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Chapter 14: Problem 95

The reaction \(2 \mathrm{NO}_{2} \longrightarrow 2 \mathrm{NO}+\mathrm{O}_{2}\) has the rate constant \(k=0.63 \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Based on the units for \(k\), is the reaction first or second order in \(\mathrm{NO}_{2}\) ? (b) If the initial concentration of \(\mathrm{NO}_{2}\) is \(0.100 \mathrm{M}\), how would you determine how long it would take for the concentration to decrease to \(0.025 \mathrm{M}\) ?

Short Answer

Expert verified
(a) The reaction is second order in NO鈧�, as the units of the rate constant k are M鈦宦� s鈦宦�. (b) To determine the time taken for the concentration to decrease, use the rate equation for a second-order reaction: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\). Plug in the given concentrations and solve for t: \(t = \frac{3}{0.63} = 4.76 s\). So, it takes 4.76 seconds for the concentration of NO鈧� to decrease from 0.100 M to 0.025 M.

Step by step solution

01

Determine the order of the reaction

Given that the rate constant k = 0.63 M鈦宦� s鈦宦�. The units of the rate constant can help us determine the order of the reaction with respect to NO鈧�. For a reaction of the form: \(aA \longrightarrow products\) The rate of the reaction is given by: \(rate = k[A]^n\) where n is the order of the reaction and [A] is the concentration of the reactant A. For a first-order reaction (n=1), the units of the rate constant k would be s鈦宦�. For a second-order reaction (n=2), the units of the rate constant k would be M鈦宦� s鈦宦�. The units of the given rate constant are M鈦宦� s鈦宦�, so the reaction is second order in NO鈧�.
02

Write the rate equation for the reaction

The rate equation for a second-order reaction is as follows: \(\frac{1}{[A]_t} - \frac{1}{[A]_0} = kt\) where [A]鈧� is the initial concentration of the reactant, [A]鈧� is the concentration at time t, and k is the rate constant.
03

Calculate the time taken for the concentration to decrease

We are given that the initial concentration of NO鈧� [A]鈧� is 0.100 M, and we need to determine the time required for the concentration to decrease to 0.025 M, which is [A]鈧�. Using the rate equation from Step 2, we can write: \(\frac{1}{0.025} - \frac{1}{0.100} = 0.63 \times t\) To find t, first solve the equation within the brackets: \(\frac{1}{0.025} - \frac{1}{0.100} = \frac{4 - 1}{1} = 3\) Next, divide both sides by the rate constant 0.63: \(t = \frac{3}{0.63}\)
04

Calculate the time value

Now we can calculate the value of t: \(t = \frac{3}{0.63} = 4.76 s\) Thus, it takes 4.76 seconds for the concentration of NO鈧� to decrease from 0.100 M to 0.025 M in the given reaction.

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Key Concepts

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

Understanding Second-Order Reactions
In chemical kinetics, the term "second-order reaction" refers to a reaction where the rate depends on the concentration of one reactant raised to the second power, or on the concentrations of two different reactants each raised to the first power. This is written as:
  • Rate = k[A]^2 for a single reactant
  • Rate = k[A][B] for two reactants
The order of a reaction is crucial as it tells us how the concentration of reactants affects the speed of the reaction.
In the original exercise, we see that the rate constant k has units of M鈦宦� s鈦宦�, which is characteristic of a second-order reaction.
This is because, for second-order reactions, the rate constant must have these units to maintain dimensional consistency in the rate equation: Rate = k[A]虏 (for one reactant).
Therefore, understanding the unit of the rate constant is pivotal in determining the reaction order.
The Role of the Rate Constant
The rate constant, denoted as k, is an essential part of the rate equation for any reaction. It is a proportionality factor that links the reaction rate to the concentrations of the reactants.
The value of k is influenced by several factors, including temperature and the presence of a catalyst.
In the second-order reaction from our example, the rate constant is 0.63 M鈦宦� s鈦宦�.
These units tell us that the reaction speed is dependent on the square of the concentration of NO鈧�. This means, as NO鈧� concentration decreases, the reaction rate also decreases at a quadratic rate.
The calculation of reaction time utilizing this rate constant involves substituting values into the integrated rate law formula for second-order reactions. This helps to determine how the concentration of a reactant changes over time, which is crucial in kinetic studies.
Concentration Change Over Time
The change in concentration of a reactant over time is a central part of chemical kinetics.
For second-order reactions, the relationship between the concentration at given times and the rate constant is described by the formula:
  • \( \frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \)
This formula allows us to predict how long it will take for a reactant to fall from one concentration to another.
The example in our exercise demonstrated this by using initial and final concentrations of NO鈧�, with a calculated time of 4.76 seconds for the concentration to decrease from 0.100 M to 0.025 M.
By substituting into this formula, we can easily solve for time t, offering a precise understanding of concentration dynamics over the course of the reaction.
Overall, mastering these calculations is fundamental in predicting reaction behavior, which is vital in various scientific and industrial applications.

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Most popular questions from this chapter

Heterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure 14.24, are subject to "poisoning," which shuts down their catalytic ability. Compounds of sulfur are often poisons. Suggest a mechanism by which such compounds might act as poisons.

A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{~B}]^{2}\). (a) If \([\mathrm{A}]\) is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(A\) and \(B\) ? What is the overall reaction order? (c) What are the units of the rate constant?

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products?

The decomposition reaction of \(\mathrm{N}_{2} \mathrm{O}_{5}\) in carbon tetrachloride is \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \longrightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\). The rate law is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\). At \(64^{\circ} \mathrm{C}\) the rate constant is \(4.82 \times 10^{-3} \mathrm{~s}^{-1}\). (a) Write the rate law for the reaction. (b) What is the rate of reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.0240 \mathrm{M}\) ? (c) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is doubled to \(0.0480 \mathrm{M}\) ? (d) What happens to the rate when the concentration of \(\mathrm{N}_{2} \mathrm{O}_{5}\) is halved to \(0.0120 \mathrm{M}\) ?

The following kinetic data are collected for the initial rates of a reaction \(2 \mathrm{X}+\mathrm{Z} \longrightarrow\) products: $$ \begin{array}{llll} \hline \text { Experiment } & {[\mathrm{X}]_{0}(M)} & {[\mathrm{Z}]_{0}(M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.25 & 0.25 & 4.0 \times 10^{1} \\ 2 & 0.50 & 0.50 & 3.2 \times 10^{2} \\ 3 & 0.50 & 0.75 & 7.2 \times 10^{2} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(X\) is \(0.75 M\) and that of \(Z\) is \(1.25 M ?\)
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