/*! 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 39 How are integrated rate laws use... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

How are integrated rate laws used to determine reaction order? What is the reaction order in each of these cases? (a) A plot of the natural logarithm of [reactant] vs. time is linear. (b) A plot of the inverse of [reactant] vs. time is linear. (c) [reactant] vs. time is linear.

Short Answer

Expert verified
(a) first-order (b) second-order (c) zero-order

Step by step solution

01

Understand Integrated Rate Laws

Integrated rate laws relate the concentration of a reactant to time for different orders of reaction. They help to determine the order by analyzing the time-dependence of concentration.
02

Identify the Plot for Each Reaction Order

Each order of reaction has a characteristic linear plot:- First-order: A plot of the natural logarithm of the concentration of the reactant \(\text{ln[reactant]}\) vs. time is linear.- Second-order: A plot of the inverse of the concentration of the reactant \(1/[\text{reactant}]\) vs. time is linear.- Zero-order: A plot of the concentration of the reactant \([\text{reactant}]\) vs. time is linear.
03

Determine the Reaction Order for Each Case

(a) A plot of \(\text{ln[reactant]}\) vs. time being linear indicates a first-order reaction. (b) A plot of \(1/[\text{reactant}]\) vs. time being linear indicates a second-order reaction. (c) A plot of \([\text{reactant}]\) vs. time being linear indicates a zero-order reaction.

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 91Ó°ÊÓ!

Key Concepts

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

reaction order
Understanding the **reaction order** is crucial in chemical kinetics. It tells us how the rate of a reaction depends on the concentration of the reactants. The reaction order is determined experimentally and can be zero, first, second, or even fractional. To determine the reaction order of a reaction, scientists use integrated rate laws. These laws link the concentration of a reactant to time, allowing us to identify patterns in how the concentration changes. By plotting the concentration in different ways, as indexed by time, one can visually determine which order the reaction follows.
  • Integrated rate laws help identify the reaction order.
  • Each reaction order has a distinct linear plot.
first-order reaction
A **first-order reaction** is characterized by the fact that the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction is given by:
\(\text{ln}([A]) = -kt + \text{ln}([A]_0)\)
where \([A]_0\) is the initial concentration, \(t\) is time, and \(k\) is the rate constant. For a first-order reaction, a plot of the natural logarithm of the reactant concentration (ln[reactant]) vs. time (\text t\text t) will yield a straight line. If a given reaction's plot shows this linear relationship, it can be confirmed as a first-order reaction.
  • Rate depends linearly on reactant concentration.
  • Integrated rate law: \(\text{ln}([A]) = -kt + \text{ln}([A]_0)\).
  • Characteristic plot: linear \text{ln}([reactant]) vs. time.
second-order reaction
In a **second-order reaction**, the rate is proportional to either the square of the concentration of one reactant or the product of the concentrations of two reactants. The integrated rate law for a second-order reaction involving one reactant is:
\(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\)
where \([A]_0\) is the initial concentration, \(t\) is time, and \(k\) is the rate constant. For a second-order reaction, a plot of the inverse of the reactant concentration (1/[reactant]) vs. time will be a straight line. Observing this linear relationship confirms the reaction as second-order.
  • Rate depends on the square of a single reactant's concentration or the product of two concentrations.
  • Integrated rate law: \(\frac{1}{[A]} = kt + \frac{1}{[A]_0}\).
  • Characteristic plot: linear \(\frac{1}{[reactant]}\) vs. time.
zero-order reaction
A **zero-order reaction** is one where the rate of reaction is constant and does not depend on the concentration of the reactants. The integrated rate law for a zero-order reaction is:
\([A] = -kt + [A]_0\)
where \([A]_0\) is the initial concentration, \(t\) is time, and \(k\) is the rate constant. For a zero-order reaction, a plot of the reactant concentration ([reactant]) vs. time will be linear. This linear decrease in concentration over time confirms the reaction as zero-order.
  • Rate is independent of reactant concentration.
  • Integrated rate law: \([A] = -kt + [A]_0\).
  • Characteristic plot: linear [reactant] vs. time.
kinetics
**Kinetics** is the branch of chemistry that deals with the rates of chemical reactions and the factors that affect these rates. A deep understanding of kinetics is essential for fields such as pharmaceuticals, environmental science, and engineering, as it helps in optimizing reactions. Factors affecting reaction rates include:
  • Concentration of reactants
  • Temperature
  • Presence of a catalyst
  • Surface area of solid reactants
  • Nature of reactants
By studying kinetics, we can also understand the mechanism of reactions, providing a detailed step-by-step view of how reactants convert to products.

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

Give two reasons to measure initial rates in a kinetics study.

The overall equation and rate law for the gas-phase decomposition of dinitrogen pentoxide are \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad\) rate \(=k\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]\) Which of the following can be considered valid mechanisms for the reaction? I One-step collision II \(2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow 2 \mathrm{NO}_{3}(g)+2 \mathrm{NO}_{2}(g) \quad[\) slow \(]\) \(2 \mathrm{NO}_{3}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)+2 \mathrm{O}(g)\) [fast] \(2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)\) [fast] III \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g)\) [fast] \(\mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow 3 \mathrm{NO}_{2}(g)+\mathrm{O}(g) \quad\) [slow] \(\mathrm{NO}_{3}(g)+\mathrm{O}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad[\) fast \(]\) \(\mathrm{IV} 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)+\mathrm{N}_{2} \mathrm{O}_{3}(g)+3 \mathrm{O}(g) \quad[\) fast \(]\) \(\mathrm{N}_{2} \mathrm{O}_{3}(g)+\mathrm{O}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) [slow] \(2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g) \quad[\) fast \(]\) \(\mathrm{V} \quad 2 \mathrm{~N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{N}_{4} \mathrm{O}_{10}(g)\) [slow] \(\mathrm{N}_{4} \mathrm{O}_{10}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)\) [fast]

Express the rate of this reaction in terms of the change in concentration of each of the reactants and products: $$ \mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g}) $$ When [B] is decreasing at \(0.5 \mathrm{~mol} / \mathrm{L}\) -s, how fast is \([\mathrm{A}]\) decreasing?

How does an increase in pressure affect the rate of a gasphase reaction? Explain.

A biochemist studying the breakdown of the insecticide DDT finds that it decomposes by a first-order reaction with a halflife of 12 yr. How long does it take DDT in a soil sample to decrease from 275 ppbm to \(10 .\) ppbm (parts per billion by mass)?

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.