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

Short Answer

Expert verified
The rate law is \( \text{Rate} = k[B]^2[C] \). The rate stays the same when \( [A] \) is tripled, quadruples when \( [B] \) is doubled, triples when \( [C] \) is tripled, is multiplied by 8 when all are doubled, and reduces to 1/8 when all are halved.

Step by step solution

01

Determine the Rate Law

The problem states that the reaction is zero order in \( \mathrm{A} \), second order in \( \mathrm{B} \), and first order in \( \mathrm{C} \). This means the rate law can be written in the form \( \text{Rate} = k[A]^0[B]^2[C]^1 \). Simplifying, this becomes \( \text{Rate} = k[B]^2[C] \) since \( [A]^0 = 1 \).
02

Tripling [A]

Since the reaction is zero order in \( \mathrm{A} \), tripling \( [A] \) will not affect the rate of the reaction. Therefore, the rate remains the same.
03

Doubling [B]

The reaction is second order in \( \mathrm{B} \). According to the rate law, if \( [B] \) is doubled, the rate is multiplied by \( 2^2 \). Thus, the new rate will be four times the original rate.
04

Tripling [C]

The reaction is first order in \( \mathrm{C} \). Therefore, if \( [C] \) is tripled, the rate will be multiplied by \( 3^1 \), resulting in a rate that is three times the original.
05

Doubling All Reactants

When all reactant concentrations are doubled, \( [A]^0 \) remains unaffected. The rate for \( \mathrm{B} \) will multiply by \( 2^2 \) (which is 4), and for \( \mathrm{C} \) will multiply by \( 2^1 \) (which is 2). Therefore, the overall factor by which the rate changes is \( 4 \times 2 = 8 \).
06

Halving All Reactants

Halving all reactant concentrations results in no change for \( [A]^0 \), but for \( [B] \), the rate is multiplied by \( (0.5)^2 \) (which is 0.25), and for \( [C] \), is multiplied by \( 0.5^1 \) (which is 0.5). Hence, the overall factor is \( 0.25 \times 0.5 = 0.125 \), reducing the rate to \( \frac{1}{8} \).

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Key Concepts

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

Rate of Reaction
The rate of a reaction is a measure of how quickly a chemical reaction occurs. In other words, it is the change in concentration of a reactant or product per unit time. For any given reaction, the rate is expressed using a rate law, which shows how the rate depends on the concentration of the reactants. For our hypothetical reaction involving substances A, B, and C, the rate law is determined by their reaction orders. This defines how each component affects the rate when its concentration changes. To summarize:
  • The rate of reaction is a vital metric for understanding reaction speed.
  • Expressed as: Rate = 饾憳 [A]鈦 [B]岬 [C]岬, where k is the rate constant and n, m, o are the orders of reaction for each component.
  • Changes in concentrations of reactants can directly affect the reaction rate based on this law.
Understanding the rate helps predict how long a reaction will take under different conditions.
Reaction Order
The reaction order indicates the power to which the concentration of a reactant is raised in the rate law. It tells us how changes in a reactant's concentration influence the reaction rate. For example, in our hypothetical reaction: - It is zero order in A, meaning changes in A's concentration have no effect on reaction rate. - It is second order in B, so the rate quadruples if the concentration of B is doubled. - It is first order in C, meaning the rate triples if the concentration of C is tripled. Reaction order can be:
  • Zero order: Rate is constant regardless of concentration.
  • First order: Rate is directly proportional to concentration.
  • Second order: Rate is proportional to the square of the concentration.
By understanding reaction orders, we can control and predict the rates of chemical reactions during experiments or industrial processes.
Concentration Effect
The concentration effect describes how the concentration of reactants impacts the rate of a chemical reaction. In our hypothetical scenario: - Tripling the concentration of A does not affect the rate because its reaction order is zero. - Doubling the concentration of B increases the rate fourfold, given its second-order dependence. - Tripling the concentration of C increases the rate threefold due to its first-order nature. The overarching effect is that reaction rate will change more dramatically with higher order reactants:
  • Higher order means a more significant effect on rate from a change in concentration.
  • Zero order reactants do not alter rate but remain critical in determining overall reaction completion.
  • Understanding these effects enables precise control over chemical processes by manipulating reactant concentrations.
Grasping the concentration effect allows chemists to tailor reactions to be efficient and effective for desired outcomes.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the rates of chemical reactions and the factors affecting them. It provides insights into reaction mechanisms and influences decision-making in industrial and laboratory settings. Important points include: - Kinetics investigates how different conditions like temperature and concentration influence reaction rates. - Reaction rates offer clues about reaction steps and intermediates, helping to determine the pathway of reactions. - Chemical kinetics considers variables such as activation energy and catalysts, which can change the speed of reactions significantly. In our exercise, chemical kinetics principles allow us to understand why certain changes in reactant concentrations have specific effects on reaction rate. For industrial applications, this knowledge ensures efficiency by optimizing conditions to achieve desired speeds and yields. Learning about kinetics helps in engineering better chemical processes and understanding natural phenomena.

