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Rate constants for the decomposition of gaseous dinitrogen pentoxide are \(3.7 \times 10^{-5} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\) and \(1.7 \times 10^{-3} \mathrm{~s}^{-1}\) at \(55^{\circ} \mathrm{C}\). $$ 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$ (a) What is the activation energy for this reaction in \(\mathrm{kJ} / \mathrm{mol}\) ? (b) What is the rate constant at \(35^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
(a) 101.1 kJ/mol, (b) 2.17 脳 10鈦宦 s鈦宦.

Step by step solution

01

Use the Arrhenius Equation

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

Convert Temperatures to Kelvin

Convert 25掳C and 55掳C to Kelvin: \( 25^{\circ} \text{C} = 298 \text{ K} \) and \( 55^{\circ} \text{C} = 328 \text{ K} \).
03

Calculate Activation Energy

The Arrhenius equation for two temperatures, \( T_1 \) and \( T_2 \), is \( \ln \left( \frac{k_2}{k_1} \right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \). Substituting the given values: \( \ln \left( \frac{1.7 \times 10^{-3}}{3.7 \times 10^{-5}} \right) = \frac{E_a}{8.314} \left( \frac{1}{298} - \frac{1}{328} \right) \). Calculate \( E_a \).
04

Solve for Activation Energy

Calculate the natural logarithm: \( \ln \left( \frac{1.7 \times 10^{-3}}{3.7 \times 10^{-5}} \right) = \ln \left( 45.946 \right) \approx 3.827 \). The temperature term is \( \left( \frac{1}{298} - \frac{1}{328} \right) = 0.0003363 \). Solving for \( E_a \): \( 3.827 = \frac{E_a}{8.314} \times 0.0003363 \). Thus, \( E_a = \frac{3.827}{0.0003363} \times 8.314 \approx 101113 \) J/mol, or 101.1 kJ/mol.
05

Determine Rate Constant at 35掳C

Use the Arrhenius equation again to estimate the rate constant \( k \) at 35掳C. Convert 35掳C to 308 K. Substitute known values: \( \ln(\frac{k_{35}}{3.7 \times 10^{-5}}) = \frac{(-101113 J/mol)}{8.314 J/mol\cdot K} \left( \frac{1}{298} - \frac{1}{308} \right) \).
06

Solve for Rate Constant

The temperature difference term becomes \( \frac{1}{298} - \frac{1}{308} = 0.00033557 \). Using \( \ln(k_{35}) = \ln(3.7 \times 10^{-5}) + \frac{101113}{8.314} \times 0.00033557 \), solve for \( k_{35} \). You find \( \ln(k_{35}) \approx -10.21 + 4.080 \approx -6.13 \), so \( k_{35} \approx e^{-6.13} \approx 2.17 \times 10^{-3} \) s\(^{-1}\).

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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 fundamental formula in chemistry used to describe the relationship between the rate constant and temperature, and it provides a way to understand the effect of temperature on the speed of a chemical reaction. Mathematically, the equation is written as:\[ k = A e^{-E_a/(RT)} \]where:
  • \( k \) is the rate constant, which gives us an idea of how fast a reaction proceeds.
  • \( A \) is the pre-exponential factor (also known as the frequency factor), which represents the number of times that molecules collide correctly per second.
  • \( E_a \) is the activation energy, the minimum energy required for the reaction to take place.
  • \( R \) is the universal gas constant, \(8.314 \, \text{J/mol K} \).
  • \( T \) is the temperature in Kelvin.
Understanding this equation helps us calculate how the rate of reaction changes with different temperatures. It shows that even a small increase in temperature can lead to a significant increase in the rate constant due to the exponential relationship.
Rate constant
The rate constant \( k \) is a crucial component of reaction kinetics, as it quantifies the speed of a chemical reaction at a given temperature. Its units depend on the order of the reaction. For a first-order reaction like the decomposition of dinitrogen pentoxide, the units are \( \text{s}^{-1} \).

