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

Short Answer

Expert verified
(a) If the concentration of A is doubled, the rate will not change, as A does not appear in the rate law. The rate constant will not change, as it is independent of the reactant concentrations. (b) The reaction order for A is 0, and for B is 2. The overall reaction order is 2. (c) The units of the rate constant k for this reaction are \(\frac{1}{L \cdot \text{mol} \cdot s}\).

Step by step solution

01

(a) Effect of doubling [A] on the rate

: The given rate law is: Rate \(=k[\mathrm{B}]^{2}\). Notice that the concentration of A (\([\mathrm{A}]\)) is not present in the rate law. Therefore, if we double the concentration of A, there will be no effect on the rate. The rate will remain the same.
02

(a) Effect of doubling [A] on the rate constant

: The rate constant (k) is independent of the concentrations of the reactants. It depends only on temperature and the nature of the reactants. As a result, the rate constant will not change when the concentration of A is doubled.
03

(b) Reaction orders for A and B

: We can determine the reaction orders for A and B by looking at the exponents of their concentrations in the rate law. The given rate law is: Rate \(=k[\mathrm{B}]^{2}\). The concentration of A is not present in the rate law, which means its reaction order is 0. The concentration of B has an exponent of 2, so the reaction order for B is 2.
04

(b) Overall reaction order

: The overall reaction order is the sum of the reaction orders for each reactant (A and B). In this case, the overall reaction order is \(0 + 2 = 2\).
05

(c) Units of the rate constant

: To determine the units of the rate constant (k), we can rearrange the rate law equation and express k as follows: \(k=\frac{\text{Rate}}{[\mathrm{B}]^{2}}\). In general, the reaction rate has units of moles/liter*time (\(\frac{\text{mol}}{L \cdot s}\)). The concentration of B is in moles/liter (\(\frac{\text{mol}}{L}\)) with an exponent of 2. Now, dividing the rate units by the concentration units squared, we get \(k\) in the units of \(\frac{1}{L \cdot \text{mol} \cdot s}\). The units of the rate constant k for this reaction are \(\frac{1}{L \cdot \text{mol} \cdot s}\).

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Key Concepts

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

Understanding Reaction Rate
In chemistry, the reaction rate is a measure of how fast reactants are transformed into products over time. It can be thought of as the speed of a chemical reaction. When teaching this concept, it's helpful to compare it to something relatable like the speed of a car—just as a car's speed determines how quickly it reaches a destination, the reaction rate determines how quickly reactants become products. For instance, if a reaction has a high rate, it means that substances react quickly, whereas a low rate indicates a slower process.

Let's consider the example from the exercise: A + B → C, where the rate law is given by Rate = k[B]2. In this scenario, the rate refers to how fast the product C is formed as a result of reactants A and B combining. Since the rate law does not include concentration of A ([A]), it suggests that A's concentration does not influence the speed at which product C is produced.
The Significance of the Rate Constant
The rate constant, denoted by 'k' in the rate law equation, is a proportionality factor that relates the concentrations of reactants to the reaction rate. It's important to clarify that the rate constant is, as its name suggests, constant for a given reaction at a fixed temperature. This means that it does not change with varying concentrations of reactants.

In our example, regardless of whether the amount of reactant A is doubled or halved, the rate constant k remains unaffected. It is only the temperature or the intrinsic nature of reactants that can alter the value of k. Understanding this helps students grasp why the constant is pivotal for predicting reaction rates under specific conditions.
Determining Reaction Order
The reaction order is an exponent that shows how the rate is affected by the concentration of each reactant. It is obtained by observing the exponents of the reactants' concentrations in the rate equation. In the exercise, the rate law shows only [B] raised to the second power, indicating that the reaction rate varies with the square of the concentration of B. Therefore, B has a reaction order of 2.

Since [A] does not appear in the rate law, it has a reaction order of 0, meaning changes in A's concentration do not affect the rate. The overall reaction order is found by summing the orders of all reactants involved. In our example, that would be 0 (for A) + 2 (for B) = 2. This combined reaction order offers insights into the sensitivity of the reaction rate to changes in reactant concentrations, which is crucial for both theoretical studies and practical applications, such as in industrial chemical processes.
Exploring Concentration Effect
The concentration effect refers to the impact that changes in the concentration of reactants have on the reaction rate. In general, an increase in the concentration of a reactant leads to an increased reaction rate, because there are more reactant particles available to collide and form products.

However, as illustrated by our exercise, this is not always the case. The rate law Rate = k[B]2 suggests that only the concentration of B affects the reaction rate for this particular chemical process. Doubling [A] does not alter the rate, clearly demonstrating that the concentration effect is highly specific to the underlying reaction mechanism. It's essential for students to recognize that the concentration effect is not universal and depends on the reaction's individual rate law; this understanding forms the basis for predicting and controlling rates in chemical reactions.

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Most popular questions from this chapter

Suppose that a certain biologically important reaction is quite slow at physiological temperature \(\left(37^{\circ} \mathrm{C}\right)\) in the absence of a catalyst. Assuming that the collision factor remains the same, by how much must an enzyme lower the activation energy of the reaction to achieve a \(1 \times 10^{5}\)-fold increase in the reaction rate?

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene $\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\(. A \)0.0400 \mathrm{M}\( solution of \)\mathrm{C}_{5} \mathrm{H}_{6}$ was monitored as a function of time as the reaction $2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}$ proceeded. The following data were collected: $$ \begin{array}{cc} \hline \text { Time (s) } & {\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)} \\\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{array} $$ Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, $\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\( versus time, and \)1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]$ versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

Many metallic catalysts, particularly the precious-metal ones, are often deposited as very thin films on a substance of high surface area per unit mass, such as alumina \(\left(\mathrm{Al}_{2} \mathrm{O}_{3}\right)\) or silica \(\left(\mathrm{SiO}_{2}\right)\). (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How does the surface area affect the rate of reaction?

(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?

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) (d) \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{NH}_{3}(g)\)

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