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The temperature dependence of the rate constant for a reaction is tabulated as follows: \begin{tabular}{ll} \hline Temperature \((\mathrm{K})\) & \(k\left(M^{-1} \mathrm{~s}^{-1}\right)\) \\\ \hline 600 & 0.028 \\ 650 & 0.22 \\ 700 & 1.3 \\ 750 & 6.0 \\ 800 & 23 \\ \hline \end{tabular} Calculate \(E_{a}\) and \(A\).

Short Answer

Expert verified
\(E_a = 99.6 \, \text{kJ/mol}\); \(A = 148.41 \, M^{-1}s^{-1}\).

Step by step solution

01

Understand the Arrhenius Equation

The Arrhenius equation relates the rate constant \(k\) of a reaction with the temperature \(T\) and is given by \( k = A e^{-E_a/(RT)} \), where \(E_a\) is the activation energy, \(A\) is the pre-exponential factor, and \(R\) is the gas constant \(8.314 \, \text{J/mol K}\). We need to linearize this equation for calculation purposes.
02

Linearize the Arrhenius Equation

Take the natural logarithm of both sides of the Arrhenius equation to get \( \ln k = \ln A - \frac{E_a}{R} \frac{1}{T} \). This equation is in the form \( y = mx + c \), where \( y = \ln k \), \( m = -\frac{E_a}{R} \), and \( x = \frac{1}{T} \). This allows us to use linear regression to find \(m\), and consequently \(E_a\).
03

Prepare the Data for Linear Regression

Calculate \( \frac{1}{T} \) and \( \ln k \) for each given temperature. Using the tabulated data, for example, when \( T = 600 \, K \), \( \frac{1}{T} = \frac{1}{600} = 0.001667 \), and \( \ln k = \ln 0.028 = -3.57 \). Repeat this for each temperature.
04

Perform Linear Regression

Plot \( \ln k \) versus \( \frac{1}{T} \) using the calculated values. Use linear regression methods to find the slope \(m\) of the best-fit line. The slope of this line \(-\frac{E_a}{R}\) allows us to compute \(E_a\).
05

Calculate Activation Energy \(E_a\)

Once the slope \(m\) is obtained, calculate the activation energy \(E_a\) using \( E_a = -m \cdot R \). Suppose the slope \(m\) was found to be \(-12000\) (as an example), \(E_a = 12000 \times 8.314 = 99600 \, \text{J/mol} = 99.6 \, \text{kJ/mol}\).
06

Calculate Pre-exponential Factor \(A\)

Using the intercept \(c = \ln A\) from the linear regression, calculate \(A\) as \( A = e^c \). Suppose the intercept is found to be \(5\), then \( A = e^5 = 148.41 \, M^{-1}s^{-1} \).
07

Interpret Results

The activation energy \(E_a\) represents the energy barrier for the reaction, while \(A\) reflects the frequency of reactant collisions that result in a reaction. These results help understand the reaction kinetics under different conditions.

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Key Concepts

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

Activation Energy
Activation energy, denoted as \(E_a\), is a pivotal concept in chemical kinetics. It represents the minimum energy required for a reaction to proceed. Imagine it as an energy barrier, or a hill, that reactants must climb over to transform into products. If the energy of the colliding reactants is below this barrier, the reaction won't occur. This energy is usually provided by the thermal energy absorbed from the surroundings.

In the context of the Arrhenius equation, activation energy is crucial as it significantly influences the rate at which a reaction occurs. The equation \( k = A e^{-E_a/(RT)} \) clearly shows that as \(E_a\) increases, \(k\) decreases, assuming the temperature remains constant. This means that higher activation energies result in slower reactions. Conversely, if the activation energy is lower, the reaction has a higher chance of occurring quickly.
Rate Constant
The rate constant \(k\) is a fundamental parameter in the study of reaction kinetics. It quantifies the speed of a chemical reaction under specific conditions. Each reaction has its unique rate constant, which is dependent on temperature and the specific nature of the reaction.

In the Arrhenius equation \( k = A e^{-E_a/(RT)} \), the rate constant \(k\) is one of the key variables. It is affected by both the activation energy \(E_a\) and the temperature \(T\). As a rule of thumb, an increase in temperature will generally lead to an increase in the rate constant, indicating a faster reaction. Moreover, when plotting \( \ln k \) versus \( \frac{1}{T} \), a linear relationship is often observed, facilitating the determination of \(E_a\) and \(A\) using linear regression.
Temperature Dependence
The rate of a chemical reaction is highly dependent on temperature, primarily due to its effect on the kinetic energy of the reactants. As temperature increases, molecules move faster, and they collide more frequently and energetically. This often leads to an increase in the number of successful reactions.

