Problem 89 - Chemistry: An Atoms-Focused Approach

Question

Activation Energy of Smog-Forming Reactions The initial step in the formation of smog is the reaction between nitrogen and oxygen. At the temperatures indicated, values of the rate constant of the reaction $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{NO}(g)$$ are as follows: $$\begin{array}{cc}T(\mathrm{K}) & k\left(M^{-1 / 2} \mathrm{s}^{-1}\right) \\\2000 & 318 \\\\\hline 2100 & 782 \\\\\hline 2200 & 1770 \\\\\hline 2300 & 3733 \\\\\hline 2400 & 7396 \\\\\hline\end{array}$$ a. Calculate the activation energy of the reaction. b. Calculate the frequency factor for the reaction. c. Calculate the value of the rate constant at ambient temperature, \(T=300 \mathrm{K}\).

Step-by-Step Solution

Answer

The activation energy of the reaction is approximately 141,610 J/mol. b. What is the frequency factor for the reaction? The frequency factor for the reaction is approximately \(1.24 \times 10^{13} M^{-1/2}s^{-1}\). c. What is the rate constant of the reaction at ambient temperature (300 K)? The rate constant at ambient temperature (300 K) is approximately \(8.34 \times 10^{-17} M^{-1/2}s^{-1}\).

Step 1: Choose two data points

We can choose any two data points from the given rate constants and their corresponding temperatures. For the demonstration, let's consider the first two data points: $$T_1 = 2000 K, k_1 = 318 M^{-1/2}s^{-1}$$ $$T_2 = 2100 K, k_2 = 782 M^{-1/2}s^{-1}$$

Step 2: Use the ratio of rate constants with Arrhenius equation

We can relate the rate constants and temperatures using the Arrhenius equation. Divide both equations and solve for \(E_a\): $$\frac{k_2}{k_1} = \frac{Ae^{-\frac{E_a}{R T_2}}}{Ae^{-\frac{E_a}{RT_1}}}$$ Solve for \(E_a\): $$E_a = R \frac{T_1 T_2}{T_2 - T_1} \ln(\frac{k_2}{k_1})$$

Step 3: Calculate the activation energy

Plug in values and solve for \(E_a\): $$E_a = 8.314\frac{2000(2100)}{2100 - 2000} \ln(\frac{782}{318})$$ $$E_a \approx 141610 J \cdot mol^{-1}$$ b. Calculate the frequency factor for the reaction.

Step 1: Use one of the data points and E_a to solve for A

Now that we have the activation energy, we can use any of the data points with the Arrhenius equation to solve for the frequency factor, A. Let's use the first data point again: $$k_1 = Ae^{-\frac{E_a}{RT_1}}$$ Solve for A: $$A = k_1 e^{\frac{E_a}{RT_1}}$$

Step 2: Calculate the frequency factor

Plug in values and solve for A: $$A = 318 e^{\frac{141610}{8.314(2000)}}$$ $$A \approx 1.24 \times 10^{13} M^{-1/2}s^{-1}$$ c. Calculate the value of the rate constant at ambient temperature, T = 300 K.

Step 1: Use activation energy, frequency factor, and T to find k

Now that we have the activation energy and frequency factor, we can calculate the rate constant at ambient temperature (300 K) using the Arrhenius equation: $$k = Ae^{-\frac{E_a}{RT}}$$ Plug in values and solve for k: $$k = 1.24 \times 10^{13} e^{-\frac{141610}{8.314(300)}}$$ $$k \approx 8.34 \times 10^{-17} M^{-1/2}s^{-1}$$ So, the rate constant at ambient temperature of 300 K is \(8.34 \times 10^{-17} M^{-1/2}s^{-1}\).

Key Concepts

Frequency Factor

The frequency factor, also known as the pre-exponential factor, is an important parameter in the Arrhenius equation. It represents the number of times that reactants collide with the correct orientation and sufficient energy to result in a reaction.
This factor is denoted by the symbol \( A \) and is unique to each chemical reaction. It provides insight into the likelihood of molecules overcoming the activation barrier without any thermal energy input.

To determine the frequency factor, one can rearrange the Arrhenius equation to solve for \( A \):

• \( A = k \cdot e^{\frac{E_a}{RT}} \)

where:

• \( k \) is the rate constant.
• \( E_a \) is the activation energy.
• \( R \) is the universal gas constant.
• \( T \) is the temperature in Kelvin.

In the provided example, after calculating the activation energy, the frequency factor was found using a rate constant at a given temperature, showing a value of approximately \( 1.24 \times 10^{13} \, M^{-1/2}s^{-1} \).
This reflects the effective number of collisions that lead to a successful reaction per unit time.

Arrhenius Equation

The Arrhenius equation is the centerpiece of chemical kinetics, enabling us to relate the rate constant of a reaction to the temperature and activation energy. This equation is given by:

• \( k = A e^{-\frac{E_a}{RT}} \)

where:

• \( k \) is the rate constant.
• \( A \) is the frequency factor, as discussed earlier.
• \( E_a \) represents the activation energy required to start the reaction.
• \( R \) is the gas constant, approximately \( 8.314 \, J/mol\, K \).
• \( T \) is the absolute temperature in Kelvin.

This equation allows us to see how sensitive the reaction rate is to changes in temperature.

By plotting the natural logarithm of the rate constant, \( \ln(k) \), against the inverse of the temperature, \( 1/T \), one can obtain a straight line for many reactions. The slope of this line is linked to the activation energy. This method is often employed to find \( E_a \), offering a straightforward way to visualize kinetic data.

Rate Constant

The rate constant \( k \) is a crucial part of the Arrhenius equation and quantifies the speed of a chemical reaction at a particular temperature. It depends on both the frequency factor and the temperature-dependent exponential term involving the activation energy.

In practice, the rate constant is determined experimentally and can vary significantly with temperature. According to the Arrhenius equation, as the temperature increases, so does the rate constant, indicating a faster reaction.

Consider that the rate constant at 300 K was estimated to be \( 8.34 \times 10^{-17} \, M^{-1/2}s^{-1} \). This small value suggests that at a much lower ambient temperature, the reaction proceeds noticeably slower compared to the higher temperatures provided in the original data.
Understanding how the rate constant behaves with temperature is essential for controlling reaction conditions in practical scenarios, such as in industrial processes or when minimizing the formation of undesirable products like smog.