/*! 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 43 Assume that the chemical reactio... [FREE SOLUTION] | 91影视

91影视

Assume that the chemical reaction is reactants \(\rightarrow\) products. Determine the rate constant and order from the concentration-time dependence. $$ \begin{array}{cc} \text { Time (s) } & \text { [Reactant] (M) } \\ \hline 0 & 0.250 \\ 1 & 0.216 \\ 2 & 0.182 \\ 3 & 0.148 \\ 4 & 0.114 \\ 5 & 0.080 \end{array} $$

Short Answer

Expert verified
The reaction is first-order with a rate constant approximately equal to 0.080 s鈦宦.

Step by step solution

01

Determine the type of reaction order

To determine the order of the reaction, compare the changes in concentration with time. Take note of the concentration changes and examine if they fit a zero-order, first-order, or second-order rate equation. Zero-order reactions show constant decrease in concentration over time, first-order reactions exhibit exponential decay, and second-order reactions involve an inverse relationship with concentration.
02

Test for a First-Order Reaction

Calculate the natural logarithm of the concentration of reactant at each time point. If the plot of ln([Reactant]) versus time is linear, the reaction is first-order. Thus, calculate ln(0.250), ln(0.216), ln(0.182), ln(0.148), ln(0.114),ln(0.080) and plot these values against corresponding time points.
03

Check Linear Fit for First-Order Reaction

Using the calculated values from Step 2, plot them to see if a straight line is formed. Calculate the slope of this line, which indicates the negative of the rate constant (k) for a first-order reaction if the plot is linear.
04

Test for a Second-Order Reaction

If the data does not show linearity for a first-order reaction, check for second-order reaction. Plot 1/[Reactant] versus time. If the resulting graph is linear, the reaction is second-order.
05

Check Linear Fit for Second-Order Reaction

Plot the calculated values from Step 4 and verify if the points form a straight line. Calculate the slope of this line which should represent the rate constant (k) if it is a second-order reaction.
06

Determine Order and Rate Constant

Based on the linearity of the plots, confirm the order of the reaction. If the first-order plot was linear, the slope of that graph is the negative rate constant. If the second-order plot was linear, the positive slope is the rate constant. Conduct the verification and write the resulting order and rate constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Reaction Order
Reaction order is a crucial concept in understanding how the concentration of reactants affects the rate of a chemical reaction. It depicts the power to which the concentration of a reactant is raised in the rate equation. Reaction orders are typically integers like zero, one, or two, though they can also be fractions in some cases.

The order of a reaction determines the relationship between the concentration of the reactants and the rate of the reaction. There are three main types of reaction orders you might encounter:
  • Zero-order: The rate of reaction is constant and does not depend on the concentration of the reactant. The concentration decreases linearly over time.
  • First-order: The rate of reaction is directly proportional to the concentration of one reactant. When plotted as the natural logarithm of concentration versus time, it results in a straight line.
  • Second-order: The rate depends on the square of the concentration of one reactant or the product of concentrations of two reactants. This order shows a linear relationship when plotting the inverse of concentration versus time.
In the exercise, by plotting the concentration data, we determine the order by examining which plot results in a straight line, either zero, first, or second-order reaction behavior.
Rate Constant
The rate constant, denoted by the symbol \( k \), is a proportionality factor in the rate equation of a chemical reaction. It provides a link between the reaction rate and the reactants' concentrations, essential for predicting how fast a reaction will occur under given conditions.

Each reaction order has its own form of the rate constant:
  • For zero-order reactions, the rate constant has units of concentration/time (e.g., M/s).
  • For first-order reactions, it has units of 1/time (e.g., s\(^{-1}\)).
  • For second-order reactions, the rate constant is expressed in units of 1/(concentration 脳 time) (e.g., M\(^{-1}\)s\(^{-1}\)).
The value of \( k \) is determined experimentally through the slope of the linear plot produced when testing the reaction order. In the given exercise, if the first-order plot is linear, the negative slope represents the rate constant. If the second-order plot turns out to be linear, the positive slope would be \( k \). Thus, identifying not only the rate order but finding \( k \) is integral for understanding the kinetics of the reaction.
Concentration-Time Dependence
Concentration-time dependence explains how reactant concentrations diminish over time throughout the course of a reaction. It is modeled by different equations depending on the order of the reaction. Understanding this relationship is essential for predicting how long a reaction will take to complete or reach a specific conversion level.

