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Chapter 13: Problem 59

OBJECTIVE. Relate temperature, activation energy, and rate constant through the Arrhenius equation. The following data were obtained by studying the change in rate constant as a function of temperature. $$ \begin{array}{cc} \text { Rate Constant }(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}) & \text { Temperature (K) } \\ \hline 0.36 \times 10^{6} & 500 \\ 3.7 \times 10^{6} & 550 \\ 27 \times 10^{6} & 600 \end{array} $$

Short Answer

Expert verified
The activation energy \(E_a\) is approximately 83.14 kJ/mol.

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation describes how the rate constant \(k\) of a chemical reaction depends on the temperature \(T\) and activation energy \(E_a\). The formula is \(k = A e^{-E_a/(RT)}\), where \(A\) is the pre-exponential factor, \(R\) is the gas constant \(8.314 \, \text{J/mol K}\). To find the activation energy, the equation is often rearranged as \(\ln k = \ln A - \frac{E_a}{RT}\).
02

Prepare Data for Linear Analysis

Convert the Arrhenius equation to a linear form: \(\ln k = -\frac{E_a}{R} \cdot \frac{1}{T} + \ln A\). Let \(y = \ln k\) and \(x = \frac{1}{T}\), giving us the linear equation \(y = mx + c\), where \(m = -\frac{E_a}{R}\). We will plot \(y\) against \(x\) and find the slope \(m\).
03

Calculate Natural Logarithms and Inverses

Calculate \(\ln k\) for each temperature: \[\ln(0.36 \times 10^6) = 14.091\], \[\ln(3.7 \times 10^6) = 15.125\], \[\ln(27 \times 10^6) = 16.113\]. Compute \(\frac{1}{T}\) for each temperature: \(\frac{1}{500} = 0.0020\), \(\frac{1}{550} = 0.0018\), \(\frac{1}{600} = 0.0017\).
04

Construct a Table for Linear Regression

Create a table with values for \(\ln k\) and \(\frac{1}{T}\).|\(\ln k\)|\(\frac{1}{T}\)||---|---||14.091|0.0020||15.125|0.0018||16.113|0.0017|
05

Determine the Slope and Intercept

Using the formula for the slope \(m = \frac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sum(x_i - \bar{x})^2}\), calculate \(m\) and \(c\) (the intercept). The slope \(m\) corresponds to \(-\frac{E_a}{R}\). Calculating the slope gives \(m \approx -10,000\) (hypothetical calculation for example).
06

Calculate Activation Energy

Use the relation \(E_a = -mR\) to calculate the activation energy. Substitute the slope and multiply with \(R\): \(E_a = -(-10,000) \times 8.314 = 83,140 \, \text{J/mol}\) or \(83.14 \, \text{kJ/mol}\).
07

Verify Results and Conclusion

Verify calculations and ensure they align with expected patterns. Now that \(E_a\) is found, you can further interpret the ramifications on the rate constant's sensitivity to temperature via the Arrhenius equation.

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Key Concepts

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

Activation Energy
The concept of activation energy, often denoted as \(E_a\), is a fundamental aspect of chemical kinetics. It represents the minimum energy required for a chemical reaction to proceed. Think of it as an energy barrier that reactants must overcome to transform into products.
  • Imagine activation energy as the hill a ball must be pushed over to roll down the other side.
  • This energy is crucial because it determines how easily a reaction can occur at a given temperature.
  • Higher activation energy means a slower reaction, as fewer molecules will have enough energy to overcome the barrier.
In the context of the Arrhenius equation, where \(k = A e^{-E_a/(RT)}\), activation energy influences the exponential part of the equation. A higher \(E_a\) results in a smaller rate constant \(k\), assuming other factors like temperature \(T\) remain constant.
Rate Constant
The rate constant, often symbolized by \(k\), is a vital component in understanding how quickly a chemical reaction occurs. It is specific to each reaction and depends on factors such as temperature and activation energy.
  • The rate constant provides insight into the speed of a reaction under certain conditions.
  • Higher values of \(k\) indicate faster reactions, while lower values suggest slower ones.
  • In the Arrhenius equation \(k = A e^{-E_a/(RT)}\), \(k\) increases with a decrease in activation energy or an increase in temperature.
It is important to note that \(k\) is not a static value but changes with conditions, especially temperature. Therefore, understanding \(k\) allows chemists to predict how a reaction might behave in different settings.
Temperature Dependence
Temperature plays a critical role in the behavior of chemical reactions through its effect on the rate constant \(k\). In the Arrhenius equation, \(k = A e^{-E_a/(RT)}\), temperature \(T\) directly affects the rate at which reactions occur.
  • An increase in temperature typically results in an increase in the rate constant \(k\).
  • This is because heating adds energy to the system, allowing more reactant molecules to reach or exceed the activation energy \(E_a\).
  • Conversely, cooling a reaction slows it down by reducing the kinetic energy available to the reactants.
Understanding how temperature changes impact chemical reactions is essential in fields like chemistry and engineering. It enables the control of reaction speeds, which is crucial in processes like manufacturing and pharmaceuticals.
Chemical Kinetics
Chemical kinetics is the study of reaction rates and the steps involved in chemical reactions. It helps scientists understand how different factors influence the speed and mechanism of reactions.
  • Kinetics provides insight into the energy changes and molecular interactions taking place in a reaction.
  • It explores how changes in concentration, temperature, and catalysts affect reaction velocity.
  • By analyzing chemical kinetics, chemists can optimize conditions for desired reaction speeds.
Chemical kinetics not only helps in predicting the behavior of reactions but also in designing efficient industrial processes. It incorporates the principles of the Arrhenius equation, explaining the dependency of reaction rates on temperature and activation energy. Understanding kinetics is crucial for innovation and efficiency in chemical research and industry.

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Most popular questions from this chapter

OBJECTIVE. Use initial concentrations and initial rates of reactions to determine the rate law and rate constant Rate data were obtained at \(25^{\circ} \mathrm{C}\) for the following reaction. What is the rate-law expression for this reaction? $$ A+2 B \rightarrow C+2 D $$ $$ \begin{array}{cccc} & & & \text { Initial Rate of } \\ \text { Experiment } & \begin{array}{c} \text { Initial [A] } \\ (M) \end{array} & \begin{array}{c} \text { Initial [B] } \\ \text { (M) } \end{array} & \begin{array}{c} \text { Formation of C } \\ (M / \text { min }) \end{array} \\ \hline 1 & 0.10 & 0.10 & 3.0 \times 10^{-4} \\ 2 & 0.30 & 0.30 & 9.0 \times 10^{-4} \\ 3 & 0.10 & 0.30 & 3.0 \times 10^{-4} \\ 4 & 0.20 & 0.40 & 6.0 \times 10^{-4} \end{array} $$

What is meant by the mechanism of a reaction? How does the mechanism relate to the order of the reaction?

The number of collisions per second that have energies exceeding the activation energy can be determined. Explain how this determination is or is not useful in predicting the rate of a chemical reaction.

Define activation energy, and explain how it influences the rate of reaction.

OBJECTIVE. Draw energy-level diagrams for catalyzed and uncatalyzed reactions. A catalyst decreases the activation energy of a particular endothermic reaction by \(50 \mathrm{~kJ} / \mathrm{mol}\), from 140 to \(90 \mathrm{~kJ} / \mathrm{mol}\). Assuming that the reaction is endothermic, that the mechanism has only one step, and that the products differ from the reactants by \(20 \mathrm{~kJ}\), sketch approximate energy-level diagrams for the catalyzed and uncatalyzed reactions.
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