Question

For the reaction $$ \mathrm{NO}+\mathrm{O}_{3} \rightarrow \mathrm{NO}_{2}+\mathrm{O}_{2} $$ the second order rate constant has a value of \(1.8 \times 10^{-14} \mathrm{molecules}^{-1} \mathrm{~cm}^{3} \mathrm{~s}^{-1}\) at \(25^{-} \mathrm{C}\). The concentration of NO in a relatively clean atmosphere is \(0.10 \mathrm{ppbv}\) and that of \(\mathrm{O}_{3}\) is \(15 \mathrm{ppbv}\). Convert these two concentrations into units of molecules \(\mathrm{cm}^{-3}\). Calculate the rate of the NO oxidation using units of molecules \(\mathrm{cm}^{-3} \mathrm{~s}^{-1}\). Show how the rate law may be expressed in pseudo first order terms and calculate the corresponding pseudo first order rate constant.

Short answer

The rate of NO oxidation is \(1.63 \times 10^7 \text{ molecules cm}^{-3} \text{ s}^{-1}\), and the pseudo first order rate constant is \(6.64 \times 10^{-3} \text{ s}^{-1}\).

Step-by-step solution

Convert ppbv to molecules per cm³

In atmospheric chemistry, 1 ppbv (parts per billion by volume) is equivalent to \(2.46 \times 10^{10}\) molecules per \(\text{cm}^{3}\). For NO, \( 0.10 \text{ ppbv} \times 2.46 \times 10^{10}\) gives \(2.46 \times 10^9 \text{ molecules cm}^{-3}\). For \( \text{O}_3\), \( 15 \text{ ppbv} \times 2.46 \times 10^{10}\) results in \(3.69 \times 10^{11} \text{ molecules cm}^{-3}\).

Calculate the rate of NO oxidation

The rate law for the reaction is: \[ \text{rate} = k[\text{NO}][\text{O}_3] \]. Plugging in the values: \(k = 1.8 \times 10^{-14} \text{ molecules}^{-1} \text{ cm}^{3} \text{ s}^{-1}\), \([\text{NO}] = 2.46 \times 10^9 \text{ molecules cm}^{-3}\), and \([\text{O}_3] = 3.69 \times 10^{11} \text{ molecules cm}^{-3}\), we calculate the rate: \[ \text{rate} = 1.8 \times 10^{-14} \times 2.46 \times 10^9 \times 3.69 \times 10^{11} \]. \(\text{Rate} = 1.63 \times 10^7 \text{ molecules cm}^{-3} \text{ s}^{-1}\).

Express the rate law in pseudo first order terms

Pseudo first order conditions apply when the concentration of one reactant is in vast excess and can be assumed constant. Here, we assume \([\text{O}_3]\) is in excess. The rate law becomes: \[ \text{rate} = k'[\text{NO}] \], where \(k' = k[\text{O}_3]\). Given \([\text{O}_3] = 3.69 \times 10^{11} \text{ molecules cm}^{-3}\), \(k' = 1.8 \times 10^{-14} \times 3.69 \times 10^{11} = 6.64 \times 10^{-3} \text{ s}^{-1}\).

Calculate the pseudo first order rate constant

The pseudo first order rate constant, \(k'\), was found in the previous step: \(k' = 6.64 \times 10^{-3} \text{ s}^{-1}\). This is the effective rate constant assuming \([\text{O}_3]\) remains constant over the reaction process.

Key concepts

Reaction Rate Calculation

Understanding how to calculate the reaction rate is fundamental in atmospheric chemistry. The reaction rate tells us how quickly a reactant is transformed into a product in a given time and volume. For a reaction involving gases, this is often expressed in terms of the change in concentration of a substance per unit time.

For example, consider our reaction:\[\text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2.\]
The rate at which nitric oxide (NO) oxidizes to nitrogen dioxide (NO₂) depends on two main factors:

• The rate constant \(k\), which is specific to the reaction and conditions like temperature.
• The concentrations of the reactants \(\text{NO}\) and \(\text{O}_3\).In this case, the rate constant \(k\) is given as \(1.8 \times 10^{-14} \text{ molecules}^{-1} \text{ cm}^{3} \text{ s}^{-1}\).

Using the rate law \(\text{rate} = k[\text{NO}][\text{O}_3]\), we substitute the concentrations measured in molecules cm³ to compute the rate. Calculating reaction rates requires careful attention to units, as atmospheric concentrations often need conversion from typical measurements like parts per billion by volume (ppbv) into molecules per cubic centimeter to use in the rate equation.

Second Order Reaction

A second order reaction depends on the concentration of two reactants. The rate of the reaction is proportional to the product of these concentrations, meaning that if either concentration doubles, the rate will also double.

In the given reaction, \(\text{NO} + \text{O}_3 \rightarrow \text{NO}_2 + \text{O}_2\), the reaction is second order because it involves two molecules reacting—one molecule of nitric oxide (NO) with one molecule of ozone (O₃). The mathematical expression for such a reaction rate is:\[\text{rate} = k[\text{NO}][\text{O}_3].\]
Here, both reactant concentrations impact the rate of change equally. Understanding these dependencies helps in predicting how changes in atmospheric concentrations will affect the rate of photochemical smog formation, for example. In environmental systems, reactions are often second order due to the collision required between distinct reactant molecules.

Pseudo First Order Kinetics

Pseudo first order kinetics is an important simplification tool in chemical kinetics. It applies when one reactant is present in such vast excess that its concentration effectively remains constant during the reaction. The reaction then behaves as if it were first order with respect to the other reactant.

For the given oxidation of NO by ozone, if \([\text{O}_3]\) is significantly larger than \([\text{NO}]\), it can be treated as constant. This simplifies the rate law:\[\text{rate} = k'[\text{NO}],\]
where \(k' = k[\text{O}_3].\)
This makes the complex second order reaction look like a simpler first order one, where only the concentration of NO is variable. Such simplification allows easier analysis and understanding of the reaction kinetics over vast atmospheric regions.

Unit Conversion in Chemistry

Unit conversion plays a crucial role in chemistry, especially when dealing with concentrations that influence reaction rates. Atmospheric chemists often need to convert standard concentration units into those directly usable in rate equations.

For example, concentrations of trace gases in the atmosphere are often given in parts per billion by volume (ppbv). To perform calculations involving reaction rates, these must be converted into molecules per cubic centimeter (molecules cm⁻³). This conversion factor in atmospheric conditions is roughly \(2.46 \times 10^{10}\) molecules per cm³ per ppbv.

• Therefore, for NO with a concentration of 0.10 ppbv, the conversion is:

\(0.10 \text{ ppbv} \times 2.46 \times 10^{10} = 2.46 \times 10^9 \text{ molecules cm}^{-3}.\)
Similarly, for O₃ with a concentration of 15 ppbv, it becomes:\(15 \text{ ppbv} \times 2.46 \times 10^{10} = 3.69 \times 10^{11} \text{ molecules cm}^{-3}.\)
By efficiently converting units, chemists can apply these values directly into reaction rate calculations, ensuring accuracy in their results.