In the context of the Arrhenius equation, a higher rate constant indicates a faster reaction at the corresponding temperature. Conversely, a lower rate constant suggests a slower reaction. By comparing the rate constants at different temperatures, as seen in our example, we can understand how temperature affects the reaction speed.

In practical applications, knowing the rate constant is useful for predicting how long a reaction will take to reach completion or to achieve a desired concentration of products.
Temperature conversion
Temperature conversion is an important step in calculations involving chemical reactions, especially when using the Arrhenius equation. The temperature in the Arrhenius equation must be in Kelvin, as it is the absolute scale that aligns with how energy processes are calculated in thermodynamics.

Converting temperatures from Celsius to Kelvin is straightforward:
  • Add 273.15 to the Celsius temperature.
For instance, 25掳C becomes 298 K and 55掳C converts to 328 K. Similarly, for the unknown rate constant calculation at 35掳C, we convert it to 308 K. This uniformity in temperature scales helps in maintaining accuracy and consistency in chemical calculations.
Reaction kinetics
Reaction kinetics is the study of rates of chemical processes and the factors affecting them. It explores how different variables such as temperature, pressure, concentration, and presence of catalysts can influence the speed of a reaction.

This helps chemists and engineers design and optimize processes for the synthesis of products or for understanding natural processes. In our case, examining the kinetics of dinitrogen pentoxide breakdown involves analyzing how rate constants change with temperature, as determined by the Arrhenius equation.

Understanding reaction kinetics allows for comprehensive control over reaction conditions, enabling better yield and efficiency in industrial processes, or simply ensuring that reactions occur safely and predictably in laboratory conditions or in everyday life scenarios.

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Most popular questions from this chapter

The rate constant for the decomposition of gaseous \(\mathrm{NO}_{2}\) to \(\mathrm{NO}\) and \(\mathrm{O}_{2}\) is \(4.7 /(\mathrm{M} \cdot \mathrm{s})\) at \(383^{\circ} \mathrm{C}\). Consider the decomposition of a sample of pure \(\mathrm{NO}_{2}\) having an initial pressure of \(746 \mathrm{~mm} \mathrm{Hg}\) in a \(5.00 \mathrm{~L}\) reaction vessel at \(383^{\circ} \mathrm{C}\). (a) What is the order of the reaction? (b) What is the initial rate of formation of \(\mathrm{O}_{2}\) in \(\mathrm{g} /(\mathrm{L} \cdot \mathrm{s})\) ? (c) What is the mass of \(\mathrm{O}_{2}\) in the vessel after a reaction time of \(1.00 \mathrm{~min} ?\)

In acidic aqueous solution, the purple complex ion \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}^{2+}\) undergoes a slow reaction in which the bromide ion is replaced by a water molecule, yielding the pinkishorange complex ion \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5}\left(\mathrm{H}_{2} \mathrm{O}\right)^{3+}\) : The reaction is first order in \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}^{2+}\), the rate constant at \(25^{\circ} \mathrm{C}\) is \(6.3 \times 10^{-6} \mathrm{~s}^{-1}\), and the initial concentration of \(\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{Br}^{2+}\) is \(0.100 \mathrm{M}\).

Beginning with the integrated rate law, derive a general equation for the half-life of a zeroth-order reaction of the type \(\mathrm{A} \rightarrow\) Products. How does the length of each half-life compare with the length of the previous one? Make the same comparison for firstorder and second-order reactions.

When the temperature of a gas is raised by \(10{ }^{\circ} \mathrm{C}\), the collision frequency increases by only about \(2 \%\), but the reaction rate increases by \(100 \%\) (a factor of 2 ) or more. Explain.

A certain first-order reaction has a rate constant of \(1.0 \times 10^{-3} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). (a) If the reaction rate doubles when the temperature is increased to \(35^{\circ} \mathrm{C}\), what is the activation energy for this reaction in \(\mathrm{kJ} / \mathrm{mol} ?\) (b) What is the \(E_{a}\) (in \(\mathrm{kJ} / \mathrm{mol}\) ) if the same temperature change causes the rate to triple?

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