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Chapter 14: Problem 37

Calculate the activation energy, \(E_{a}\) for the reaction $$ 2 \mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \rightarrow 4 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ from the observed rate constants: \(k\) at \(25^{\circ} \mathrm{C}=\) \(3.46 \times 10^{-5} \mathrm{s}^{-1}\) and \(k\) at \(55^{\circ} \mathrm{C}=1.5 \times 10^{-3} \mathrm{s}^{-1}\).

Short Answer

Expert verified
The activation energy, \(E_a\), is 101.0 kJ/mol.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius Equation relates the rate constant \(k\) to the temperature \(T\) and activation energy \(E_a\): \[ k = A e^{-E_a / (RT)} \] where \(A\) is the frequency factor, \(R\) is the universal gas constant \(8.314 \, \text{J/mol} \, \text{K}\), and \(T\) is the temperature in Kelvin.
02

Convert Temperatures to Kelvin

Convert temperatures from degrees Celsius to Kelvin by adding 273.15. For 25掳C: \[ T_1 = 25 + 273.15 = 298.15 \, \text{K} \] and for 55掳C: \[ T_2 = 55 + 273.15 = 328.15 \, \text{K} \].
03

Set Up the Arrhenius Equation Ratio

Using the points \((k_1,T_1)\) and \((k_2,T_2)\), the ratio of the two Arrhenius equations can be written as: \[ \frac{k_2}{k_1} = \frac{A e^{-E_a / (R T_2)}}{A e^{-E_a / (R T_1)}} \, \] which simplifies to \[ \ln\left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \].
04

Substitute Known Values

Substitute \(k_1 = 3.46 \times 10^{-5} \, \text{s}^{-1}\), \(k_2 = 1.5 \times 10^{-3} \, \text{s}^{-1}\), \(T_1 = 298.15 \, \text{K}\), \(T_2 = 328.15 \, \text{K}\), and \(R = 8.314 \, \text{J/mol} \, \text{K}\) into the simplified equation. Calculate \( \ln\left(\frac{k_2}{k_1}\right) \) and \( \frac{1}{T_2} - \frac{1}{T_1} \).
05

Solve for Activation Energy \(E_a\)

Calculate \( \ln\left(\frac{1.5 \times 10^{-3}}{3.46 \times 10^{-5}}\right)\) which simplifies to \( \ln(43.352) = 3.768 \). Then compute \(\frac{1}{328.15} - \frac{1}{298.15} = -3.036 \times 10^{-4} \, \text{K}^{-1}\). Solve the equation: \[ 3.768 = \frac{-E_a}{8.314}(-3.036 \times 10^{-4}) \] which results in \(E_a = 101.0 \, \text{kJ/mol}\).

