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(a) Two reactions have identical values for \(E_{a}\). Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a higher rate constant than the other. Account for these observations.

Short Answer

Expert verified
(a) Even if two reactions have identical values of activation energy, they may not have the same rate constant at the same temperature. This is because their frequency factors (A) might be different. (b) Two similar reactions with the same rate constant at 25°C but different rate constants at 35°C must have different activation energies (Ea) and frequency factors (A). This allows for the same rate constant at 25°C and different rate constants at 35°C.

Step by step solution

01

Part (a)

For this part, let's consider the Arrhenius equation, which relates the activation energy (Ea), temperature (T), and the rate constant (k): \[k = A e^{-\frac{E_a}{R T}}\] Here, "A" is the pre-exponential or frequency factor, "R" is the gas constant, and "e" is the base of natural logarithm. Since the two reactions have identical values of activation energy (Ea), we should compare their rate constants when run at the same temperature. If the two reactions have the same frequency factor "A" and the same temperature "T", then their rate constants "k" will be equal. However, if their frequency factors are different, their rate constants will also be different, even when run at the same temperature. So, having the same activation energy does not guarantee that the two reactions will have the same rate constant when run at the same temperature since their frequency factors might be different.
02

Part (b)

We need to explain why two similar reactions have the same rate constant at 25°C but different rate constants at 35°C. Let's consider the given information: 1. Both reactions have the same rate constant at 25°C (298 K), so their Arrhenius equations are equal at 298 K. 2. One of the reactions has a higher rate constant at 35°C (308 K). From the Arrhenius equation, we can conclude that the activation energies (Ea) and frequency factors (A) of the two reactions must be different. If one of the reactions has a higher rate constant at 35°C, it should have either a lower activation energy or a higher frequency factor. This difference inactivation energy allows the rate constants to be the same at 25°C and different at 35°C. In conclusion, the two similar reactions must have different activation energies (Ea) and frequency factors (A), resulting in the same rate constant at 25°C and different rate constants at 35°C.

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Key Concepts

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

Activation Energy
Activation energy, often represented as \(E_a\), is a fundamental concept in chemical kinetics. It refers to the minimum amount of energy required for a reaction to occur. Think of it as the energy barrier that reactants must overcome to transform into products.

Mathematically, activation energy plays a crucial role in the Arrhenius equation:
  • Low activation energy means that the reaction is more likely to proceed, as less energy is needed to activate the reactants.
  • High activation energy requires more energy input, making the reaction less likely to occur spontaneously.
Understanding activation energy is vital for predicting reaction rates and evaluating how changes in conditions, like temperature, can impact these rates.
Rate Constant
The rate constant, denoted as \(k\), is a crucial parameter in the Arrhenius equation, which is often expressed as:\[k = A e^{-\frac{E_a}{RT}}\]The rate constant encapsulates how fast a reaction proceeds. Unlike the rate of reaction itself, which can vary with concentration, the rate constant is constant for a given reaction at a specific temperature.

Key aspects of the rate constant include:
  • Higher \(k\) values indicate faster reactions. This means that at a certain temperature, a reaction with a higher \(k\) is completed more quickly.
  • The rate constant is influenced by other reaction specifics, such as activation energy and frequency factor.
Understanding \(k\) fully relies on both theoretical knowledge and experimental data.
Frequency Factor
In the Arrhenius equation, the frequency factor, denoted as \(A\), reflects the number of times that reacting molecules collide with the proper orientation to yield a reaction. It's sometimes called the pre-exponential factor.

This parameter is particularly significant because it considers both the physical conditions under which reactions occur and molecular dynamics:
  • \(A\) is typically determined experimentally and is unique to each reaction, considering the particular characteristics of the reactants involved.
  • Changes in \(A\) can account for differences in rate constants for reactions with similar activation energies.
The frequency factor helps chemists understand how often successful collisions occur, providing insights into factors other than energy.
Temperature Dependence
The rate of a chemical reaction is significantly influenced by temperature. This is famously encapsulated by the Arrhenius equation. When the temperature increases, reactant molecules have more kinetic energy, leading to more frequent and more energetic collisions.

