/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 100 The decomposition of \(\mathrm{N... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 100

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 NO鈧�(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 NO鈧�(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鈰卆tm/mol鈰匥), 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 k鈧� is the rate constant at 273 K (given as \(2.3 \times 10^{-12}\; \text{L/mol}\cdot\text{s}\)), E鈧� is the activation energy (given as \(111 \times 10^3\; \text{J/mol}\)), R is the gas constant in J/mol鈰匥 (8.314 J/mol鈰匥), 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 NO鈧�(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 NO鈧�(g) to decrease from C鈧� to C鈧�: \(\int_{C_1}^{C_2} \frac{d[\text{NO}_2]}{[\text{NO}_2]^2} = -\int_{0}^{t} k \; dt\) Solving the integral, we get: \(\left[-\frac{1}{[\text{NO}_2]}\right]_{C_1}^{C_2} = -k(t - 0)\) Substitute the values of C鈧�, C鈧�, 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 NO鈧�(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 NO鈧� 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 NO鈧� 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 NO鈧� over time.

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Most popular questions from this chapter

The activation energy of a certain uncatalyzed biochemical reaction is \(50.0 \mathrm{~kJ} / \mathrm{mol}\). In the presence of a catalyst at \(37^{\circ} \mathrm{C}\), the rate constant for the reaction increases by a factor of \(2.50 \times 10^{3}\) as compared with the uncatalyzed reaction. Assuming the frequency factor \(A\) is the same for both the catalyzed and uncatalyzed reactions, calculate the activation energy for the catalyzed reaction.

Chemists commonly use a rule of thumb that an increase of \(10 \mathrm{~K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C}\) ?

Each of the statements given below is false. Explain why. a. The activation energy of a reaction depends on the overall energy change \((\Delta E)\) for the reaction. b. The rate law for a reaction can be deduced from examination of the overall balanced equation for the reaction. c. Most reactions occur by one-step mechanisms.

One mechanism for the destruction of ozone in the upper atmosphere is $$ \begin{array}{ll} \mathrm{O}_{3}(g)+\mathrm{NO}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) & \text { Slov } \\ \mathrm{NO}_{2}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \text { Fast } \\ \hline \end{array} $$ Overall reaction \(\mathrm{O}_{3}(\mathrm{~g})+\mathrm{O}(\mathrm{g}) \longrightarrow 2 \mathrm{O}_{2}(\mathrm{~g})\) a. Which species is a catalyst? b. Which species is an intermediate? c. \(E_{\mathrm{a}}\) for the uncatalyzed reaction $$ \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{O}_{2} $$ is \(14.0 \mathrm{~kJ} . E_{\mathrm{a}}\) for the same reaction when catalyzed is \(11.9 \mathrm{~kJ}\). What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction at \(25^{\circ} \mathrm{C}\) ? Assume that the frequency factor \(A\) is the same for each reaction.

The mechanism for the reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon dioxide is thought to be $$ \begin{aligned} \mathrm{NO}_{2}+\mathrm{NO}_{2} \longrightarrow \mathrm{NO}_{3}+\mathrm{NO} & \text { Slow } \\ \mathrm{NO}_{3}+\mathrm{CO} \longrightarrow \mathrm{NO}_{2}+\mathrm{CO}_{2} & \text { Fast } \end{aligned} $$ Write the rate law expected for this mechanism. What is the overall balanced equation for the reaction?
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