Problem 91 - Chemistry: An Atoms-Focused Approach

Question

The kinetics of the reaction between chlorine dioxide and ozone are relevant to the study of atmospheric ozone destruction. The value of the rate constant for the reaction between chlorine dioxide and ozone was measured at four temperatures between 193 and \(208 \mathrm{K}\). The results were as follows: $$\begin{array}{cc}T(\mathrm{K}) & k\left(M^{-1} \mathrm{s}^{-1}\right) \\\193 & 34.0 \\\\\hline 198 & 62.8 \\\\\hline 203 & 112.8 \\\\\hline 208 & 196.7 \\\\\hline\end{array}$$ a. Calculate the values of the activation energy and the frequency factor for the reaction. b. What is the value of the rate constant higher in the stratosphere where \(T=245 \mathrm{K} ?\)

Step-by-Step Solution

Answer

Question: Determine the activation energy and frequency factor for the reaction between chlorine dioxide and ozone based on the given data points. Also, calculate the rate constant at 245 K. Answer: The activation energy (Ea) for the reaction is 47.84 kJ/mol, and the frequency factor (A) is 4.91 x 10^8 M^-1s^-1. The rate constant at 245 K is 1301.5 M^-1s^-1.

Step 1: Understand the Arrhenius Equation

The Arrhenius equation is given by: $$k = A\exp\left(-\frac{E_a}{RT}\right),$$ where \(k\) is the rate constant, \(A\) is the frequency factor, \(E_a\) is the activation energy, \(R\) is the gas constant (8.314 J/mol·K), and \(T\) is the temperature in Kelvin.

Step 2: Linearize the Arrhenius Equation

We can make the Arrhenius equation linear by taking the natural logarithm of both sides: $$\ln k = \ln A - \frac{E_a}{RT},$$ which is in the form of a linear equation y = mx + b, where y = \(\ln k\), x = \(\frac{1}{T}\), m = \(-\frac{E_a}{R}\), and b = \(\ln A\).

Step 3: Create a Table of Data Points

From the exercise, we have the following data points: | \(T(K)\) | \(k (M^{-1}s^{-1})\) | |--------|---------------------| | 193 | 34.0 | | 198 | 62.8 | | 203 | 112.8 | | 208 | 196.7 | For each data point, calculate the corresponding values of \(T^{-1}\) and \(\ln k\): | \(T(K)\) | \(T^{-1} (\frac{1}{K})\) | \(k (M^{-1}s^{-1})\) | \(\ln k\) | |--------|-------------------------|---------------------|---------| | 193 | 0.005181 | 34.0 | 3.526361| | 198 | 0.005051 | 62.8 | 4.138363| | 203 | 0.004926 | 112.8 | 4.724276| | 208 | 0.004808 | 196.7 | 5.280421|

Step 4: Perform a Linear Regression

Using the linear regression method, calculate the slope (m), and the y-intercept (b) of the linear equation. There are many methods to perform linear regression, such as using Excel, online tools, or by hand. m = -5754.45 b = 19.9920

Step 5: Calculate Activation Energy and Frequency Factor

Recall that m= \(-\frac{E_a}{R}\) and b = \(\ln A\). Therefore, $$E_a = -mR = (-5754.45) * (8.314 \;\mathrm{J/mol \cdot K}) = 47843.90 \;\mathrm{J/mol}$$ $$A = \exp{b} = \exp{(19.9920)} = 4.91\times10^8 M^{-1}s^{-1}$$ So, \(E_a = 47.84\;\mathrm{kJ/mol}\) and \(A = 4.91\times10^8 M^{-1}s^{-1}\).

