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Chapter 12: Problem 4

How does temperature affect \(k\), the rate constant? Explain.

Short Answer

Expert verified
The rate constant (k) and temperature (T) in a chemical reaction are related through the Arrhenius equation: \(k = Ae^{-Ea/RT}\), where A is the pre-exponential factor, Ea is the activation energy, R is the universal gas constant, and T is the temperature in Kelvin. As temperature increases, the rate constant (k) generally increases, signifying an increased reaction rate. This is because a higher temperature allows a larger fraction of molecules to have enough energy to overcome the activation energy barrier. Moreover, reactions with higher activation energies are more sensitive to changes in temperature, as even a small change in temperature can significantly impact the rate constant (k).

Step by step solution

01

Understand the Rate Constant

The rate constant (k) is a proportionality constant that appears in the rate law of a chemical reaction. It reflects the likelihood of a successful collision between reactant molecules that leads to the formation of products. The magnitude of the rate constant is influenced by various factors such as temperature, concentration, and the activation energy of the reaction.
02

Introduce the Arrhenius Equation

The Arrhenius equation is a mathematical formula that describes the relationship between the rate constant (k) and temperature (T) in a chemical reaction. The equation is: \[k = Ae^{-Ea/RT}\] where: - k is the rate constant - A is the pre-exponential factor or the frequency factor, which reflects the likelihood of a successful molecular collision - Ea is the activation energy required for the reaction to occur (in joules per mole, J/mol) - R is the universal gas constant (8.314 J/mol路K) - T is the temperature in Kelvin
03

Explain the Temperature Dependence

As temperature increases, the fraction of molecules with enough energy to overcome the activation energy barrier also increases. According to the Arrhenius equation, a higher temperature results in a larger value of the rate constant, k. This means that the reaction rate generally increases as the temperature rises. Moreover, the exponential term in the Arrhenius equation, \(e^{-Ea/RT}\), indicates that even a small change in temperature can significantly impact the rate constant.
04

Refer to the Activation Energy

Activation energy (Ea) is the minimum amount of energy required for a reaction to occur. A higher activation energy means that a larger fraction of the reactant molecules need to gain energy to successfully react. The higher the activation energy, the more sensitive the rate constant is to changes in temperature. In summary, the relationship between the rate constant (k) and temperature (T) can be described by the Arrhenius equation. As temperature increases, so does the rate constant, leading to an increased reaction rate. The effect of temperature on the rate constant is more pronounced for reactions with higher activation energies.

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Key Concepts

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

Rate Constant
The rate constant, often represented as \(k\), is a key element in the rate law of a chemical reaction. The value of \(k\) provides insights into how quickly a reaction proceeds. Essentially, it is a measure of the probability that reactant molecules will collide with enough energy to transform into products. This constant is not static and can be influenced by several factors, including:
  • Temperature: As temperature increases, typically so does the rate constant, indicating a faster reaction rate.
  • Concentration: Changes in the concentration of reactants can also affect \(k\).
  • Activation Energy: Higher activation energies reduce the likelihood of successful collisions, affecting the rate constant.
The rate law, which includes the rate constant, allows chemists to predict how changing conditions might alter the speed of a reaction. Overall, being familiar with the rate constant is crucial for understanding reaction dynamics.
Temperature Dependence
The link between temperature and the rate constant \(k\) is captured by the Arrhenius equation: \[k = Ae^{-Ea/RT}\]This equation shows how sensitive \(k\) is to temperature changes. When the temperature increases, more molecules within the reaction have sufficient energy to overcome the activation energy barrier, resulting in a higher rate constant. Temperature influences reaction rates in the following ways:
  • Increasing Temperature: Raises the kinetic energy of molecules, which means more molecules can collide with adequate energy, thus increasing \(k\).
  • Exponential Relationship: The term \(e^{-Ea/RT}\) in the Arrhenius equation implies that small increases in temperature can cause significant changes in \(k\) due to the exponential nature of the relationship.
Understanding this temperature dependence is critical, as it allows for the control and optimization of reactions in various industries, from pharmaceuticals to energy production.
Activation Energy
Activation energy, represented by \(Ea\), is the minimum energy barrier that must be overcome for a chemical reaction to proceed. It's like a threshold the molecules need to overcome to transform into products.This concept is fundamental for understanding reaction kinetics because:
  • Barrier to Reaction: A higher \(Ea\) implies that fewer molecules have enough energy to reaction, hence slowing down the reaction rate.
  • Temperature Sensitivity: Reactions with high activation energies are more sensitive to temperature changes, according to the Arrhenius equation. This means these reactions will have more pronounced increases in rate with temperature increases.
Ultimately, knowing the activation energy helps chemists predict how resistant a reaction is to changes in temperature and how likely it is to occur under given conditions. It explains why some reactions require heating to proceed effectively, as increased temperatures help substances surpass the activation energy threshold.

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Most popular questions from this chapter

The rate law for the reaction $$ 2 \mathrm{NOBr}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ at some temperature is $$ \text { Rate }=-\frac{\Delta[\mathrm{NOBr}]}{\Delta t}=k[\mathrm{NOBr}]^{2} $$ a. If the half-life for this reaction is \(2.00 \mathrm{~s}\) when \([\mathrm{NOBr}]_{0}=\) \(0.900 M\), calculate the value of \(k\) for this reaction. b. How much time is required for the concentration of \(\mathrm{NOBr}\) to decrease to \(0.100 \mathrm{M}\) ?

Two isomers (A and B) of a given compound dimerize as follows: $$ \begin{aligned} &2 \mathrm{~A} \stackrel{k_{\mathrm{i}}}{\longrightarrow} \mathrm{A}_{2} \\ &2 \mathrm{~B} \stackrel{\mathrm{k}_{4}}{\longrightarrow} \mathrm{B}_{2} \end{aligned} $$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be \(0.250 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C}\). In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C}\), where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} M\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} M .\) It was found that after each reaction had progressed for \(3.00 \mathrm{~min}\), \([\mathrm{A}]=3.00[\mathrm{~B}]\). In this case the rate laws are defined as $$ \begin{aligned} &\text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{~A}]^{2} \\\ &\text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{~B}]^{2} \end{aligned} $$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after \(3.00 \mathrm{~min}\). b. Calculate the value of \(k_{2}\). c. Calculate the half-life for the experiment involving \(\mathrm{A}\).

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\), with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{aligned} &k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{~s}^{-1} \\ &k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{~s}^{-1} \end{aligned} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) \(4.00[\mathrm{~B}] ?\)

A certain first-order reaction is \(45.0 \%\) complete in \(65 \mathrm{~s}\). What are the values of the rate constant and the half-life for this process?

You and a coworker have developed a molecule that has shown potential as cobra antivenin (AV). This antivenin works by binding to the venom (V), thereby rendering it nontoxic. This reaction can be described by the rate law $$ \text { Rate }=k[\mathrm{AV}]^{1}[\mathrm{~V}]^{1} $$ You have been given the following data from your coworker: $$ \begin{aligned} [\mathrm{V}]_{0} &=0.20 \mathrm{M} \\ [\mathrm{AV}]_{0} &=1.0 \times 10^{-4} \mathrm{M} \end{aligned} $$ A plot of \(\ln [\mathrm{AV}]\) versus \(t\) (s) gives a straight line with a slope of \(-0.32 \mathrm{~s}^{-1}\). What is the value of the rate constant \((k)\) for this reaction?
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