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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 central concept in chemical kinetics, providing insight into the temperature dependence of reaction rates. This equation is given by \[ k = A e^{-E_a / (RT)} \]where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.
  • The frequency factor \(A\) represents the frequency of collisions with the correct orientation for a reaction to occur.
  • Activation energy \(E_a\) is the minimum energy that must be provided for a chemical reaction to occur.
  • The universal gas constant \(R\) is a constant with the value \(8.314 \, \text{J/mol} \, \text{K}\).
Understanding the Arrhenius equation allows us to predict how the reaction rate \(k\) changes with temperature \(T\). When the activation energy \(E_a\) is high, the reaction rate is more sensitive to temperature changes. In contrast, with a lower \(E_a\), reactions will proceed more readily even at lower temperatures.
rate constant
The rate constant \(k\) is a critical parameter in the field of chemistry, determining the speed at which a chemical reaction occurs. It quantifies the rate of a reaction and is derived from the Arrhenius equation:
  • Higher values of \(k\) suggest faster reactions as they represent a greater number of successful collisions per unit time.
  • By analyzing changes in \(k\) at different temperatures, we can better understand a reaction's kinetics and calculate the activation energy \(E_a\).
  • In the equation \(\ln(k) = \ln(A) - \frac{E_a}{RT}\), adjusting temperature \(T\) affects the value of \(k\), demonstrating the relationship between heat and reaction speed.
Rate constants have units dependent on the overall reaction order (e.g., \(\text{s}^{-1}\) for first order), indicating how quickly or slowly a reaction will proceed under certain conditions.
temperature conversion
Temperature conversion is essential when dealing with chemical kinetics, especially in the context of the Arrhenius equation, where temperature appears in the exponential term. Temperature in chemistry is typically noted in Kelvin.
  • The Kelvin scale is used because it starts at absolute zero, providing an absolute measure of temperature.
  • To convert from degrees Celsius to Kelvin, simply add 273.15: \[ T_{\text{K}} = T_{\text{掳C}} + 273.15 \]
  • Using Kelvin allows equations like the Arrhenius equation to remain consistent and accurate across all scientific calculations.
In our calculation, converting 25掳C and 55掳C to Kelvin allowed precise use of the Arrhenius equation, ensuring the calculation of activation energy \(E_a\) was accurate.
universal gas constant
The universal gas constant \(R\) is a fundamental constant in both physics and chemistry. It plays a crucial role in the Arrhenius equation by linking the energy scale to temperature in Kelvin.
  • The value of \(R\) is \(8.314 \, \text{J/mol} \, \text{K}\), making it a pivotal factor in calculations involving moles and temperature.
  • It appears in many important equations, such as the ideal gas law \(PV = nRT\), and is also used to relate activation energy to the temperature and rate constant in kinetic analyses.
  • Because it is expressed in Joules per mole per Kelvin, \(R\) ensures that various thermodynamic and kinetic calculations remain dimensionally consistent.
In calculating activation energy \(E_a\) using the Arrhenius equation, \(R\) provides the connection between the energy units and thermal jiggling of reactant molecules, allowing scientists to better understand reactions' temperature dependencies.

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Most popular questions from this chapter

We know that the decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first- order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ with a half-life of 245 minutes at 600 K. If you begin with a partial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) of \(25 \mathrm{mm}\) Hg in a 1.0 -L flask, what is the partial pressure of each reactant and product after 245 minutes? What is the partial pressure of each reactant and product after 12 hours?

The decomposition of phosphine, \(\mathrm{PH}_{3}\), proceeds according to the equation $$ \mathrm{PH}_{3}(\mathrm{g}) \rightarrow^{1 / 4} \mathrm{P}_{4}(\mathrm{g})+3 / 2 \mathrm{H}_{2}(\mathrm{g}) $$ It is found that the reaction has the following rate equation: Rate \(=k\left[\mathrm{PH}_{3}\right] .\) The half-life of \(\mathrm{PH}_{3}\) is 37.9 seconds at \(120^{\circ} \mathrm{C}\) (a) How much time is required for three fourths of the \(\mathrm{PH}_{3}\) to decompose? (b) What fraction of the original sample of \(\mathrm{PH}_{3}\) remains after 1.00 minute?

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures, a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. The rate constant for this reaction is \(3.40 \mathrm{L} / \mathrm{mol} \cdot \mathrm{min} .\) Determine the time needed for the concentration of \(\mathrm{NO}_{2}\) to decrease from \(2.00 \mathrm{mol} / \mathrm{L}\) to \(1.50 \mathrm{mol} / \mathrm{L}\).

Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2}\), decomposes slowly in aqueous solution according to the following reaction: $$ \mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The reaction follows the experimental rate law $$ \text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} $$ (a) What is the apparent order of the reaction in a pH buffered solution? (In a pH buffered solution, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is a constant.) (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. (Note that when writing the expression for \(K,\) the equilibrium constant, \(\left.\left[\mathrm{H}_{2} \mathrm{O}\right] \text { is not involved. See Chapter } 15 .\right)\) Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 $$ \mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{\overleftarrow{k_{2}^{\prime}}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} $$ \(\mathrm{NO}_{2} \mathrm{NH}_{3}^{+} \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+} \quad\) (rate-limiting step) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\underset{(\text { rapid equilibrium })}{\mathrm{H}_{3} \mathrm{O}^{+}}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \stackrel{k_{5}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rate-limiting step) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (d) Based on the experimental rate law, will the reaction rate increase or decrease if the pH of the solution is increased?
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