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The decomposition of \(\mathrm{NO}_{2}(g)\) occurs by the following bimolecular elementary reaction: $$ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ The rate constant at \(273 \mathrm{~K}\) is \(2.3 \times 10^{-12} \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\), and the activation energy is \(111 \mathrm{~kJ} / \mathrm{mol}\). How long will it take for the concentration of \(\mathrm{NO}_{2}(g)\) to decrease from an initial partial pressure of \(2.5\) atm to \(1.5\) atm at \(500 . \mathrm{K}\) ? Assume ideal gas behavior.

Short Answer

Expert verified
It will take approximately \(1.04 \times 10^7\) seconds for the concentration of NO₂(g) to decrease from an initial partial pressure of 2.5 atm to 1.5 atm at 500 K, assuming ideal gas behavior.

Step by step solution

01

Convert partial pressures to concentrations

We will first convert the given initial and final partial pressures of NO₂(g) to initial and final concentrations. Using the ideal gas law: \(C = \frac{P}{RT}\) where C is the concentration, P is the partial pressure, R is the ideal gas constant (0.08206 L⋅atm/mol⋅K), and T is the temperature. Initial concentration: \(C_{1} = \frac{2.5\;\text{atm}}{(0.08206\;\text{L}\cdot\text{atm/mol} \cdot \text{K})(500\;\text{K})} = 0.06091\;\text{mol/L}\) Final concentration: \(C_{2} = \frac{1.5\;\text{atm}}{(0.08206\;\text{L}\cdot\text{atm/mol} \cdot \text{K})(500\;\text{K})} = 0.03654\;\text{mol/L}\)
02

Find the rate constant at 500 K

We will use the Arrhenius equation to find the rate constant (k) at 500 K: \(k = k_0 e^{\frac{-E_a}{RT}}\) where k₀ is the rate constant at 273 K (given as \(2.3 \times 10^{-12}\; \text{L/mol}\cdot\text{s}\)), Eₐ is the activation energy (given as \(111 \times 10^3\; \text{J/mol}\)), R is the gas constant in J/mol⋅K (8.314 J/mol⋅K), and T is the temperature at which the reaction takes place (500 K). \(k = (2.3 \times 10^{-12}) e^{\frac{-(111 \times 10^3)}{(8.314)(500)}} = 9.33 \times 10^{-10}\; \text{L/mol}\cdot\text{s}\)
03

Write the rate expression

Given a bimolecular elementary reaction: \(2\;\text{NO}_2 (g) \longrightarrow 2\;\text{NO} (g) + \text{O}_2 (g)\) The rate expression for the decomposition of NO₂(g) is: \(-\frac{d[\text{NO}_2]}{dt} = k[\text{NO}_2]^2\)
04

Integrate the rate expression and solve for time

Now, we will integrate the rate expression to find the time (t) taken for the concentration of NO₂(g) to decrease from C₁ to C₂: \(\int_{C_1}^{C_2} \frac{d[\text{NO}_2]}{[\text{NO}_2]^2} = -\int_{0}^{t} k \; dt\) Solving the integral, we get: \(\left[-\frac{1}{[\text{NO}_2]}\right]_{C_1}^{C_2} = -k(t - 0)\) Substitute the values of C₁, C₂, and k: \(\left[-\frac{1}{0.03654} + \frac{1}{0.06091}\right] = -(9.33 \times 10^{-10})(t)\) Solve for t: \(t = \frac{-1}{9.33 \times 10^{-10}}\left[-\frac{1}{0.03654} + \frac{1}{0.06091}\right]\) \(t = 1.04 \times 10^7\; \text{s}\) So, it will take approximately \(1.04 \times 10^7\) seconds for the concentration of NO₂(g) to decrease from an initial partial pressure of 2.5 atm to 1.5 atm at 500 K, assuming ideal gas behavior.

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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 understanding reaction rates and kinetics. It refers to the minimum amount of energy that reactant molecules must possess to overcome the energy barrier for a reaction to occur. The higher the activation energy, the slower the reaction, as fewer molecules will have enough energy to react at a given temperature.
  • This concept is central because it indicates how easily a reaction can be initiated.
  • Higher activation energies imply more energy is needed, leading to slower reaction rates.
The Arrhenius equation can be used to relate activation energy to the rate constant, showing how temperature affects the reaction rate. When the temperature increases, more molecules have sufficient energy to surpass the activation energy, thereby accelerating the reaction.
Bimolecular Reaction
A bimolecular reaction involves two reactant molecules coming together to react. Such reactions are denoted as second-order reactions because the rate at which the reaction occurs depends on the product of the concentrations of the two reacting species.
  • The rate of a bimolecular reaction can be expressed as: \[- rac{d[ ext{Reactant}]}{dt} = k[ ext{A}][ ext{B}],\]where \(k\) is the rate constant, and \( [A] \) and \( [B] \) are the concentrations of the reactants.
  • In the given exercise, \(2 ext{NO}_2 ightarrow 2 ext{NO} + ext{O}_2\)is a bimolecular reaction where two molecules of NO₂ come together.
These reactions require correct orientation and sufficient energy for successful collisions, exemplifying why activation energy is critical.
Rate Constant
The rate constant, represented by \(k\), is a determinant of the reaction rate under specified conditions. It is affected by temperature and is unique to every reaction. The Arrhenius equation provides a formula to calculate how the rate constant changes with temperature:\[k = k_0 e^\frac{-E_a}{RT},\]where \(k_0\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the gas constant, and \(T\) is the temperature.
  • The temperature dependency of \(k\) means reactions typically proceed faster at higher temperatures since more molecules have energy exceeding the activation energy.
  • In the example, finding \(k\) at 500 K was crucial for understanding how fast the decomposition of NO₂ will happen under those conditions.
Ideal Gas Law
The ideal gas law is a fundamental principle in chemistry, connecting pressure, volume, temperature, and amount of gas. It states that:\[PV = nRT,\]where \(P\) is the pressure, \(V\) is the volume, \(n\) is the amount of gas in moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin.
In the context of this exercise, the ideal gas law was pivotal for converting pressures into concentrations:\[C = \frac{P}{RT}\]
  • This conversion is essential because reaction rates are calculated based on concentrations rather than pressures.
  • The assumption of ideal gas behavior simplifies calculations, allowing students to focus on the kinetic aspects without detailed consideration of intermolecular forces.
Understanding these relationships provided the necessary groundwork for calculating changes in the concentration of NO₂ over time.

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Most popular questions from this chapter

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\), with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l} k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{~s}^{-1} \\ k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{~s}^{-1} \end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{~B}] ?\)

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

Define what is meant by unimolecular and bimolecular steps. Why are termolecular steps infrequently seen in chemical reactions?

Consider the following initial rate data for the decomposition of compound \(\mathrm{AB}\) to give \(\mathrm{A}\) and \(\mathrm{B}\) : Determine the half-life for the decomposition reaction initially having \(1.00 M \mathrm{AB}\) present.

The reaction $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{~K}\) and \(y\) -intercept of \(33.5\). Assume \(k\) has units of \(\mathrm{s}^{-1}\). a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\). c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\).

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