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Chapter 14: Problem 8

A reaction has the experimental rate equation Rate \(=k[\mathrm{A}]^{2} .\) How will the rate change if the concentration of A is tripled? If the concentration of A is halved?

Short Answer

Expert verified
Tripling [A] increases the rate by 9 times; halving [A] decreases the rate by 4 times.

Step by step solution

01

Understand the Rate Equation

The given rate equation is \(\text{Rate} = k[A]^2\), where \(k\) is the rate constant, and \([A]\) is the concentration of \(\text{A}\). This tells us how the reaction rate depends on the concentration of \(\text{A}\).
02

Rate Change When Concentration is Tripled

If the concentration of \(\text{A}\) is tripled, the new concentration, \([A']\), is \(3[A]\). Substitute \([A']\) into the rate equation: \(\text{New Rate} = k(3[A])^2 = k \, 9[A]^2 = 9 \, \text{Rate}.\) Therefore, the rate of the reaction increases by a factor of 9.
03

Rate Change When Concentration is Halved

If the concentration of \(\text{A}\) is halved, the new concentration, \([A']\), is \([A]/2\). Substitute \([A'/2]\) into the rate equation: \(\text{New Rate} = k\left(\frac{[A]}{2}\right)^2 = k \, \frac{[A]^2}{4} = \frac{1}{4} \, \text{Rate}.\) Therefore, the rate of the reaction decreases by a factor of 4.

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Key Concepts

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

Rate Equation
In chemical kinetics, the rate equation is a mathematical expression that describes how the rate of a reaction changes with the concentration of reactants. It's like a recipe card that tells us exactly how the ingredients (reactants) combine to affect the final cooking time (reaction rate). Let's break it down with an example: for the reaction given in the exercise, the rate equation is \[\text{Rate} = k [\text{A}]^2\]This equation tells us a few things:
  • The reaction rate is directly proportional to the square of the concentration of A. This means if you increase or decrease the concentration of A, the rate will change in a squared manner.
  • \(k\) is the rate constant, which we鈥檒l discuss later. It remains constant as long as the temperature does not change.
By understanding this equation, we can predict how the rate will change when the concentration of A is varied.
Concentration Effects
The concentration of a reactant can greatly influence the rate of a reaction. In the given rate equation, \[\text{Rate} = k [\text{A}]^2\]the concentration of A is the main factor affecting the rate. Let's see how does this effect show in two different scenarios:
  • When A is Tripled: If the concentration of A is tripled, i.e., it becomes \(3[A]\), substitute it into the equation to get:\[\text{New Rate} = k (3[A])^2 = k \cdot 9[A]^2\]This means the reaction rate becomes 9 times quicker. The squaring in the rate equation makes the effects dramatic.
  • When A is Halved: If the concentration of A is halved, i.e., it becomes \([A]/2\), substitute it into the equation to find:\[\text{New Rate} = k \left(\frac{[A]}{2}\right)^2 = k \cdot \frac{[A]^2}{4}\]Therefore, the rate is a quarter of the original rate or 4 times slower.
These calculations show how even small adjustments in concentration can lead to significant changes in reaction rate.
Reaction Rate
The reaction rate refers to how quickly a reaction occurs. It's like measuring how fast you are driving a car; speed varies with different conditions. Similarly, the reaction rate varies with changes in reactant concentrations as shown in the rate equation \[\text{Rate} = k [\text{A}]^2\]The factors that influence reaction rate include:
  • Concentration of Reactants: As illustrated earlier, the rate depends on the concentration of reactant A.
  • Temperature: Although not part of this specific exercise, increasing temperature generally increases reaction rates.
  • Catalysts: These substances can speed up a reaction without being consumed.
Thus, understanding these factors helps in controlling how fast or slow a reaction proceeds.
Rate Constant
The rate constant, denoted as \(k\), is a crucial part of the rate equation \[\text{Rate} = k [\text{A}]^2\]It is a numerical value that quantifies the speed of a reaction under certain conditions, excluding the influence of concentrations.
  • Nature: \(k\) remains constant for a given reaction at a fixed temperature. However, as temperature changes, \(k\) changes, because increased kinetic energy results in a faster reaction.
  • Units: The units of \(k\) depend on the overall order of the reaction. In our case, with a second-order reaction, the units would be \(\text{M}^{-1}\,\text{s}^{-1}\), highlighting its dependency to normalize the concentration effect.
Understanding the rate constant helps determine the intrinsic reactivity of the reaction happening, independent of how concentrated the reactants might be. It's a piece of the puzzle in measuring how efficient a reaction truly is.

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Most popular questions from this chapter

When heated to a high temperature, cyclobutane, \(\mathrm{C}_{4} \mathrm{H}_{8},\) decomposes to ethylene: $$ \mathrm{C}_{4} \mathrm{H}_{8}(\mathrm{g}) \rightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{g}) $$ The activation energy, \(E_{x}\) for this reaction is \(260 \mathrm{kJ} / \mathrm{mol} .\) At \(800 \mathrm{K},\) the rate constant \(k=\) \(0.0315 \mathrm{s}^{-1} .\) Determine the value of \(k\) at \(850 \mathrm{K}\).

The decomposition of nitrogen dioxide at a high temperature $$ \mathrm{NO}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g})+1 / 2 \mathrm{O}_{2}(\mathrm{g}) $$ is second-order in this reactant. (a) Determine the rate constant for this reaction if it takes 1.76 min for the concentration of \(\mathrm{NO}_{2}\) to fall from 0.250 mol/L to 0.100 mol/L. (b) If the chemical equation is written as $$ 2 \mathrm{NO}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) $$ what is the value of the rate constant?

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) is first-order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \begin{array}{c} \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right] \text { where } k=0.17 / \mathrm{hr} \end{array} $$ (a) What is the rate of decomposition when \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.010 \mathrm{M} ?\) (b) What is the half-life of the reaction? (c) If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a flask is \(0.050 \mathrm{atm},\) what is the pressure of all gases (i.e., the total pressure) in the flask after the reaction has proceeded for one half-life?

The following statements relate to the reaction for the formation of HI: $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left|\mathrm{I}_{2}\right| $$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Identify which of the following statements are incorrect. If the statement is incorrect, rewrite it to be correct. (a) Reactions are faster at a higher temperature because activation energies are lower. (b) Rates increase with increasing concentration of reactants because there are more collisions between reactant molecules. (c) At higher temperatures, a larger fraction of molecules have enough energy to get over the activation energy barrier. (d) Catalyzed and uncatalyzed reactions have identical mechanisms.
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