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Chapter 8: Problem 38

For a reaction that follows the general rate law, Rate \(=\mathrm{k}[\mathrm{A}][\mathrm{B}]^{2}\), what will happen to the rate of reaction if the concentration of \(\mathrm{A}\) is increased by a factor of \(3.00\) ? The rate will a. Increase by a factor of \(6.00\). b. Decrease by a factor of \(1 / 6.00\). c. Decrease by a factor of \(1 / 3.00\). d. Increase by a factor of \(3.00\).

Short Answer

Expert verified
d. Increase by a factor of 3.00.

Step by step solution

01

Understand the rate law

The given rate law is \( \text{Rate} = k[\text{A}][\text{B}]^2 \). This means that the reaction rate is directly proportional to the concentration of \( \text{A} \) raised to the power of 1 and the concentration of \( \text{B} \) raised to the power of 2.
02

Set up the initial rate equation

Using the rate law, we can set up the initial rate equation as \( \text{Rate}_1 = k[\text{A}]_1[\text{B}]_1^2 \), where \( [\text{A}]_1 \) and \( [\text{B}]_1 \) are the initial concentrations of \( \text{A} \) and \( \text{B} \) respectively.
03

Adjust concentration of A

We are given that the concentration of \( \text{A} \) is increased by a factor of 3. This means the new concentration \( [\text{A}]_2 = 3[\text{A}]_1 \).
04

Write the new rate equation

Substitute the new concentration of \( [\text{A}]_2 \) into the rate law to get the new rate: \( \text{Rate}_2 = k(3[\text{A}]_1)[\text{B}]_1^2 \).
05

Simplify the new rate equation

Simplifying the new rate, we get \( \text{Rate}_2 = 3k[\text{A}]_1[\text{B}]_1^2 \).
06

Determine the factor of change

Comparing the new rate equation \( 3k[\text{A}]_1[\text{B}]_1^2 \) with the initial rate equation \( k[\text{A}]_1[\text{B}]_1^2 \), the factor by which the rate increases is 3. Therefore, the rate increases by a factor of 3.

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Key Concepts

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

Reaction Kinetics
Reaction kinetics is an important area of chemistry that studies the rate at which chemical reactions occur. This branch analyzes the factors that influence reaction speeds, like temperature, pressure, and concentrations of reactants.
Understanding kinetics can help us manipulate reaction conditions to alter the speed of a reaction according to our needs.
  • Each reaction has a unique rate law, which shows how the rate depends on the concentration of the reactants.
  • In a rate law, coefficients of reactants indicate their order in the reaction, affecting how changes in concentration impact the rate.
For instance, consider the rate law of a reaction expressed as \( ext{Rate} = k[A][B]^2 \). Here, brackets denote the concentrations of reactants \( A \) and \( B \). The rate constant \( k \) is unique for every reaction at a given temperature.
Concentration Effect
The concentration of reactants is a crucial factor in reaction kinetics. When the concentration of a reactant varies, the rate of the reaction tends to change as well.
This phenomenon is known as the concentration effect. It helps us predict how the speed of a chemical reaction will change when the amounts of substances involved vary.
  • An increase in the concentration of a reactant generally increases the rate of reaction. More molecules available for collision lead to a higher frequency of fruitful interactions.
  • For example, if the concentration of \( A \) in a reaction given by \( ext{Rate} = k[A][B]^2 \) is tripled, the reaction rate will increase by the same factor associated with \( A\).
This direct relationship between concentration change and rate is a key aspect of kinetic studies, offering insights into how reactions can be controlled and optimized.
Order of Reaction
In chemical kinetics, the order of a reaction provides essential information about the relationship between concentration and reaction rate. This concept helps to determine the influence of each reactant on the reaction rate.
  • The order with respect to a specific reactant corresponds to the exponent of its concentration in the rate law.
  • In \( ext{Rate} = k[A][B]^2 \), the reaction is first order in \( A \) and second order in \( B \).
The total order of the reaction is the sum of the orders of all reactants. In our current example, it would be third order overall. Understanding the order helps in predicting how changes in concentrations affect the rate. For instance, if a reaction is first order in \( A \), doubling \( [A] \) will double the reaction rate.
Knowing the order is vital not only for academic purposes but also for practical applications, such as developing methods to enhance the efficiency of industrial reactions.

