Question

Peroxyacetyl nitrate (PAN) undergoes thermal decomposition as follows: $$ \mathrm{CH}_{3}(\mathrm{CO}) \mathrm{OONO}_{2} \longrightarrow \mathrm{CH}_{3}(\mathrm{CO}) \mathrm{OO}+\mathrm{NO}_{2} $$ The rate constant is \(3.0 \times 10^{-4} \mathrm{~s}^{-1}\) at \(25^{\circ} \mathrm{C}\). At the boundary between the troposphere and stratosphere, where the temperature is about \(-40^{\circ} \mathrm{C},\) the rate constant is reduced to \(2.6 \times 10^{-7} \mathrm{~s}^{-1}\). (a) Calculate the activation energy for the decomposition of PAN. (b) What is the half-life of the reaction (in min) at \(25^{\circ} \mathrm{C} ?\)

Short answer

(a) The activation energy is approximately 120.6 kJ/mol. (b) The half-life at 25°C is about 38.5 minutes.

Step-by-step solution

Understand the Arrhenius Equation

The Arrhenius equation relates the rate constant \(k\) of a reaction to the temperature \(T\): \(k = A e^{-E_a/(RT)}\), where \(A\) is the pre-exponential factor, \(E_a\) is the activation energy, and \(R\) is the universal gas constant (\(8.314 \text{ J/mol·K}\)). In this problem, we want to calculate \(E_a\) using the given rate constants at two different temperatures.

Rearrange the Arrhenius Equation

Taking the natural logarithm of both sides of the Arrhenius equation, we get \(\ln k = \ln A - \frac{E_a}{RT}\). For two different rate constants \(k_1\) and \(k_2\) at temperatures \(T_1\) and \(T_2\) respectively, the equation becomes: \[\ln\left(\frac{k_2}{k_1}\right) = \frac{-E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)\].

Convert Temperatures to Kelvin

Given temperatures are in degrees Celsius: \(25^{\circ} \mathrm{C}\) and \(-40^{\circ} \mathrm{C}\). Convert these to Kelvin: - \(T_1 = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\)- \(T_2 = -40^{\circ} \mathrm{C} + 273.15 = 233.15 \mathrm{K}\).

Solve for Activation Energy \(E_a\)

Substitute \(k_1 = 3.0 \times 10^{-4} \mathrm{s}^{-1}\), \(k_2 = 2.6 \times 10^{-7} \mathrm{s}^{-1}\), and the temperatures into the rearranged Arrhenius equation:\[\ln\left(\frac{2.6 \times 10^{-7}}{3.0 \times 10^{-4}}\right) = \frac{-E_a}{8.314} \left(\frac{1}{233.15} - \frac{1}{298.15}\right)\].Calculate \(E_a\): - \(\ln\left(0.000003 / 0.0003\right) = -E_a/8.314 \times (0.00429 - 0.00336)\)- Solving yields \(E_a \approx 120.6 \text{ kJ/mol}\).

Use Half-Life Formula for First-Order Reactions

For a first-order reaction, the half-life is given by \(t_{1/2} = \frac{0.693}{k}\).Given \(k = 3.0 \times 10^{-4} \mathrm{s}^{-1}\) at \(25^{\circ} \mathrm{C}\), the half-life is:\[t_{1/2} = \frac{0.693}{3.0 \times 10^{-4} \mathrm{s}^{-1}}\].

Convert Half-Life from Seconds to Minutes

Calculate the half-life from Step 5 and convert it to minutes:\[t_{1/2} = 2310 \text{ seconds} = \frac{2310}{60} \text{ minutes} \approx 38.5 \text{ minutes}\].

Key concepts

Activation Energy

Understanding Activation Energy is crucial for grasping how reactions occur and how fast they progress. It is the minimum energy that the reacting species must possess for a reaction to take place. Think of Activation Energy, often represented as \( E_a \), as an energy barrier that the reactants need to overcome to transform into products.

When the energy is provided to the reactants, they gain enough energy to reach this threshold and the reaction can proceed. In the case of peroxyacetyl nitrate (PAN) decomposition, we calculated \( E_a \) using known rate constants at two temperatures and the Arrhenius equation. By comparing rate constants at different temperatures, we were able to derive \( E_a \) and found it to be approximately 120.6 kJ/mol. This value tells us about the difficulty of breaking the chemical bonds in PAN during its decomposition. The higher the activation energy, the slower the reaction will be at a given temperature, assuming other factors like concentration remain constant.

Arrhenius Equation

The Arrhenius Equation provides a formula that relates the rate constant \( k \) of a chemical reaction to the temperature \( T \) and activation energy \( E_a \). It is expressed as: \[k = A e^{-E_a/(RT)}\]where \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

The equation is vital because it helps us predict how changes in temperature affect the reaction rate. By taking the natural logarithm of both sides of the Arrhenius equation, we can rearrange it in a linear form suitable for calculating \( E_a \) using experimental data from rate constants at different temperatures. For instance, with the PAN decomposition, applying the linear form of the equation allowed us to solve for \( E_a \) using two known rate constants and their corresponding temperatures.
- Helps us understand temperature dependence of reaction rates- Useful for determining activation energy by measuring rate constants at different temperatures - Shows rigor in chemical kinetics in explaining reaction mechanismsKnowing the Arrhenius equation enables chemists to control and optimize reactions by adjusting conditions under which reactions proceed.

Half-Life Calculation

The half-life of a reaction gives us the time required for half of the reactant to be consumed. For first-order reactions, where the rate is directly proportional to the concentration of one reactant, it is particularly straightforward to calculate. The formula for half-life \( t_{1/2} \) in such reactions is: \[t_{1/2} = \frac{0.693}{k}\]where \( k \) is the rate constant.

In the case of PAN at 25°C, we calculated \( t_{1/2} \) using a rate constant of \( 3.0 \times 10^{-4} \text{ s}^{-1} \). This calculation resulted in a half-life of approximately 38.5 minutes. Understanding half-life helps us gauge how long a reaction will take to proceed to a significant extent, which is invaluable in both laboratory and industrial settings.
- Simple yet powerful tool for predicting reaction progress- Helps in designing reactions and processes with safety and efficiency- Provides a clearer perspective on reaction kinetics by indicating how quickly reactants get used up over timeBy mastering how to calculate and interpret half-life, we can better manage reactions in everyday chemical processes, from medical applications to manufacturing.