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 for each of the following if the concentrations are expressed in moles per liter and the time in seconds? a. rate of a chemical reaction b. rate constant for a zero-order rate law c. rate constant for a first-order rate law d. rate constant for a second-order rate law e. rate constant for a third-order rate law

Short Answer

Expert verified
a. rate of a chemical reaction: \(M/s\) b. rate constant for a zero-order rate law: \(M/s\) c. rate constant for a first-order rate law: \(s^{-1}\) d. rate constant for a second-order rate law: \(M^{-1}s^{-1}\) e. rate constant for a third-order rate law: \(M^{-2}s^{-1}\)

Step by step solution

01

Identifying the required units for the rate of a chemical reaction

The rate of a chemical reaction is given by the change in concentration of a reactant or product per unit time. Since the concentration is expressed in moles/L (M) and time in seconds (s), the units for the rate of a chemical reaction will simply be the concentration units (moles/L) divided by the time units (seconds), resulting in M/s.
02

Identifying the required units for a zero-order rate law

A zero-order reaction has a rate law in the form of: Rate = k[A]^0, where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. Since [A]^0 is equal to 1, the rate law simplifies to Rate = k. Therefore, the units for the rate constant (k) in a zero-order rate law will be the same as the units for the rate of a chemical reaction, which is M/s.
03

Identifying the required units for a first-order rate law

A first-order reaction has a rate law in the form of: Rate = k[A], where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. To determine the units for k, we can rearrange the rate law: k = Rate / [A]. Since Rate has the units M/s and [A] has the units M, the units for k in a first-order rate law will be (M/s)/(M), which simplifies to 1/s or s^{-1}.
04

Identifying the required units for a second-order rate law

A second-order reaction has a rate law in the form of: Rate = k[A]^2, where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. To determine the units for k, we can rearrange the rate law: k = Rate / ([A]^2). Since Rate has the units M/s and [A]^2 has the units M^2, the units for k in a second-order rate law will be (M/s)/(M^2), which simplifies to L/mol*s or M^{-1}s^{-1}.
05

Identifying the required units for a third-order rate law

A third-order reaction has a rate law in the form of: Rate = k[A]^3, where Rate is the rate of the reaction, k is the rate constant, and [A] is the concentration of a reactant. To determine the units for k, we can rearrange the rate law: k = Rate / ([A]^3). Since Rate has the units M/s and [A]^3 has the units M^3, the units for k in a third-order rate law will be (M/s)/(M^3), which simplifies to L^2/mol^2*s or M^{-2}s^{-1}. In conclusion, the units for each parameter are as follows: a. rate of a chemical reaction: M/s b. rate constant for a zero-order rate law: M/s c. rate constant for a first-order rate law: s^{-1} d. rate constant for a second-order rate law: M^{-1}s^{-1} e. rate constant for a third-order rate law: M^{-2}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!

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 rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text {Rate} =-\frac{\Delta[\text { NOBr }]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is 2.00 s when \([\mathrm{NOBr}]_{0}=\) \(0.900 M,\) calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of NOBr to decrease to 0.100\(M ?\)

Provide a conceptual rationale for the differences in the halflives of zero-, first-, and second-order reactions.

How does temperature affect k, the rate constant? Explain.

Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{array}{l}{2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} A_{2}} \\ {2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2}}\end{array} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\) . In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C},\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} M\) It was found that after each reaction had progressed for \(3.00 \mathrm{min},[\mathrm{A}]=3.00[\mathrm{B}]\) . In this case the rate laws are defined as $$ \begin{array}{l}{\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2}} \\ {\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2}}\end{array} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 \(\mathrm{min}\) . b. Calculate the value of \(k_{2}\) . c. Calculate the half-life for the experiment involving A.

The decomposition of NH3 to N2 and H2 was studied on two surfaces: Without a catalyst, the activation energy is 335 \(\mathrm{kJ} / \mathrm{mol}\) . a. Which surface is the better heterogeneous catalyst for the decomposition of \(\mathrm{NH}_{3} ?\) Why? b. How many times faster is the reaction at 298 \(\mathrm{K}\) on the W surface compared with the reaction with no catalyst present? Assume that the frequency factor \(A\) is the same for each reaction. c. The decomposition reaction on the two surfaces obeys a rate law of the form $$ |text {Rate} =k \frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{H}_{2}\right]} $$ How can you explain the inverse dependence of the rate on the \(\mathrm{H}_{2}\) concentration?

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