Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

What are the units of \(k\) for each type of reaction? a. first-order reaction b. second-order reaction c. zero-order reaction

Short Answer

Expert verified
For a first-order reaction, the units of k are s^-1. For a second-order reaction, the units of k are M^-1·s^-1. For a zero-order reaction, the units of k are M·s^-1.

Step by step solution

01

Understanding Rate Laws and Units of k

The rate of a chemical reaction can be described by its rate law, which shows how the rate depends on the concentration of the reactants. The rate constant 'k' has different units depending on the order of the reaction, which is determined by the sum of the exponents of the concentration terms in the rate law equation.
02

Units of k for a First-Order Reaction

For a first-order reaction with rate law equation rate = k[A], the units of k are 1/time, since the reaction rate has units of concentration/time and [A] is in concentration units. This compensates for the concentration term, leaving units of inverse time. Common units are s^-1.
03

Units of k for a Second-Order Reaction

For a second-order reaction, with a rate law rate = k[A]^2 or rate = k[A][B], the units of k are concentration^-1·time^-1. This is because we need to cancel out the units of concentration squared in the rate equation, so common units for k are M^-1·s^-1 (where M is molarity).
04

Units of k for a Zero-Order Reaction

For a zero-order reaction, the rate law is rate = k, which implies that the rate of reaction is independent of the concentration of reactants. Therefore, the units for k are concentration/time, as k alone must have the same units as the rate of the reaction. Common units are M·s^-1.

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 Vaia!

Key Concepts

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

Chemical Reaction Rates
Chemical reaction rates represent the speed at which a chemical reaction occurs. In simple terms, they tell us how fast reactants are being converted into products in a given reaction. These rates can be influenced by various factors, including the concentration of reactants, temperature, presence of a catalyst, and surface area.

The rate of a chemical reaction is typically expressed as the change in concentration of a reactant (or product) per unit time. For instance, if the concentration of a reactant decreases by 1 moldm-3 over a time period of 1 second, the reaction rate would be 1 moldm-3s-1. Understanding the reaction rate is crucial because it helps chemists control conditions to either speed up or slow down reactions, depending on the desired outcome.
First-Order Reaction
In a first-order reaction, the rate is directly proportional to the concentration of one reactant. This means that if you were to double the concentration of that reactant, the rate of the reaction would also double. Mathematically, the rate law for a first-order reaction is often written as \( rate = k[A] \), where \( [A] \) represents the concentration of reactant A, and \( k \) is the rate constant.

The units for the rate constant \( k \) in a first-order reaction are reciprocal seconds \( s^{-1} \), which account for the change in concentration over time without any concentration unit because the reaction rate is divided by the concentration of the reactant.
Second-Order Reaction
A second-order reaction is one where the rate is proportional to the square of the concentration of one reactant or to the product of the concentrations of two reactants. In mathematical terms, its rate law can be given as either \( rate = k[A]^2 \) or \( rate = k[A][B] \), where \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

The units for the rate constant \( k \) in a second-order reaction are concentration-1 multiplied by time-1. This typically equates to \( M^{-1}·s^{-1} \) when dealing with solutions. The units here come from the necessity to cancel out the concentration squared term in the rate equation to ensure the rate itself has the correct units of concentration per time.
Zero-Order Reaction
Unlike first and second-order reactions, a zero-order reaction has a rate that is independent of the concentration of the reactant(s). This means that changing the concentration of the reactant does not affect the rate of reaction. The rate law for a zero-order reaction can be simply written as \( rate = k \), with \( k \) symbolizing the rate constant.

Consequently, the units for the rate constant \( k \) in a zero-order reaction must reflect the overall reaction rate's units, which are typically given as concentration per unit time, such as \( M·s^{-1} \). The units here directly represent the rate of change in concentration, without any dependency on reactant concentration.
Rate Laws
Rate laws are mathematical expressions that relate the rate of a reaction to the concentration of its reactants. Each rate law is specific to a particular reaction and is determined experimentally. They are fundamentally important because they provide insights into the steps involved in chemical reactions, and they allow chemists to predict the effects of various conditions on the rate of a reaction.

In context to the units of the rate constant \( k \), rate laws justify why these units differ among zero, first, and second-order reactions. For instance, the units must compensate for concentration terms in the law so that the overall rate equation has the correct units. Rate laws also underscore the relationship between reaction mechanisms and their kinetics, lending critical clues in the study of how reactions proceed.

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

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and has a rate constant of \(1.42 \times 10^{-4} \mathrm{~s}^{-1}\) at a certain temperature. a. What is the half-life for this reaction? b. How long will it take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(25 \%\) of its initial concentration? c. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.00 \mathrm{M}\), how long will it take for the concentration to decrease to \(0.78 \mathrm{M} ?\) d. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(0.150 \mathrm{M},\) what is the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after \(2.00 \times 10^{2} \mathrm{~s}\) ? After \(5.00 \times 10^{2} \mathrm{~s} ?\)

The energy of activation for the decomposition of \(2 \mathrm{~mol}\) of \(\mathrm{HI}\) to \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) in the gas phase is \(185 \mathrm{~kJ}\). The heat of formation of \(\mathrm{HI}(g)\) from \(\mathrm{H}_{2}(g)\) and \(\mathrm{I}_{2}(g)\) is \(-5.65 \mathrm{~kJ} / \mathrm{mol} .\) Find the energy of activation for the reaction of \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) and \(1 \mathrm{~mol}\) of \(\mathrm{I}_{2}\) to form 2 mol of HI in the gas phase.

Consider the reaction: $$ \mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\frac{1}{2} \mathrm{O}_{2}(g) $$ The tabulated data were collected for the concentration of \(\mathrm{NO}_{2}\) as a function of time: $$ \begin{array}{cc} \text { Time (s) } & {\left[\mathrm{NO}_{2}\right] \text { (M) }} \\ \hline 0 & 1.000 \\ \hline 10 & 0.951 \\ \hline 20 & 0.904 \\ \hline 30 & 0.860 \\ \hline 40 & 0.818 \\ \hline 50 & 0.778 \\ \hline 60 & 0.740 \\ \hline 70 & 0.704 \\ \hline 80 & 0.670 \\ \hline 90 & 0.637 \\ \hline 100 & 0.606 \\ \hline \end{array} $$ a. What is the average rate of the reaction between 10 and 20 s? Between 50 and 60 s? b. What is the rate of formation of \(\mathrm{O}_{2}\) between 50 and \(60 \mathrm{~s}\) ?

What are enzymes? What is the active site of an enzyme? What is a substrate?

A chemical reaction is endothermic and has an activation energy that is twice the value of the enthalpy change of the reaction. Draw a diagram depicting the energy of the reaction as it progresses. Label the position of the reactants and products and indicate the activation energy and enthalpy of reaction.

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.

Sign-up for free