 minutes?

When heated, cyclopropane is converted to propene (Example \(14.5) .\) Rate constants for this reaction at \(470^{\circ} \mathrm{C}\) and \(510^{\circ} \mathrm{C}\) are \(k=\) \(1.10 \times 10^{-4} \mathrm{s}^{-1}\) and \(k=1.02 \times 10^{-3} \mathrm{s}^{-1},\) respec- tively. Determine the activation energy, \(E_{a}\), from these data.
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