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Most popular questions from this chapter

For this reaction \(\mathrm{X}^{-}+\mathrm{OH}^{-} \rightarrow \mathrm{X}^{-}+\mathrm{XO}^{-}\)in an aque- ons medium, the rate of the reaction is given as \(\frac{\left(\mathrm{d}\left(\mathrm{XO}^{-}\right)\right.}{\mathrm{dt}}=\mathrm{K} \frac{\left[\mathrm{X}^{-}\right]\left[\mathrm{XO}^{-}\right]}{\left[\mathrm{OH}^{-}\right]}\) The overall order for this reaction is a. Zero b. 1 c. \(-1\) d. \(1 / 2\)

The rate law has the form; Rate \(=\mathrm{k}[\mathrm{A}][\mathrm{B}]^{3 / 2}\), can the reaction be an elementary process? a. yes b. no c. may be yes or no d. can not be predicted

The rate constant of a reaction is \(1.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(50^{\circ} \mathrm{C}\) and \(4.5 \times 10^{7} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). What is the value of activation energy? a. \(2.2 \times 10^{3} \mathrm{~J} \mathrm{~mol}^{-1}\) b. \(2300 \mathrm{~J} \mathrm{~mol}^{-1}\) c. \(2.2 \times 10^{4} \mathrm{~J} \mathrm{~mol}^{-1}\) d. \(220 \mathrm{~J} \mathrm{~mol}^{-1}\)

Two reactions \(\mathrm{X} \rightarrow\) Products and \(\mathrm{Y} \rightarrow\) products have rate constant \(\mathrm{k}_{\mathrm{x}}\) and \(\mathrm{k}_{\mathrm{Y}}\) at temperature \(\mathrm{T}\) and activation energies \(\mathrm{E}_{\mathrm{x}}\) and \(\mathrm{E}_{\mathrm{Y}}\) respectively. If \(\mathrm{k}_{\mathrm{x}}>\) \(\mathrm{k}_{\mathrm{r}}\) and \(\mathrm{E}_{\mathrm{x}}<\mathrm{E}_{\mathrm{Y}}\) and assuming that for both the reaction is same, then a. At lower temperature \(\mathrm{k}_{\mathrm{Y}}>\mathrm{k}_{\mathrm{x}}\) b. At higher temperature \(\mathrm{k}_{\mathrm{x}}\) will be greater than \(\mathrm{k}_{\mathrm{y}}\) c. At lower temperature \(\mathrm{k}_{\mathrm{x}}\) and \(\mathrm{k}_{\mathrm{Y}}\) will be close to each other in magnitude d. At temperature rises, \(\mathrm{k}_{\mathrm{x}}\) and \(\mathrm{k}_{\mathrm{Y}}\) will be close to each other in magnitude

The rate constant for the reaction, \(2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightarrow 4 \mathrm{NO}_{2}+\mathrm{O}_{2}\) is \(3.0 \times 10^{-4} \mathrm{~s}^{-1}\). If start made with \(1.0 \mathrm{~mol} \mathrm{~L}^{-1}\) of \(\mathrm{N}_{2} \mathrm{O}_{5}\), calculate the rate of formation of \(\mathrm{NO}_{2}\) at the moment of the reaction when concentration of \(\mathrm{O}_{2}\) is \(0.1 \mathrm{~mol} \mathrm{~L}^{-1}\). a. \(1.2 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) b. \(3.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) c. \(9.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\) d. \(4.8 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}\)
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