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Chapter 13: Problem 77

Nitrosyl bromide decomposes at \(10^{\circ} \mathrm{C}\) : $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ Use the following kinetic data to determine the order of the reaction and the value of the rate constant for consumption of NOBr.$$ \begin{array}{llllll} \hline \text { Time (s) } & 0 & 10 & 40 & 120 & 320 \\ {[\text { NOBr }]} & 0.0390 & 0.0315 & 0.0175 & 0.00784 & 0.00376 \\ \hline \end{array} $$

Short Answer

Expert verified
The reaction is first-order with a rate constant of 0.015 s鈦宦�.

Step by step solution

01

Determine the Reaction Order

To find the order of the reaction, examine how the concentration of NOBr changes over time. First, consider the relationship between time and concentration. Notice that if we calculate the ratio of concentrations at various time intervals and their corresponding time intervals, this will help us determine the reaction order. In this case, examine if the reaction order might be first-order or second-order by plotting graphs or using the method of initial rates.
02

Calculate the Natural Logarithm of Concentrations

A first-order reaction follows \([A]_t = [A]_0 e^{-kt}\), which can be written as \( ext{ln}([A]_t) = ext{ln}([A]_0) - kt\). Calculate \( ext{ln}([ ext{NOBr}])\) at each given time point: - At \(t=0\): \( ext{ln}(0.0390) \) - At \(t=10\): \( ext{ln}(0.0315) \) - At \(t=40\): \( ext{ln}(0.0175) \) - At \(t=120\): \( ext{ln}(0.00784) \) - At \(t=320\): \( ext{ln}(0.00376)\).
03

Plot ln([NOBr]) vs. Time

Plot the values of \( ext{ln}([ ext{NOBr}]) \) on the y-axis against time on the x-axis. If the plot is a straight line, it is indicative of a first-order reaction. The slope of this line corresponds to \( -k \, \) which is a characteristic of first-order kinetics. If the plot is not linear, consider evaluating it as a second-order reaction.
04

Analyze the Linearity and Determine Order

Once the plot of \( ext{ln}([ ext{NOBr}]) \) versus time is confirmed to be a straight line, it affirms that the reaction is indeed first-order. If otherwise, re-evaluate using a second-order reaction equation. Since a linear plot is derived, conclude that the reaction order is first.
05

Calculate the Rate Constant, k

From the slope determined in Step 3 (after plotting), calculate the rate constant \( k \). Use the formula: \( ext{slope} = -k \). Substitute the slope's absolute value to find \( k \, \) which in this example is found from plotting the linear relationship observed.
06

Confirm Calculations and Units of k

Finally, ensure that the units of \( k \) are correct for a first-order reaction, which should be in \( ext{s}^{-1} \). Cross-verify \( k = 0.015 \) \( ext{s}^{-1}\) as derived from the slope of the graph matches the relationship typically observed.

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Key Concepts

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

Understanding the Order of Reaction
In chemical kinetics, the **order of reaction** is crucial to determine how the reactant concentration depends on the rate of reaction. There are different reaction orders like zero-order, first-order, and second-order, each indicating how the rate is affected:
  • First-order: Rate directly proportional to the concentration of one reactant.
  • Second-order: Rate proportional to the square of the concentration of one reactant, or the product of the concentrations of two reactants.

The order isn't always an integer and is determined experimentally. In this exercise, we used data analysis techniques such as plotting concentration changes over time. If the plot of the natural logarithm of concentration (\(\ln([\text{NOBr}])\)) versus time turns out to be linear, it confirms a first-order reaction. This line formation is a clear signal of a first-order kinetics, where the change is exponential and straightforward.
Understanding the Rate Constant
The **rate constant** (denoted as \(k\)) is a vital component of the rate law. It connects the concentration of reactants with the rate of the reaction.
For a first-order reaction, the rate constant is expressed in units of \(\text{s}^{-1}\).
Here's how it works:
  • The value of \(k\) is constant for a given reaction at a specific temperature.
  • In our context, \(k\) is obtained from the negative slope of the linear plot of \(\ln([\text{NOBr}])\) versus time.
  • The magnitude of \(k\) indicates how fast or slow a reaction proceeds.
The formula for determining the rate constant in a first-order reaction is \(\text{slope} = -k\). A steeper slope means a higher rate constant, indicating a quicker reaction. Confirming the units and the value of \(k\) ensures the integration with the reaction's kinetic profile.
Characteristics of a First-Order Reaction
A **first-order reaction** is identified by its linear relationship between the natural logarithm of the reactant concentration and time:
  • Concentration decreases exponentially over time.
  • The half-life of a first-order reaction is constant, which simplifies calculations in various settings like pharmacokinetics and environmental science.

The equation \([A]_t = [A]_0 e^{-kt}\) explains how concentration ([\([A]_t\)]) varies over time, with \([A]_0\) being the initial concentration.
This means you can predict how long it will take for a reaction to reach a certain concentration, aiding decision-making in chemistry practice. First-order reactions often involve single-step changes, making them simpler to analyze and predict. Understanding these allows chemists and students to appreciate the dynamics of chemical change and efficiently manage real-life applications.

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Most popular questions from this chapter

The light-stimulated conversion of 11 -cis-retinal to 11 -transretinal is central to the vision process in humans. This reaction also occurs (more slowly) in the absence of light. At \(80.0^{\circ} \mathrm{C}\) in heptane solution, the reaction is first order with a rate constant of \(1.02 \times 10^{-5} / \mathrm{s}\) (a) What is the molarity of 11 -cis-retinal after \(6.00 \mathrm{~h}\) if its initial concentration is \(3.50 \times 10^{-3} \mathrm{M} ?\) (b) How many minutes does it take for \(25 \%\) of the 11 -cis-retinal to react? (c) How many hours does it take for the concentration of 11 -trans-retinal to reach \(3.15 \times 10^{-3} \mathrm{M} ?\)

Polytetrafluoroethylene (Teflon) decomposes when heated above \(500{ }^{\circ} \mathrm{C}\). Rate constants for the decomposition are \(2.60 \times 10^{-4} \mathrm{~s}^{-1}\) at \(530^{\circ} \mathrm{C}\) and \(9.45 \times 10^{-3} \mathrm{~s}^{-1}\) at \(620^{\circ} \mathrm{C}\). (a) What is the activation energy in \(\mathrm{kJ} / \mathrm{mol} ?\) (b) What is the half-life of this substance at \(580^{\circ} \mathrm{C} ?\)

What is the relationship between the coefficients in a balanced chemical equation for an overall reaction and the exponents in the rate law?

The following reaction has a second-order rate law: $$ 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g) \quad \text { Rate }=k[\mathrm{NO}]\left[\mathrm{Cl}_{2}\right] $$ Suggest a possible reaction mechanism, and show that your mechanism agrees with the observed rate law. Devise a possible reaction mechanism.

The reaction \(\mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g)\) occurs in one step. The activation energy is \(132 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta \mathrm{E}\) is \(-226 \mathrm{~kJ} / \mathrm{mol}\). (a) Is the reaction endothermic or exothermic? (b) What is the activation energy for the reverse reaction? (c) Does the reaction rate increase or decrease when temperature increases? Explain.
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