What is the half-life for the decomposition of \(\mathrm{O}_{3}\) when the concentration of \(\mathrm{O}_{3}\) is \(2.35 \times 10^{-6} \mathrm{M}\) ? The rate constant for this second-order reaction is \(50.4 \mathrm{L} \mathrm{mol}^{-1} \mathrm{h}^{-1}\).

Short Answer

Expert verified
The half-life for the decomposition of \( \mathrm{O}_3 \) is approximately 8440.68 hours.

Step by step solution

01

Understand Half-Life for Second-Order Reactions

The half-life for a second-order reaction is given by the equation: \( t_{1/2} = \frac{1}{k[A]_0} \), where \( k \) is the rate constant and \( [A]_0 \) is the initial concentration of the reactant.
02

Substitute the Given Values

Substitute the given rate constant \( k = 50.4 \, \text{L mol}^{-1} \text{h}^{-1} \) and the initial concentration of \( \mathrm{O}_3 \), \( [A]_0 = 2.35 \times 10^{-6} \, \text{M} \) into the half-life equation.
03

Calculate the Half-Life

Using the substituted values, calculate the half-life: \( t_{1/2} = \frac{1}{(50.4 \, \text{L mol}^{-1} \text{h}^{-1})(2.35 \times 10^{-6} \, \text{M})} \).
04

Perform the Calculation

Now calculate the numerical value of the half-life: \( t_{1/2} = \frac{1}{(50.4) \times (2.35 \times 10^{-6})} = \frac{1}{0.00011844 \, \text{L mol}^{-1} \text{h}^{-1}} \approx 8440.68 \, \text{h} \).

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is a branch of physical chemistry that concerns the rates at which chemical reactions occur. Its study involves understanding how different variables—such as temperature, pressure, reactant concentration, and the presence of a catalyst—affect the speed of a reaction. The rate of a chemical reaction is usually expressed in terms of the change in concentration of reactants or products per unit of time. For students tackling kinetic problems, grasping the fundamental concept of reaction rates is crucial. It’s this rate that governs how fast a product is formed or a reactant is consumed.

In the case of second-order reactions, the rate is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. Differentiating between reaction orders is vital because it determines the mathematical relationships that allow us to predict reaction behavior and calculate essential parameters like half-life—a concept we explore in the context of the given exercise.
Reaction Rate Constant
The reaction rate constant, denoted as 'k', is a proportionality factor that provides a quantitative measure of the speed of a chemical reaction. For a given reaction at a particular temperature, the rate constant ties the reaction rate to the concentration(s) of the reactant(s) as described by the rate law. It is affected by environmental conditions, notably temperature, and in the case of heterogeneous reactions, surface area.

When solving kinetics problems, understanding the implications of the rate constant is essential. For instance, in a second-order reaction, a larger 'k' value means a faster reaction, as you can see from the inverse relationship in the second-order half-life equation. As demonstrated in the methodology provided for calculating the half-life of ozone decomposition, the rate constant is indispensable in predicting how long it takes for half of the reactant to react.
Reactant Concentration
Reactant concentration, often signified by square brackets \( [A] \), plays a pivotal role in the kinetics of a reaction. In simple terms, it's the measure of how much of a substance is present in a given volume of solution. The change in this concentration over time is what chemists refer to as the reaction rate. As you advance in chemistry, you'll often come across concentration expressed in molarity (M), or moles per liter (mol/L).

For second-order reactions specifically, the reaction rate depends on the concentration of one reactant raised to the second power, or the product of two reactant concentrations. This distinctive feature is reflected in the unique form of the half-life equation for these reactions. In practice, as the reactant concentration decreases, the rate of reaction slows down - a concept vividly illustrated by the exercise in question. Students should note that keeping track of concentration changes is fundamental for accurate reaction rate analysis and predictions.

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Most popular questions from this chapter

How much and in what direction will each of the following affect the rate of the reaction: \(\mathrm{CO}(g)+\mathrm{NO}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{NO}(g)\) if the rate law for the reaction is rate \(=k\left[\mathrm{NO}_{2}\right]^{2} ?\) (a) Decreasing the pressure of \(\mathrm{NO}_{2}\) from 0.50 atm to 0.250 atm. (b) Increasing the concentration of CO from \(0.01 M\) to \(0.03 M\).

The hydrolysis of the sugar sucrose to the sugars glucose and fructose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) follows a first-order rate law for the disappearance of sucrose: rate \(=k\left[\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right]\) (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.) (a) In neutral solution, \(k=2.1 \times 10^{-11} \mathrm{s}^{-1}\) at \(27^{\circ} \mathrm{C}\) and \(8.5 \times 10^{-11} \mathrm{s}^{-1}\) at \(37^{\circ} \mathrm{C} .\) Determine the activation energy, the frequency factor, and the rate constant for this equation at \(47^{\circ} \mathrm{C}\) (assuming the kinetics remain consistent with the Arrhenius equation at this temperature). (b) When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the concentration of sucrose is \(1.65 \times 10^{-7} \mathrm{M}\). How long will it take the solution to reach equilibrium at \(27^{\circ} \mathrm{C}\) in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible. (c) Why does assuming that the reaction is irreversible simplify the calculation in part (b)?

Write the rate law for each of the following elementary reactions: (a) \(\mathrm{O}_{3} \stackrel{\text { sunlight }}{\longrightarrow} \mathrm{O}_{2}+\mathrm{O}\) (b) \(\mathrm{O}_{3}+\mathrm{Cl} \longrightarrow \mathrm{O}_{2}+\mathrm{ClO}\) (c) \(\mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2}\) (d) \(\mathrm{O}_{3}+\mathrm{NO} \longrightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}\) (e) \(\mathrm{NO}_{2}+\mathrm{O} \longrightarrow \mathrm{NO}+\mathrm{O}_{2}\)

Nitrogen and oxygen react at high temperatures. What will happen to the concentrations of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO}\) at equilibrium if a catalyst is added?

Account for the increase in reaction rate brought about by a catalyst.

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