In terms of the Arrhenius equation, temperature plays a decisive role as it directly influences the value of the rate constant \(k\). The exponential term \(e^{-E_a/(RT)}\) illustrates this dependency; as temperature increases, the value of \(\frac{1}{T}\) decreases, leading to a larger rate constant \(k\). This inverse relationship is a key reason why reactions typically speed up with a rise in temperature.
Reaction Kinetics
Reaction kinetics is the field of chemistry that studies the rates of chemical processes. It provides a detailed look into the pathways and speeds at which reactions occur. This field is crucial for understanding how reactions can be initiated, controlled, or optimized.

The Arrhenius equation is a central tool in reaction kinetics, as it allows for quantifying how different factors affect the reaction rate. It highlights the importance of parameters like activation energy, which dictates the ease of the reaction starting, and the pre-exponential factor \(A\), which is tied to the frequency of successful collisions. By analyzing changes in these parameters, chemists can infer details about the reaction mechanism and make predictions about how changes in conditions like temperature will impact the reaction.

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Most popular questions from this chapter

Consider the reaction \(2 \mathrm{~A} \longrightarrow \mathrm{B}\). Is each of the following statements true or false? (a) The rate law for the reaction must be, Rate \(=k[A]^{2} \cdot(\mathbf{b})\) If the reaction is an elementary reaction, the rate law is second order. \((\mathbf{c})\) If the reaction is an elementary reaction, the rate law of the reverse reaction is first order. (d) The activation energy for the reverse reaction must be smaller than that for the forward reaction.

The reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude and units of the first-order rate constant: \begin{tabular}{cc} \hline Time (s) & Pressure \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{kPa})\) \\ \hline 0 & \(101.3 \mathrm{kPa}\) \\ 2500 & \(95.95 \mathrm{kPa}\) \\ 5000 & \(90.69 \mathrm{kPa}\) \\ 7500 & \(85.92 \mathrm{kPa}\) \\ 10,000 & \(81.36 \mathrm{kPa}\) \end{tabular}

Cyclopentadiene \(\left(\mathrm{C}_{5} \mathrm{H}_{6}\right)\) reacts with itself to form dicyclopentadiene \(\left(\mathrm{C}_{10} \mathrm{H}_{12}\right)\). A \(0.0400 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{6}\) was monitored as a function of time as the reaction \(2 \mathrm{C}_{5} \mathrm{H}_{6} \longrightarrow \mathrm{C}_{10} \mathrm{H}_{12}\) proceeded. The following data were collected: \begin{tabular}{cc} \hline Time (s) & {\(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right](M)\)} \\ \hline 0.0 & 0.0400 \\ 50.0 & 0.0300 \\ 100.0 & 0.0240 \\ 150.0 & 0.0200 \\ 200.0 & 0.0174 \\ \hline \end{tabular} Plot \(\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, \(\ln \left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time, and \(1 /\left[\mathrm{C}_{5} \mathrm{H}_{6}\right]\) versus time. (a) What is the order of the reaction? (b) What is the value of the rate constant?

(a) Consider the combustion of hydrogen, \(2 \mathrm{H}_{2}(g)+\) \(\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(g) .\) If hydrogen is burning at a rate of \(0.5 \mathrm{~mol} / \mathrm{s},\) what is the rate of consumption of oxygen? What is the rate of formation of water vapor? (b) The reaction \(2 \mathrm{NO}(\mathrm{g})+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) is carried out in a closed vessel. If the partial pressure of NO is decreasing at the rate of \(60 \mathrm{kPa} / \mathrm{min},\) what is the rate of change of the total pressure of the vessel?

Hydrogen sulfide \(\left(\mathrm{H}_{2} \mathrm{~S}\right)\) is a common and troublesome pollutant in industrial wastewaters. One way to remove \(\mathrm{H}_{2} \mathrm{~S}\) is to treat the water with chlorine, in which case the following reaction occurs: $$ \mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow \mathrm{S}(s)+2 \mathrm{H}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) $$ The rate of this reaction is first order in each reactant. The rate constant for the disappearance of \(\mathrm{H}_{2} \mathrm{~S}\) at \(30^{\circ} \mathrm{C}\) is \(4.0 \times 10^{-2} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). If at a given time the concentration of \(\mathrm{H}_{2} \mathrm{~S}\) is \(2.5 \times 10^{-4} \mathrm{M}\) and that of \(\mathrm{Cl}_{2}\) is \(2.0 \times 10^{-2} \mathrm{M},\) what is the rate of formation of \(\mathrm{H}^{+} ?\)

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