Question

The conversion of \(\mathrm{NO}_{2}(g)\) to \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) can occur through the following reaction: $$\mathrm{NO}_{2}(g) \rightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ The activation energy for this reaction is \(111 \mathrm{kJ} \mathrm{mol}^{-1}\) and the pre-exponential factor is \(2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1}\). Assume that these quantities are temperature independent. a. What is the rate constant for this reaction at \(298 \mathrm{K} ?\) b. What is the rate constant for this reaction at the tropopause where \(\mathrm{T}=225 \mathrm{K}\).

Short answer

a. At 298 K, the rate constant \(k\) is approximately \(9.91 \times 10^{-12} \mathrm{M}^{-1} \mathrm{s}^{-1}\). b. At 225 K, the rate constant \(k\) is approximately \(2.57 \times 10^{-14} \mathrm{M}^{-1} \mathrm{s}^{-1}\).

Step-by-step solution

Calculate k at 298K

We are given the activation energy (Ea = 111 kJ mol⁻¹) and the pre-exponential factor (A = 2.0 x 10⁻⁹ M⁻¹ s⁻¹). At 298 K, we can calculate the rate constant (k) using the Arrhenius equation: \(k = Ae^{-\frac{E_{a}}{RT}}\) where R = 8.314 J mol⁻¹ K⁻¹ and T = 298 K. To use compatible units, we need Ea in J mol⁻¹, so Ea = 111,000 J mol⁻¹. $$k = 2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1} e^{-\frac{111,000 \mathrm{J} \mathrm{mol}^{-1}}{(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1})(298 \mathrm{K})}}$$ Calculate the exponent and then solve for k.

Calculate k at 225K

Now we follow the same steps as before but with T = 225 K. \(k = Ae^{-\frac{E_{a}}{RT}}\) $$k = 2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1} e^{-\frac{111,000 \mathrm{J} \mathrm{mol}^{-1}}{(8.314 \mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1})(225 \mathrm{K})}}$$ Calculate the exponent and then solve for k.

Key concepts

Activation Energy

Activation energy is a term that defines the minimum amount of energy necessary for a chemical reaction to occur. Think of this as a barrier that reactants need to overcome to transform into products. The concept of activation energy is crucial for understanding why some reactions proceed at a fast rate while others are much slower.

For example, striking a match initiates a reaction. The energy you apply via friction is enough to overcome the activation energy barrier which allows the chemicals in the match head to react and ignite. In the case of our textbook exercise, the activation energy for the conversion of \(\mathrm{NO}_{2}(g)\) to \(\mathrm{NO}(g)\) and \(\mathrm{O}_{2}(g)\) is stated to be 111 kJ/mol. This means that for this specific reaction to proceed, a significant amount of energy is needed, which impacts the rate of reaction at different temperatures.

Rate Constant

The rate constant, represented by the letter \(k\), is an essential factor in chemical kinetics, which provides the speed at which a chemical reaction proceeds. It's unique for each reaction and is influenced by conditions such as temperature and the presence of a catalyst. A higher rate constant indicates a faster reaction.

The Arrhenius equation allows us to calculate the rate constant based on the known variables such as the activation energy and the temperature. In our exercise, understanding the rate constant at different temperatures, such as \(298 K\) and \(225 K\), highlights how even a slight change in temperature can greatly affect how fast the reaction takes place. The exercise shows that with the provided activation energy and pre-exponential factor, we can solve for the rate constant at any given temperature.

Pre-exponential Factor

The pre-exponential factor, often denoted as \(A\), appears in the Arrhenius equation and is sometimes referred to as the frequency factor. It relates to the frequency of collisions and the orientation of reactant molecules that result in successful reactions. It's important because it reflects how often particles collide in the right way to react.

In simpler terms, if we think about a room full of people moving randomly—some collisions result in a handshake (successful reaction), others do not. The pre-exponential factor could represent how many handshakes happen in a set amount of time if everyone had perfect aim. In the exercise at hand, the value given for \(A\) is \(2.0 \times 10^{-9} \mathrm{M}^{-1} \mathrm{s}^{-1}\), which we use along with the activation energy to determine the rate constant for the reaction.

Chemical Kinetics

Chemical kinetics is the study of the speed or rate at which chemical reactions occur. It's a crucial field in chemistry because it not only helps us understand reaction rates but also the mechanisms behind them—how reactants become products over time. Factors like temperature, concentration, surface area, and presence of catalysts can influence these rates.

For students, the study of kinetics can often seem daunting, but it is vital for predicting how long reactions will take and under what conditions they might occur more readily. The textbook exercise involving the decomposition of \(\mathrm{NO}_{2}(g)\) serves as a practical example of applying chemical kinetics principles to understand and calculate specific aspects of a reaction such as the rate constant at different temperatures.