The reaction \(2 \mathrm{~A}+3 \mathrm{~B} \longrightarrow \mathrm{C}\) is first order with respect to \(\mathrm{A}\) and \(\mathrm{B}\). When the initial concentrations are \([\mathrm{A}]=1.6 \times 10^{-2} M\) and \([\mathrm{B}]=2.4 \times 10^{-3} M,\) the rate is \(4.1 \times 10^{-4} M / \mathrm{s} .\) Calculate the rate constant of the reaction.

Short Answer

Expert verified
The rate constant \(k\) of the reaction is approximately \(10.7 \, l/(mol \cdot s)\).

Step by step solution

01

Interpret the rate law equation

The rate of a reaction that is first order with respect to two reactants A and B can be expressed in the form \(rate = k[A][B]\), where rate is the rate of the reaction, k is the rate constant, and [A] and [B] are concentrations of the reactants A and B, respectively.
02

Substitute the given values into the rate equation

Initial concentrations are given as \([A] = 1.6 \times 10^{-2}\) M, \([B] = 2.4 \times 10^{-3}\) M and the rate as \(4.1 \times 10^{-4}\) M/s. Substitute these values into the rate equation: \(4.1 \times 10^{-4} = k(1.6 \times 10^{-2})(2.4 \times 10^{-3})\)
03

Solve for the rate constant k

Rearranging the equation to solve for k gives \(k = \frac{4.1 \times 10^{-4}}{(1.6 \times 10^{-2})(2.4 \times 10^{-3})}\), which upon calculation gives \(k = 10.7 \, l/(mol \cdot s)\)
04

Conclusion

The rate constant of the reaction is approximately \(10.7 \, l/(mol \cdot s)\).

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.

Understanding Chemical Reaction Rates
At the heart of studying chemical kinetics is understanding the rate at which chemical reactions occur. The chemical reaction rate is a measure of the speed at which reactants are transformed into products in a given reaction. It is quantified by changes in concentration of reactants or products per unit time, often expressed in moles per liter per second (M/s).
For instance, a rapid reaction might see a reactant's concentration plummet quickly, while a slow one changes little over a comparable duration. Factors such as temperature, pressure, concentration, and the presence of catalysts heavily influence these rates. When studying reaction rates, one seeks to establish a relationship between the rate and the factors affecting it, enabling the prediction of reaction behavior under different conditions.
Remember, faster rates imply quicker consummation of reactants and formation of products—essential for both understanding fundamental chemistry and optimizing industrial processes.
Deciphering the Rate Law Equation
To relate the reaction rate with reactant concentrations, the rate law equation is employed. It is a mathematical expression that describes the connection between the rate of a reaction and the concentration of its reactants. Generally described as rate = k[A]^n[B]^m, where k is the rate constant, [A] and [B] represent the molar concentrations of reactants A and B, and n and m are their respective reaction orders.
In practice, this law helps to determine the rate of a reaction under specific conditions. It is crucial to note that the rate law and its coefficients are not deduced from the balanced chemical equation but rather are determined experimentally. Knowing the rate law of a reaction is also pivotal for controlling processes in industries, such as pharmaceuticals and energy production, where precise reaction control is vital.
First Order Reaction Kinetics
When discussing reaction orders, a first order reaction is one where the rate is directly proportional to the concentration of one reactant. Mathematically, this reads as rate = k[A]. The exponent in this expression is 1, denoting the 'first order' in reactant A; the reaction rate will change in direct proportion to changes in A's concentration.
A landmark trait of first order reactions is that their rate constants have the units of s-1, as seen in the provided exercise where the calculated rate constant has the unit liter per mole per second (l/(mol·s)), owing to the concentration units involved. In addition to this, first order reactions have a characteristic half-life that is constant and does not depend on the initial concentration of reactants. This property is widely used in pharmacokinetics to determine the duration required for a drug's concentration to reduce to half its initial value within a biological system.

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

Consider the following elementary steps for a consecutive reaction $$ \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{C} $$ (a) Write an expression for the rate of change of \(\mathrm{B}\). (b) Derive an expression for the concentration of \(\mathrm{B}\) under steady- state conditions; that is, when \(\mathrm{B}\) is decomposing to \(\mathrm{C}\) at the same rate as it is formed from \(\mathrm{A}\).

