Question

The first-order rate constant for reaction of a particular organic compound with water varies with temperature as follows: \begin{tabular}{ll} \hline Temperature (K) & Rate Constant (s \(^{-1}\) ) \\ \hline 300 & \(3.2 \times 10^{-11}\) \\ 320 & \(1.0 \times 10^{-9}\) \\ 340 & \(3.0 \times 10^{-8}\) \\ 355 & \(2.4 \times 10^{-7}\) \\ \hline \end{tabular} From these data, calculate the activation energy in units of \(\mathrm{kJ} / \mathrm{mol}\)

Short answer

The activation energy for the reaction is approximately 60.983 kJ/mol.

Step-by-step solution

Write the Arrhenius equation, then linearize it

Start with the Arrhenius equation: \(k = Ae^{\frac{-E_a}{RT}}\) Where: - \(k\) is the rate constant - \(A\) is the pre-exponential factor - \(E_a\) is the activation energy - \(R\) is the gas constant (8.314 J/(mol K)) - \(T\) is the temperature in Kelvin Now, let's take the natural logarithm of both sides to linearize the equation: \(ln(k) = ln(A) - \frac{E_a}{R}\cdot\frac{1}{T}\) This now resembles the equation of a straight line (y = mx + b), where: - y = \(ln(k)\) - x = \(\frac{1}{T}\) - m = \(-\frac{E_a}{R}\) - b = \(ln(A)\) From this transformed equation, we can use linear regression to determine the slope (m) and then find the activation energy (Ea).

Calculate the natural logarithm of the rate constants and reciprocal of temperatures

First, find the values of \(ln(k)\) and \(\frac{1}{T}\) for each data point: \begin{tabular}{lll} \hline Temperature (K) & Rate Constant (s\(^{-1}\)) & \(ln(k)\) \\\ \hline 300 & \(3.2 \times 10^{-11}\) & -24.6611 \\ 320 & \(1.0 \times 10^{-9}\) & -20.7233 \\ 340 & \(3.0 \times 10^{-8}\) & -17.0388 \\ 355 & \(2.4 \times 10^{-7}\) & -15.2441 \\ \hline \end{tabular} \begin{tabular}{ll} \hline Temperature (K) & \(\frac{1}{T}\) (\(K^{-1}\)) \\\ \hline 300 & 0.003333 \\ 320 & 0.003125 \\ 340 & 0.002941 \\ 355 & 0.002817 \\ \hline \end{tabular}

Calculate the slope using linear regression

Now calculate the slope (m) using linear regression with the data in Step 2: \(m = -\frac{E_a}{R}\) We can utilize a linear regression calculator or software to obtain the slope (m). For our data, the slope is approximately -7335.

Calculate the activation energy (Ea)

Now we can use the calculated slope (m) to determine the activation energy (Ea): \(m =-\frac{E_a}{R} \Rightarrow E_a = -m\cdot R\) Ea = (-7335)(8.314 J/(mol·K)) = 60983.29 J/mol Convert the activation energy to kJ/mol: Ea = 60.983 kJ/mol Therefore, the activation energy for this reaction is approximately 60.983 kJ/mol.

Key concepts

Arrhenius equation

The Arrhenius equation is a vital tool in chemistry. It expresses the relationship between the rate constant (\(k\)) and temperature (\(T\)), thus quantifying how temperature affects reaction rates. It can be written as:\[k = Ae^{\frac{-E_a}{RT}}\]Here:

• \(k\) is the rate constant, which you can think of as a number that indicates the speed of a reaction at a given temperature.
• \(A\) is the pre-exponential factor, representing the frequency of collisions with the correct orientation.
• \(E_a\) is the activation energy, the minimum energy required for a reaction to occur.
• \(R\) is the universal gas constant, 8.314 J/(mol·K).
• \(T\) is the absolute temperature in Kelvin.

Understanding the Arrhenius equation helps in predicting how changing temperature influences the speed of reactions. Lower activation energy suggests a reaction proceeds more easily.

linear regression in chemistry

Linear regression is a statistical method used to understand the relationship between two variables. In chemistry, it's particularly useful when linearizing complex equations like the Arrhenius equation. By taking the natural logarithm of the Arrhenius equation, we rearrange it to resemble a linear equation:\[ln(k) = ln(A) - \frac{E_a}{R}\cdot\frac{1}{T}\]This transformation allows us to plot \(ln(k)\) against \(1/T\) to obtain a straight line, where:

• \(ln(k)\) is comparable to \(y\) in the line equation \(y = mx + b\).
• \(-\frac{E_a}{R}\) corresponds to the slope \(m\).

Using regression analysis, students can determine the slope of the line, which can be used to calculate the activation energy. This procedure makes abstract concepts tangible, simplifying the determination of kinetic parameters.

first-order rate constant

A first-order reaction describes a reaction where the rate depends only on the concentration of one reactant. For such reactions, the rate constant \(k\) remains constant with time but varies with temperature as seen from the Arrhenius equation. For example, in the given exercise, the rate constant changes significantly as temperature increases:

• At 300 K: \(3.2 \times 10^{-11}\) s\(^{-1}\)
• At 355 K: \(2.4 \times 10^{-7}\) s\(^{-1}\)

The unit of \(s^{-1}\) signifies a rate constant characteristic of first-order reactions. These units indicate that the rate depends on a single concentration term. This characteristic helps in predicting the time it takes for a reaction to reach completion or a certain percentage of progress.

temperature dependence of reaction rates

Temperature profoundly affects how quickly chemical reactions proceed. Generally, higher temperatures result in faster reactions due to increased molecular movements. In context, the Arrhenius equation mathematically illustrates this dependency:\[k = Ae^{\frac{-E_a}{RT}}\]The equation signifies that as temperature \(T\) rises, the exponential term becomes less negative, indicating an increased rate constant \(k\). For instance, in our exercise:- Increasing temperature from 300 K to 355 K produces a marked increase in the rate constant from \(3.2 \times 10^{-11}\) to \(2.4 \times 10^{-7}\).This change suggests molecules attain the needed activation energy more frequently. Thus, a comprehensive understanding of temperature dependence is essential, enabling chemists to control and optimize reaction conditions effectively.