A reaction \(\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) obeys the following rate law: Rate \(=k[\mathrm{B}]^{2}\) . (a) If [A] is doubled, how will the rate change? Will the rate constant change? (b) What are the reaction orders for \(\mathrm{A}\) and \(\mathrm{B} ?\) What is the overall reaction order? (c) What are the units of the rate constant?

Short Answer

Expert verified
(a) The rate will not change when [A] is doubled because the rate law doesn't depend on the concentration of A, and the rate constant won't change either, as it's a property of the reaction. (b) The reaction order for A is 0, for B is 2, and the overall reaction order is 2. (c) The units of the rate constant k are M^(-1)s^(-1).

Step by step solution

01

(a) Change in the rate and rate constant due to doubling [A]

Since the rate law of the reaction is given as Rate = k[B]^2, it's clear that the rate of the reaction is not dependent on the concentration of A. Therefore, when [A] is doubled, there will be no change in the reaction rate. The rate constant, k, is a property of the reaction and does not depend on the concentrations of reacting species; so, the rate constant will remain unchanged.
02

(b) Reaction orders for A and B and overall reaction order

The given rate law of the reaction is Rate = k[B]^2. We can see that the power to which [B] is raised is 2, making the reaction order for B equal to 2. The concentration of A doesn't appear in the rate law, indicating that the reaction order for A is 0. The overall reaction order is given by the sum of the reaction orders of all species i.e., 0 (for A) + 2 (for B) = 2. Therefore, the overall reaction order is 2.
03

(c) Units of the rate constant

To find the units of k, the rate constant, let us consider the rate law expression: Rate = k[B]^2. The units of rate are usually expressed in units of M/s (molar concentration per second). Thus, let R be in M/s, and [B]^2 be in M^2. To balance the units, k will have the following units: \[k = \frac{(M/s)}{(M^2)}\] Simplifying the equation, we get: \[k = \frac{1}{(M \cdot s)}\] So, the units of the rate constant (k) for this reaction are M^(-1)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.

Rate Constant Units
The rate constant is a crucial element in the study of chemical kinetics, and understanding its units is essential for any chemistry student. The rate constant, denoted as k, provides the proportionality factor between the reaction rate and the concentrations of the reactants raised to the power of their respective orders in the rate law equation.

In the provided exercise example, the rate law is defined as Rate = k[B]^2. Since the reaction rate (Rate) has units of molarity per second (M/s), and the concentration of reactant B raised to the second power ([B]^2) has units of molarity squared (M^2), the units for the rate constant must be such that when multiplied by M^2, the result is M/s.

To balance this equation, the units for the rate constant are determined to be the inverse of molarity times seconds, or M^(-1)s^(-1). Practically, this means for every increase by one molar concentration of B, the rate of reaction is affected by k times that increase squared, considering the second-order relationship with B.

Importance of Rate Constant Units

Knowing the units of the rate constant is not merely for academic exercises. It provides insights into the reaction's sensitivity to changes in concentration and can be indicative of the mechanism by which the reaction proceeds. Furthermore, rate constants with different units are utilized in varying reaction orders, reinforcing the importance of understanding these units to accurately interpret and predict chemical reactions.
Reaction Order
Understanding the reaction order is an important aspect of the reaction rate law in chemical kinetics. It refers to the exponent to which the concentration of a reactant is raised in the rate law equation, representing how the rate of the reaction is affected by the concentration of that particular reactant.

In our exercise, the rate law is given as Rate = k[B]^2, where the concentration of B is squared. Therefore, the reaction order with respect to B is 2. This indicates a second-order reaction in B, meaning the rate of reaction is directly proportional to the square of the concentration of B. On the other hand, the concentration of A does not affect the rate since it does not appear in the rate law, giving A a reaction order of 0.

The overall reaction order is the sum of the orders with respect to all reactants in the reaction. In this case, B has an order of 2 and A has an order of 0, so the overall reaction order is 2.

Why Does Reaction Order Matter?

Knowing the reaction order is not only key to solving problems but also to understanding how different reactants influence the speed of a reaction. It helps predict how reaction rates will change with varying reactant concentrations and can be a gateway to understanding the intricate details of reaction mechanisms.
Chemical Kinetics
At the core of understanding how and why chemical reactions occur at the rates they do lies the field of chemical kinetics. Kinetics probes the factors affecting the speed of a reaction, such as temperature, pressure, and the presence of a catalyst. By studying the rate at which reactants turn into products, chemists can deduce the detailed steps or the mechanism of a reaction.

In our exercise, we see that the concentration of B directly influences the reaction rate, thus signifying the significance of reactant concentrations in chemical kinetics. However, chemical kinetics is not limited to just concentration effects. Factors such as the physical state of the reactants, the presence of light, and the surface area of solid reactants can also play into the kinetics of a reaction.

Application of Chemical Kinetics

The practical applications of chemical kinetics are vast. For instance, in pharmaceuticals, understanding the kinetics of drug interactions helps in designing better drugs with optimal efficacy. Similarly, in environmental science, kinetics assists in modeling the breakdown of pollutants. Chemical kinetics, thus, provides a quantitative description of reaction rates, which is pivotal for controlling processes in various scientific and industrial fields.

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

A flask is charged with 0.100 mol of A and allowed to react to form \(B\) according to the hypothetical gas-phase reaction \(A(g) \longrightarrow \mathrm{B}(g) .\) The following data are collected:(a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that \(\mathrm{A}\) is cleanly converted to \(\mathrm{B}\) with no intermediates. (b) Calculate the average rate of disappearance of A for each 40 s interval in units of mol/s. (c) Which of the following would be needed to calculate the rate in units of concentration per time: (i) the pressure of the gas at each time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A?

(a) The reaction \(\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)+\frac{1}{2} \mathrm{O}_{2}(g)\) is first order. At 300 \(\mathrm{K}\) the rate constant equals \(7.0 \times 10^{-4} \mathrm{s}^{-1}\) . Calculate the half-life at this temperature. (b) If the activation energy for this reaction is \(75 \mathrm{kJ} / \mathrm{mol},\) at what temperature would the reaction rate be doubled?

As described in Exercise 14.41 , the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at 660 \(\mathrm{K}\) is \(4.5 \times 10^{-2} \mathrm{s}^{-1}\) .half-life for this reaction? (b) If you start with 0.050\(M \mathrm{I}_{2}\) at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form \(\mathrm{I}_{2}\) ?

The gas-phase reaction of NO with \(\mathrm{F}_{2}\) to form \(\mathrm{NOF}\) and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{kJ} / \mathrm{mol} .\) and a frequency factor of \(A=6.0 \times 10^{8} M^{-1} \mathrm{s}^{-1} .\) The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g)$$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\) . (b) Draw the Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule, (c) Predict the shape for the NOF molecule.Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

The \(\mathrm{NO}_{x}\) waste stream from automobile exhaust includes species such as \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) . Catalysts that convert these species to \(\mathrm{N}_{2}\) are desirable to reduce air pollution. (a) Draw the Lewis dot and VSEPR structures of NO, NO \(_{2}\) and \(\mathrm{N}_{2} .(\mathbf{b})\) Using a resource such as Table \(8.3,\) look up the energies of the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of \(\mathrm{NO}_{x}\) into \(\mathrm{N}_{2},\) describing what wavelengths of light need to be monitored as a function of time.

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