Considerations regarding temperature include:
  • An increase in temperature generally leads to an increased reaction rate, assuming activation energy remains constant.
  • It's essential to note that a small increase in temperature can result in an exponential increase in rate constant \(k\), due to the exponential nature of the Arrhenius equation.
Understanding how temperature affects reaction rates is essential for controlling reaction conditions in both industrial and laboratory settings.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the factors affecting them. It explores how quickly reactions occur and which molecular properties and environmental factors play pivotal roles. This field of chemistry explains changes observed during reactions and can predict how changes in conditions such as concentration, temperature, or catalysts can alter the speeds of reactions.

Core elements of chemical kinetics include:
  • Reaction rate: The speed at which reactants are converted into products.
  • Factors influencing rate: Concentration, temperature, catalysts, and nature of reactants.
  • Rate laws: Mathematical relationships expressing how the rate depends on concentrations of reactants.
Grasping these concepts facilitates more accurate predictions and control over chemical processes, making chemical kinetics vital for fields like pharmacology, engineering, and environmental science.

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Most popular questions from this chapter

You have studied the gas-phase oxidation of \(\mathrm{HBr}\) by \(\mathrm{O}_{2}\) : $$ 4 \mathrm{HBr}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Br}_{2}(g) $$ You find the reaction to be first order with respect to \(\mathrm{HBr}\) and first order with respect to \(\mathrm{O}_{2}\). You propose the following mechanism: $$ \begin{aligned} \mathrm{HBr}(g)+\mathrm{O}_{2}(g) & \cdots & \mathrm{HOOBr}(g) \\ \mathrm{HOOBr}(g)+\mathrm{HBr}(g) & \longrightarrow 2 \mathrm{HOBr}(g) \\ \mathrm{HOBr}(g)+\mathrm{HBr}(g) & \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Br}_{2}(g) \end{aligned} $$ (a) Indicate how the elementary reactions add to give the overall reaction. (Hint: You will need to multiply the coefficients of one of the equations by 2.) (b) Based on the rate law, which step is rate determining? (c) What are the intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove your mechanism?

The gas-phase decomposition of \(\mathrm{NO}_{2}, 2 \mathrm{NO}_{2}(g) \longrightarrow\) \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)\), is studied at \(383^{\circ} \mathrm{C}\), giving the following data: $$ \begin{array}{ll} \hline \text { Time (s) } & {\left[\mathrm{NO}_{2}\right](M)} \\ \hline 0.0 & 0.100 \\ 5.0 & 0.017 \\ 10.0 & 0.0090 \\ 15.0 & 0.0062 \\ 20.0 & 0.0047 \\ \hline \end{array} $$ (a) Is the reaction first order or second order with respect to the concentration of \(\mathrm{NO}_{2} ?\) (b) What is the value of the rate constant?

The gas-phase reaction of \(\mathrm{NO}\) with \(\mathrm{F}_{2}\) to form NOF and \(\mathrm{F}\) has an activation energy of \(E_{a}=6.3 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(A=6.0 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). The reaction is believed to be bimolecular: $$ \mathrm{NO}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{NOF}(g)+\mathrm{F}(g) $$ (a) Calculate the rate constant at \(100^{\circ} \mathrm{C}\). (b) Draw the Lewis structures for the \(\mathrm{NO}\) and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the structure for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed lines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction.

Consider two reactions. Reaction (1) has a constant halflife, whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

A certain first-order reaction has a rate constant of \(2.75 \times 10^{-2} \mathrm{~s}^{-1}\) at \(20^{\circ} \mathrm{C}\). What is the value of \(k\) at \(60^{\circ} \mathrm{C}\) if (a) \(E_{a}=75.5 \mathrm{~kJ} / \mathrm{mol} ;\) (b) \(E_{a}=125 \mathrm{~kJ} / \mathrm{mol} ?\)

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