The rate constant \((k)\) for a reaction was measured as a function of temperature. A plot of \(\ln k\) versus \(1 / T(\) in \(\mathrm{K})\) is linear and has a slope of \(-1.01 \times 10^{4} \mathrm{~K}\). Calculate the activation energy for the reaction.

Short Answer

Expert verified
The activation energy for the reaction is \(8.39 \times 10^{4} \text{ J/mol}\).

Step by step solution

01

Recall the Arrhenius Equation

The Arrhenius equation can be written in a linear form as \(\ln k = \ln A - \frac{E_a}{R}\frac{1}{T}\) where \(k\) is the rate constant, \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvins. The slope of the plot of \(\ln k\) versus \(1 / T\) is equal to \( -\frac{E_a}{R}\).
02

Use the Slope to Calculate the Activation Energy

The value given for the slope of the line is \( -1.01 \times 10^{4} \mathrm{~K} \). Since this is the value of \( -\frac{E_a}{R}\), we can rearrange and solve for the activation energy \(E_a\) using the ideal gas constant \(R = 8.314 \text{ J/(mol K)}\): \(E_a = -\text{slope} \times R\).
03

Calculate the Activation Energy

Substitute the given slope and the value of \(R\) into the equation to find the activation energy: \(E_a = -(-1.01 \times 10^{4} \mathrm{~K}) \times 8.314 \text{ J/(mol K)}\).
04

Perform the Calculation

Multiply the two numbers to get the activation energy: \(E_a = 1.01 \times 10^{4} \mathrm{~K} \times 8.314 \text{ J/(mol K)} = 8.39 \times 10^{4} \text{ J/mol}\).

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Key Concepts

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

Activation Energy
Understanding the concept of activation energy is crucial when studying chemical kinetics. Activation energy, denoted as \(E_a\), is the minimum amount of energy that reacting particles must possess for a reaction to occur. It serves as an energy barrier that reactants must overcome to transform into products.

In the context of the Arrhenius equation, activation energy is linked to the temperature sensitivity of the rate constant \(k\). When presented in a linearized Arrhenius plot, where \(\ln k\) is plotted against \(1/T\), the slope of the line provides insight into the magnitude of \(E_a\). A steeper slope suggests a higher activation energy, indicating that the rate of reaction is more sensitive to changes in temperature. By calculating the activation energy using the slope from the plot, we gain a quantitative measure of this barrier, helping us predict and control the reaction's speed.

For example, if we were given a plot with a slope of \( -1.01 \times 10^{4} \mathrm{~K}\), using the Arrhenius equation, we could determine the actual value of \(E_a\) and further comprehend how this particular reaction behaves under various temperature conditions.
Rate Constant
The rate constant, symbolized by \(k\), is a vital parameter in the Arrhenius equation representing how quickly a chemical reaction proceeds. It is specific to each reaction and can be influenced by various factors, such as temperature and the presence of a catalyst.

In mathematical terms, the rate constant connects the concentration of reactants to the rate of reaction, allowing us to quantify the reaction's speed. A larger value of \(k\) corresponds to a faster reaction under given conditions. The Arrhenius equation highlights that \(k\) is not a fixed property; rather, it changes exponentially with temperature. This relationship is encapsulated in the Arrhenius plot mentioned earlier, where a linear trend on a \(\ln k\) versus \(1/T\) graph represents the changes in \(k\) as the temperature varies.

Understanding the rate constant is not only important for predicting the progress of a reaction but also for designing processes and safety in industrial applications. For instance, knowing that the rate constant increases with temperature might inform a chemical engineer's decision to adjust reaction conditions to achieve the desired product yield efficiently.
Temperature Dependence of Reaction Rates
The temperature dependence of reaction rates is a core concept in chemical kinetics, elegantly captured by the Arrhenius equation. As temperature increases, the rate of most chemical reactions accelerates, which is chiefly due to an increased number of molecules possessing enough energy to overcome the activation energy threshold.

This thermal sensitivity is quantified in the equation by the temperature \(T\) present in the exponential factor. Higher temperatures effectively lower the fraction \(1/T\), which in turn increases the value of \(\ln k\)—demonstrating an exponential increase in the rate constant \(k\). It is this exponential relationship that underpins the practical understanding of how reaction rates escalate with temperature.

For example, in the linear form of the Arrhenius equation \(\ln k = \ln A - \frac{E_a}{R}\frac{1}{T}\), a plot of \(\ln k\) versus \(1/T\) can help us visualize and calculate how sensitive a reaction rate is to temperature changes. Temperature control is therefore a powerful tool for chemists and engineers, allowing them to manipulate reaction speeds for optimal outcomes, whether they are conducting delicate laboratory experiments or operating industrial-scale chemical processes.

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Most popular questions from this chapter

What is a catalyst? How does a catalyst increase the rate of a chemical reaction?

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) and has a rate constant of \(1.42 \times 10^{-4} \mathrm{~s}^{-1}\) at a certain temperature. a. What is the half-life for this reaction? b. How long will it take for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(25 \%\) of its initial concentration? c. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.00 \mathrm{M}\), how long will it take for the concentration to decrease to \(0.78 \mathrm{M} ?\) d. If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(0.150 \mathrm{M},\) what is the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) after \(2.00 \times 10^{2} \mathrm{~s}\) ? After \(5.00 \times 10^{2} \mathrm{~s} ?\)

Consider the reaction: \(\mathrm{C}_{4} \mathrm{H}_{8}(g) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{4}(g)\) The tabulated data were collected for the concentration of \(\mathrm{C}_{4} \mathrm{H}_{8}\) as a function of time: $$ \begin{array}{cc} \text { Time (s) } & {\left[\mathrm{C}_{4} \mathrm{H}_{8}\right] \text { (M) }} \\ \hline 0 & 1.000 \\ \hline 10 & 0.913 \\ \hline 20 & 0.835 \\ \hline 30 & 0.763 \\ \hline 40 & 0.697 \\ \hline 50 & 0.637 \\ \hline \end{array} $$ a. What is the average rate of the reaction between 0 and 10 s? Between 40 and 50 s? b. What is the rate of formation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) between 20 and \(30 \mathrm{~s}\) ?

This reaction is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}\) $$ \mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g) $$ The rate constant for the reaction at a certain temperature is \(0.053 / \mathrm{s}\) a. Calculate the rate of the reaction when \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.055 \mathrm{M}\). b. What would the rate of the reaction be at the concentration indicated in part a if the reaction were second order? Zero order? (Assume the same numerical value for the rate con- stant with the appropriate units.)

The tabulated data show the rate constant of a reaction measured at several different temperatures. Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction. $$ \begin{array}{cc} \text { Temperature (K) } & \text { Rate Constant (1/s) } \\ 300 & 0.0134 \\ \hline 310 & 0.0407 \\ \hline 320 & 0.114 \\ \hline 330 & 0.303 \\ \hline 340 & 0.757 \\ \hline \end{array} $$

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