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Chapter 14: Problem 5

You perform a series of experiments for the reaction \(\mathrm{A} \longrightarrow \mathrm{B}+\mathrm{C}\) and find that the rate law has the form rate \(=k[\mathrm{~A}]^{x}\). Determine the value of \(x\) in each of the following cases: (a) There is no rate change when \([\mathrm{A}]_{0}\) is tripled. (b) The rate increases by a factor of 9 when \([\mathrm{A}]_{0}\) is tripled. (c) When \([\mathrm{A}]_{0}\) is doubled, the rate increases by a factor of 8 . [Section 14.3\(]\)

Short Answer

Expert verified
For the given cases, we find the values of $x$ to be: (a) $x=0$ (b) $x=2$ (c) $x=3$

Step by step solution

01

(a) Find x when rate is unchanged as [A]鈧� is tripled

In this case, we are given that there's no change in rate when the initial concentration of [A] is tripled. Let the initial rate be: rate鈧� = k[ A鈧� ]^x When [A]鈧� is tripled, the new rate remains the same: rate鈧� = k[ 3A鈧� ]^x Since the rate is unchanged, rate鈧� = rate鈧�. Let's set up the equation and solve for x: k[ A鈧� ]^x = k[ 3A鈧� ]^x First, divide both sides of the equation by k to eliminate it: [ A鈧� ]^x = [ 3A鈧� ]^x Now, let's divide both sides of the equation by [A鈧乚^x: 1 = 3^x Taking the logarithm of both sides: x = log(1) / log(3) As log(1) = 0, we have x = 0.
02

(b) Find x as rate increases by a factor of 9 when [A]鈧� is tripled

In this case, the rate increases by a factor of 9 when the initial concentration of [A] is tripled. Let the initial rate be: rate鈧� = k[ A鈧� ]^x The new rate is 9 times the initial rate when [A]鈧� is tripled: rate鈧� = 9k[ A鈧� ]^x = k[ 3A鈧� ]^x Let's set up the equation with the given conditions and solve for x: 9 * k[ A鈧� ]^x = k[ 3A鈧� ]^x First, divide both sides of the equation by k to eliminate it: 9[ A鈧� ]^x = [ 3A鈧� ]^x Now, let's divide both sides of the equation by [A鈧乚^x: 9 = 3^x Taking the logarithm of both sides: x = log(9) / log(3) As log(9) = 2, we have x = 2.
03

(c) Find x as rate increases by a factor of 8 when [A]鈧� is doubled

In this case, the rate increases by a factor of 8 when the initial concentration of [A] is doubled. Let the initial rate be: rate鈧� = k[ A鈧� ]^x The new rate is 8 times the initial rate when [A]鈧� is doubled: rate鈧� = 8k[ A鈧� ]^x = k[ 2A鈧� ]^x Let's set up the equation with the given conditions and solve for x: 8 * k[ A鈧� ]^x = k[ 2A鈧� ]^x First, divide both sides of the equation by k to eliminate it: 8[ A鈧� ]^x = [ 2A鈧� ]^x Now, let's divide both sides of the equation by [A鈧乚^x: 8 = 2^x Taking the logarithm of both sides: x = log(8) / log(2) As log(8) = 3, we have x = 3. So, the values of x for cases a, b and c are 0, 2, and 3, respectively.

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Key Concepts

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

Chemical Kinetics
Understanding the pace at which chemical reactions occur is the essence of chemical kinetics. This branch of chemistry is concerned with the rates of chemical processes, which are influenced by various factors such as concentration of reactants, temperature, and presence of catalysts.

In determining the rate law like in the exercise provided, chemists experiment with varying concentrations of reactants to observe how the rate of formation of products changes. The rate law itself is a mathematical expression encapsulating this relationship, essentially predicting the speed of a reaction for given concentrations of reactants.
Reaction Rate
The reaction rate tells us how rapidly a reaction proceeds. In other words, it's a measure of the change in concentration of reactants or products per unit time. For the reaction \( A \longrightarrow B + C \) given in the exercise, the rate can be represented as the increase in concentration of products B and C over time, or the decrease in concentration of reactant A.

Exercise Improvement Advice

To facilitate understanding, when visualizing reaction rates, one might think of it as a 'speedometer' for chemical reactions, providing real-time data on how fast reactants are being converted into products. Understanding the provided step-by-step solution requires a knack for observing how changes in concentration impact this 'speed,' which is precisely what was observed in the experiment described in the exercise.
Rate Constant
At the heart of the rate law equation lies the rate constant, denoted as \( k \). This is a proportionality constant that links the rate of the reaction to the concentrations of the reactants raised to a power, often representative of the reaction order with respect to each reactant.

The rate constant is independent of the concentration of reactants, but it varies with temperature. In the exercise solutions, we see how the rate constant remains the same when deriving the reaction order \( x \) by comparing the initial and altered rates. Another key point is that the value of \( k \) aids in determining how sensitive a reaction is to changes in reactant concentration.
Concentration Dependence
Often, the rate of a chemical reaction depends heavily on the concentration of reactants. This is known as concentration dependence.

The rate law exemplifies this dependency through the exponent \( x \) placed on the reactant's concentration, as seen in the rate equation rate \( = k[A]^x \). The varying values of \( x \) in different scenarios in the exercise solutions鈥� zero when the rate doesn't change, two when the rate increases ninefold, and three when it increases eightfold鈥攊llustrate how the reaction's rate can vary with the concentration of reactants. This can tell us whether the reaction is zero, first, or second order with respect to a given reactant.

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Most popular questions from this chapter

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21{ }^{\circ} \mathrm{C}\). (a) Write out the balanced equation for the reaction catalyzed by urease. (b) Assuming the collision factor is the same for both situations, estimate the difference in activation energies for the uncatalyzed versus enzyme-catalyzed reaction.

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c),\) is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M},\) the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \longrightarrow \mathrm{B}(a q)\). A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero, and that \(A\) cleanly converts to \(B\) with no intermediates. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 10 -min interval in units of \(M / \mathrm{s}\). (c) Between \(t=10 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(M / s\) ? Assume that the volume of the solution is constant.

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B},\) what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part (a)? (c) How do the half-lives of first-order and second-order reactions differ?

(a) What is meant by the term reaction rate? (b) Name three factors that can affect the rate of a chemical reaction. (c) Is the rate of disappearance of reactants always the same as the rate of appearance of products? Explain.
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