Here's how concentration changes over time for different reaction orders:
  • For zero-order reactions, the concentration decreases steadily with time, following a simple linear relationship \([A] = [A]_0 - kt\), where \([A]_0\) is the initial concentration.
  • For first-order reactions, the concentration decreases exponentially, described by the formula \([A] = [A]_0 e^{-kt}\). The natural logarithm plot of concentration versus time gives a straight line.
  • For second-order reactions, the inverse of the concentration increases linearly over time, \(\frac{1}{[A]} = \frac{1}{[A]_0} + kt\).
By plotting concentration data over time in a manner that aligns with these relationships, we can not only identify the order of reaction but also gain insights into the nature of reactant consumption, enabling better control and optimization of chemical processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What are the units for the zero-, first-, and second-order rate constants?

OBJECTIVE. Relate temperature, activation energy, and rate constant through the Arrhenius equation. A reaction rate doubles when the temperature increases from \(25^{\circ} \mathrm{C}\) to \(40^{\circ} \mathrm{C}\). What is the activation energy?

Nitramide decomposes to water and dinitrogen monoxide. $$ \mathrm{H}_{2} \mathrm{NNO}_{2}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{N}_{2} \mathrm{O}(\mathrm{g}) $$ This reaction was studied by J. N. Br酶nsted in 1924 as part of research into the fundamental nature of acids and bases. If \(1.00 \mathrm{~L}\) of \(0.440 \mathrm{M}\) nitramide is placed in a reactor at \(20^{\circ} \mathrm{C}\), the following results are expected. (The experiment was actually performed to measure the effect of malate ion on the rate of reaction.) $$ \begin{array}{cc} \mathrm{T} \text { (min) } & P \text { (torr) } \\ \hline 0.00 & 17.54 \\ 0.50 & 18.98 \\ 1.00 & 20.09 \\ 2.00 & 21.65 \\ 4.00 & 23.46 \\ 6.00 & 24.48 \\ 8.00 & 25.14 \\ 10.00 & 25.59 \\ \text { Completion } & 27.54 \end{array} $$ What is the rate law for this reaction? (Hint: The data show the increase in concentration of a product, which differs from the other problems in this book. The problem looks more familiar if you create a decay curve. Note that the initial point represents a small amount of product, and thus a large amount of reactant. The final point represents a large amount of product and no reactant. The problem looks like any other kinetics problem if you first calculate ([final pressure \(-\) current pressure \(]\) to see the data as a decay curve). Graph (final pressure \(-\) current pressure), rather than the current pressure, as a function of time to see the data. If this is not a straight line, you can graph \(\ln (\) final pressure current pressure) and \(1 /\) (final pressure \(-\) current pressure) to determine the order of the reaction. Scientists frequently transform their data to a familiar form. These operations make it easier to interpret the data.

OBJECTIVE. Calculate the concentration-time behavior for a first-order reaction from the rate law and the rate constant. When formic acid is heated, it decomposes to hydrogen and carbon dioxide in a first-order decay. $$ \mathrm{HCOOH}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) $$ At \(550^{\circ} \mathrm{C}\), the half-life of formic acid is 24.5 minutes. (a) What is the rate constant, and what are its units? (b) How many seconds are needed for formic acid, initially \(0.15 M\), to decrease to 0.015 M?

OBJECTIVE. Relate half-life and rate constant, and calculate concentration- time behavior from the half-life of a first-order reaction. \- The hypothetical compound A decomposes in a firstorder reaction that has a half-life of \(2.3 \times 10^{2}\) seconds at \(450^{\circ} \mathrm{C}\). If the initial concentration of \(\mathrm{A}\) is \(4.32 \times 10^{-2} \mathrm{M}\) how long will it take for the concentration of \(\mathrm{A}\) to decline to \(3.75 \times 10^{-3} M ?\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.