Step 6: Calculate the Rate Constant at 245 K

Using the calculated activation energy and frequency factor, we can find the rate constant for the reaction at 245 K using the Arrhenius equation: $$k = A\exp\left(-\frac{E_a}{RT}\right)$$ $$k = (4.91\times10^8 M^{-1}*)&#\text{s}^{-1}\exp\left(-\frac{47843.90\;\mathrm{J/mol}}{(8.314\;\mathrm{J/mol\cdot K})*(245\;\mathrm{K})}\right)$$ $$k = 1301.5\;\mathrm{M^{-1}s^{-1}}$$ So, at the temperature \(T = 245\;\mathrm{K}\), the rate constant is \(1301.5\;\mathrm{M^{-1}s^{-1}}\).

Key Concepts

Arrhenius equation

The Arrhenius equation provides significant insight into reaction kinetics. It's like a bridge between chemical reactions and temperature changes. The equation is expressed as:\[ k = A\exp\left(-\frac{E_a}{RT}\right) \]Let's break this down:

• \( k \) is the rate constant. It tells us how fast the reaction occurs.
• \( A \) is the frequency factor, which we'll explore later.
• \( E_a \) is activation energy, representing the energy needed to start a reaction.
• \( R \) is the gas constant, valued at 8.314 J/mol·K.
• \( T \) is the temperature in Kelvin.

The Arrhenius equation shows that as temperature \( T \) increases, the exponential term becomes less negative, resulting in a larger \( k \). This means the reaction rate increases with temperature. Understanding this relationship helps in predicting how different conditions can influence a chemical reaction.

Activation energy

Activation energy \( E_a \) is the minimum energy required for reactants to transform into products. Think of it as a hurdle that reactants need to overcome for a reaction to proceed. When \( E_a \) is high, fewer molecules successfully overcome this energy barrier, resulting in a slower reaction. On the other hand, lower \( E_a \) values imply a faster reaction because more molecules have the necessary energy to proceed.In the context of our kinetic exercise, the value was calculated as:\[E_a = -mR = (-5754.45) \times (8.314 \; \mathrm{J/mol \cdot K}) = 47843.90 \; \mathrm{J/mol}\]This was then converted to kJ/mol, resulting in approximately \( 47.84 \; \mathrm{kJ/mol} \). This tells us the amount of energy required to initiate the reaction between chlorine dioxide and ozone. Knowing \( E_a \) helps when considering ways to modify conditions to speed up a reaction, such as adding a catalyst which would lower \( E_a \).

Rate constant

The rate constant \( k \) is a vital part of the Arrhenius equation. It reflects the speed of a chemical reaction at a given temperature. The value of \( k \) is influenced by several factors, including the intrinsic nature of the reaction, temperature, and the presence of a catalyst.From our activity:

• Higher temperatures resulted in increased \( k \) values.
• At 193 K, \( k = 34.0 \; M^{-1}\mathrm{s}^{-1} \)
• At 208 K, \( k = 196.7 \; M^{-1}\mathrm{s}^{-1} \)

This illustrates that the reaction speeds up as temperature rises. For reactions occurring in the stratosphere at 245 K, we calculated:\[ k = 1301.5 \; \mathrm{M^{-1}s^{-1}} \]This marked increase reinforces the temperature dependence shown within the Arrhenius framework. Rate constants are essential for calculating reaction rates and designing chemical processes.

Frequency factor

The frequency factor \( A \) represents the frequency of collisions between reactant molecules that result in a reaction. It is a component of the Arrhenius equation and influences the likelihood of such collisions being successful; essentially, it's about how often molecules collide in a way that leads to a reaction. The more effective these collisions, the higher the reaction rate.In our example, \( A \) was calculated to be:\[A = \exp{(19.9920)} = 4.91\times10^8 \; M^{-1}s^{-1}\]This value suggests how conducive the physical state and orientations of reactants are for reaction. Frequency factors can vary widely between reactions, but having this figure provides insights into the mechanism and efficiency of the molecular interactions happening in the reaction.Knowing \( A \) alongside \( E_a \) gives a full picture of the reaction profile, helping understand both the speed and energetics involved in chemical transformations.