Consider this elementary step: $$ X+2 Y \longrightarrow X Y_{2} $$ (a) Write a rate law for this reaction. (b) If the initial rate of formation of \(\mathrm{XY}_{2}\) is \(3.8 \times 10^{-3} \mathrm{M} / \mathrm{s}\) and the initial concentrations of \(X\) and \(Y\) are \(0.26 M\) and \(0.88 M\), what is the rate constant of the reaction?

In recent years ozone in the stratosphere has been depleted at an alarmingly fast rate by chlorofluorocarbons (CFCs). A CFC molecule such as \(\mathrm{CFCl}_{3}\) is first decomposed by UV radiation: $$ \mathrm{CFCl}_{3} \longrightarrow \mathrm{CFCl}_{2}+\mathrm{Cl} $$ The chlorine radical then reacts with ozone as follows: $$ \begin{array}{c} \mathrm{Cl}+\mathrm{O}_{3} \longrightarrow \mathrm{ClO}+\mathrm{O}_{2} \\ \mathrm{ClO}+\mathrm{O} \longrightarrow \mathrm{Cl}+\mathrm{O}_{2} \end{array} $$ (a) Write the overall reaction for the last two steps. (b) What are the roles of \(\mathrm{Cl}\) and \(\mathrm{ClO} ?\) (c) Why is the fluorine radical not important in this mechanism? (d) One suggestion to reduce the concentration of chlorine radicals is to add hydrocarbons such as ethane \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) to the stratosphere. How will this Work?

For the reaction \(X_{2}+Y+Z \longrightarrow X Y+X Z\) it is found that doubling the concentration of \(\mathrm{X}_{2}\) doubles the reaction rate, tripling the concentration of \(Y\) triples the rate, and doubling the concentration of \(Z\) has no effect. (a) What is the rate law for this reaction? (b) Why is it that the change in the concentration of \(Z\) has no effect on the rate? (c) Suggest a mechanism for the reaction that is consistent with the rate law.

Polyethylene is used in many items such as water pipes, bottles, electrical insulation, toys, and mailer envelopes. It is a polymer, a molecule with a very high molar mass made by joining many ethylene molecules (the basic unit is called a monomer) together (see p. 369 ). The initiation step is $$ \mathrm{R}_{2} \stackrel{k_{1}}{\longrightarrow} 2 \mathrm{R} \cdot \quad \text { initiation } $$ The \(\mathrm{R}\) - species (called a radical) reacts with an ethylene molecule \((\mathrm{M})\) to generate another radical $$ \mathrm{R} \cdot+\mathrm{M} \longrightarrow \mathrm{M}_{1} \cdot $$ Reaction of \(\mathrm{M}_{1}\). with another monomer leads to the growth or propagation of the polymer chain: $$ \mathrm{M}_{1} \cdot+\mathrm{M} \stackrel{k_{\mathrm{p}}}{\longrightarrow} \mathrm{M}_{2} \cdot \quad \text { propagation } $$ This step can be repeated with hundreds of monomer units. The propagation terminates when two radicals combine $$ \mathrm{M}^{\prime} \cdot+\mathrm{M}^{\prime \prime} \cdot \stackrel{k_{t}}{\longrightarrow} \mathrm{M}^{\prime}-\mathrm{M}^{\prime \prime} \quad \text { termination } $$ (a) The initiator used in the polymerization of ethylene is benzoyl peroxide \(\left[\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2}\right]:\) $$ \left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO}\right)_{2} \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COO} \cdot $$ This is a first-order reaction. The half-life of benzoyl peroxide at \(100^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~min} .\) (a) Calculate the rate constant (in \(\min ^{-1}\) ) of the reaction. (b) If the half-life of benzoyl peroxide is \(7.30 \mathrm{~h}\) or \(438 \mathrm{~min},\) at \(70^{\circ} \mathrm{C},\) what is the activation energy (in \(\mathrm{kJ} / \mathrm{mol}\) ) for the decomposition of benzoyl peroxide? (c) Write the rate laws for the elementary steps in the above polymerization process and identify the reactant, product, and intermediates. (d) What condition would favor the growth of long high-molar-mass polyethylenes?

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