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Most popular questions from this chapter

The addition of NO accelerates the decomposition of \(\mathrm{N}_{2} \mathrm{O},\) possibly by the following mechanism: $$ \begin{aligned} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) & \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) & \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this reaction? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism?

(a) In which of the following reactions would you expect the orientation factor to be more important in leading to reaction: \(\mathrm{O}_{3}+\mathrm{O} \longrightarrow 2 \mathrm{O}_{2}\) or \(\mathrm{NO}+\mathrm{NO}_{3} \longrightarrow 2 \mathrm{NO}_{2} ?\) (b) What is related to the orientation factor? Which, smaller or larger ratio of effectively oriented collisions to all possible collisions, would lead to a smaller orientation factor?

You perform a series of experiments for the reaction \(\mathrm{A} \rightarrow 2 \mathrm{~B}\) 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) The rate increases by a factor of \(6.25,\) when \([A]_{0}\) is increased by a factor of \(2.5 .\) (b) There is no rate change when \([A]_{0}\) is increased by a factor of \(4 .(\mathbf{c})\) The rate decreases by a factor of \(1 / 2\), when \([A]\) is cut in half.

Ozone in the upper atmosphere can be destroyed by the following two-step mechanism: $$ \begin{aligned} \mathrm{Cl}(g)+\mathrm{O}_{3}(g) & \longrightarrow \mathrm{ClO}(g)+\mathrm{O}_{2}(g) \\ \mathrm{ClO}(g)+\mathrm{O}(g) & \longrightarrow \mathrm{Cl}(g)+\mathrm{O}_{2}(g) \end{aligned} $$ (a) What is the overall equation for this process? (b) What is the catalyst in the reaction? (c) What is the intermediate in the reaction?

The rate of disappearance of HCl was measured for the following reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The following data were collected: \begin{tabular}{cc} \hline Time (min) & {\([\mathrm{HCl}](M)\)} \\ \hline 0.0 & 1.85 \\ 54.0 & 1.58 \\ 107.0 & 1.36 \\ 215.0 & 1.02 \\ 430.0 & 0.580 \\ \hline \end{tabular} (a) Calculate the average rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Calculate the average rate of reaction for the entire time for the data from \(t=0.0 \mathrm{~min}\) to \(t=430.0 \mathrm{~min} .(\mathbf{c})\) Which is greater, the average rate between \(t=54.0\) and \(t=215.0 \mathrm{~min}\), or between \(t=107.0\) and \(t=430.0 \mathrm{~min} ?\) (d) Graph \([\mathrm{HCl}\) versus time and determine the instantaneous rates in \(M / \min\) and \(M /\) s at \(t=75.0 \mathrm{~min}\) and \(t=250 \mathrm{~min} .\)

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