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Consider a hypothetical reaction between \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) that is firstorder in \(A\), zero order in \(B\), and second order in C. (a) Write the rate law for the reaction. (b) How does the rate change when \([\mathrm{A}]\) is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant? (d) How does the rate change when \([\mathrm{C}]\) is tripled and the other reactant concentrations are held constant? (e) By what factor does the rate change when the concentrations of all three reactants are tripled?

Short Answer

Expert verified
a) The rate law for the given reaction is: Rate = k[A][C]^2 b) The rate doubles when the concentration of A is doubled. c) There is no change in the rate when the concentration of B is tripled. d) The rate is increased by 9 times when the concentration of C is tripled. e) When all reactant concentrations are tripled, the rate of the reaction increases by a factor of 27.

Step by step solution

01

a) Writing the Rate Law

The rate law for a reaction can be given as follows, considering the dependence on various reactants: Rate = k \([A]^m [B]^n [C]^p\) Here, 'k' is the rate constant and 'm', 'n', and 'p' are the orders regarding A, B, and C respectively. Given that the reaction is first-order in A, zero-order in B, and second-order in C, the rate law becomes: Rate = k \([A]^1 [B]^0 [C]^2\) Now, let's simplify this expression: Rate = k[A][C]^2 This is the rate law for the given reaction.
02

b) Effect of Doubling the Concentration of A

Let's see how the rate changes when we double the concentration of A, keeping the concentrations of B and C constant: New Rate = k(2[A])[C]^2 As k, [C] and [C]^2 are constants, we can see that the new rate is equal to twice the original rate: New Rate = 2(k[A][C]^2) = 2(Rate) The rate doubles when the concentration of A is doubled.
03

c) Effect of Tripling the Concentration of B

Now, let's see how the rate changes when we triple the concentration of B, keeping A and C constant: New Rate = k[A](3[B])^0[C]^2 Since [B]^0 is 1, regardless of the value of B: New Rate = k[A][C]^2 = Rate There is no change in the rate when the concentration of B is tripled.
04

d) Effect of Tripling the Concentration of C

Finally, let's see the rate change when we triple the concentration of C, keeping A and B constant: New Rate = k[A](3[C])^2 New Rate = k[A] \cdot 9[C]^2 = 9(k[A][C]^2) = 9(Rate) The rate is increased by 9 times when the concentration of C is tripled.
05

e) Effect of Tripling All Reactants

Now, we'll see the rate change when all reactants are tripled: New Rate = k(3[A])(3[B])^0 (3[C])^2 New Rate = k(3[A])(1) \cdot 9([C]^2) = 27(k[A][C]^2) = 27(Rate) When all reactant concentrations are tripled, the rate of the reaction increases by a factor of 27.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the rates of chemical reactions and the factors that affect them. It is fundamental in understanding how reactions occur and provides invaluable insights for industries ranging from pharmaceuticals to manufacturing.

At its core, chemical kinetics looks at the speed of a reaction and what can speed up or slow it down. It uses rate laws, like the one in our exercise, which mathematically relate the rate of a reaction to the concentration of its reactants. By knowing the rate law, you can predict the outcome of a reaction under different conditions, which is crucial for optimizing processes in research and industry.

For our hypothetical reaction, the rate of reaction depends specifically on the concentrations of reactants A and C, with reactant B having no effect because it's zero order. This reveals that not all components in a reaction necessarily influence its rate—an important concept in selecting and controlling conditions for a specific desired reaction rate.
Reactant Concentration Effect
The concentration of reactants plays a significant role in determining the rate of a chemical reaction. According to the law of mass action, the rate is generally proportional to the concentration of the reactants. However, this relationship varies depending on the order of the reaction with respect to each reactant.

In our exercise, doubling the concentration of reactant A, which is first-order, results in doubling the reaction rate, illustrating a direct proportionality. In contrast, any changes in the concentration of B, which is zero-order, have no effect on the rate, implying that B does not participate actively in the determining step of the reaction or is present in a large excess.

Understanding how each reactant concentration affects the rate is crucial for controlling reactions. It helps chemical engineers and scientists to design processes that optimize yields and minimize waste. For students, grasping the concept of reactant concentration effect is a stepping stone towards mastering more complex kinetics problems.
Order of Reaction
The order of a reaction describes how the rate is affected by the concentration of each reactant. It's a critical component of the rate law equation and can be determined experimentally. The order is usually an integer or fraction, although it's not necessarily related to the stoichiometry of the reaction.

In the given problem, A is first-order, meaning the rate will change linearly with its concentration. Doubling A doubles the rate because its order is one. For B, the reaction is zero-order, and the concentration of B has no impact on the rate, serving as a constant rate indicator as long as B is available. On the other hand, C is second-order; thus, the rate changes with the square of its concentration. Tripling C increases the rate ninefold, as shown by the calculation in our exercise.

Students should understand that the order of reaction dictates how significantly a change in the concentration of a reactant will alter the reaction rate. This concept is foundational for predicting the kinetics of a reaction and for designing experiments to reveal the mechanisms of chemical reactions.

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Most popular questions from this chapter

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant?

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c)\), has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C}\). (b) \(\mathrm{A}\) solution of KOH in ethanol is made up by dissolving \(0.335 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(35^{\circ} \mathrm{C} ?(\mathrm{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

(a) Two reactions have identical values for \(E_{a}\). Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a higher rate constant than the other. Account for these observations.

The following mechanism has been proposed for the gas-phase reaction of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) and chlorine: Step 1: \(\mathrm{Cl}_{2}(g) \underset{k_{1}}{\stackrel{k_{1}}{\rightleftharpoons}} 2 \mathrm{Cl}(g) \quad\) (fast) Step 2: \(\mathrm{Cl}(g)+\mathrm{CHCl}_{3}(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{HCl}(g)+\mathrm{CCl}_{3}(g) \quad\) (slow) Step 3: \(\mathrm{Cl}(g)+\mathrm{CCl}_{3}(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CCl}_{4} \quad\) (fast) (a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each of the elementary reactions? (d) What is the rate-determining step? (e) What is the rate law predicted by this mechanism? (Hint: The overall reaction order is not an integer.)

Americium-241 is used in smoke detectors. It has a rate constant for radioactive decay of \(k=1.6 \times 10^{-3} \mathrm{yr}^{-1}\). By contrast, iodine- 125, which is used to test for thyroid functioning, has a rate constant for radioactive decay of \(k=0.011\) day \(^{-1}\). (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of a \(1.00-\mathrm{mg}\) sample of either isotope remains after three half-lives?

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