Question

The reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) is second order in \(\mathrm{NO}\) and first order in \(\mathrm{O}_{2} .\) When \([\mathrm{NO}]=0.040 \mathrm{M}\) and \(\left[\mathrm{O}_{2}\right]=0.035 \mathrm{M},\) the observed rate of disappearance of \(\mathrm{NO}\) is \(9.3 \times 10^{-5} \mathrm{M} / \mathrm{s} .(\mathbf{a})\) What is the rate of disappearance of \(\mathrm{O}_{2}\) at this moment? (b) What is the value of the rate constant? (c) What are the units of the rate constant? (d) What would happen to the rate if the concentration of NO were increased by a factor of \(1.8 ?\)

Short answer

(a) The rate of disappearance of O₂ at this moment is: \[Rate_{\mathrm{O}_{2}} = \frac{1}{2} \cdot (9.3 \times 10^{-5}\, \mathrm{M/s}) = 4.65 \times 10^{-5} \mathrm{M/s}\] (b) The value of the rate constant k is: \[k = \frac{9.3 \times 10^{-5}\, \mathrm{M/s}}{(0.040\, \mathrm{M})^2(0.035\, \mathrm{M})} = 2.07 \times 10^2 \mathrm{M^{-2}s^{-1}}\] (c) The units of the rate constant k are M⁻²s⁻¹. (d) The change in the rate when the concentration of NO is increased by a factor of 1.8 is: \[Rate \, Change = \frac{(1.8)^2}{1} = 3.24\] The rate will increase by a factor of 3.24 when the concentration of NO is increased by a factor of 1.8.

Step-by-step solution

1. Write the Rate Law for the Reaction

: The rate law shows the dependence of the reaction rate on the concentrations of the reactants. Given that the reaction is second order in NO and first order in O₂, we can write the rate law as: \[Rate = k[\mathrm{NO}]^2[\mathrm{O}_{2}]\]

2. Determine the Rate of Disappearance of O₂ (a)

: We are given the rate of disappearance of NO, so we will use stoichiometry to find the rate of disappearance of O₂. From the balanced reaction, we have the stoichiometric relationship: \(2\, \mathrm{NO}(g) + \mathrm{O}_{2}(g) \longrightarrow 2\, \mathrm{NO}_{2}(g)\) Using the stoichiometry, the relationship between rates of disappearance is: \[\frac{Rate_{\mathrm{O}_{2}}}{Rate_{\mathrm{NO}}} = \frac{-\Delta[\mathrm{O}_{2}]}{-2\Delta[\mathrm{NO}]}\] Now, plug in the given rate of disappearance of NO and solve for rate of disappearance of O₂: \[Rate_{\mathrm{O}_{2}} = \frac{1}{2} \cdot (9.3 \times 10^{-5}\, \mathrm{M/s})\]

3. Calculate the Value of the Rate Constant, k (b)

: Plug in the given reactant concentrations and the calculated rate of disappearance of O₂ into the rate law, and solve for k: \[9.3 \times 10^{-5}\, \mathrm{M/s} = k(0.040\, \mathrm{M})^2(0.035\, \mathrm{M})\] Solve for k, keeping in mind to report to an appropriate number of significant figures.

4. Determine the Units of the Rate Constant, k (c)

: We have the rate law in the form: \[Rate = k[\mathrm{NO}]^2[\mathrm{O}_{2}]\] We know that the rate has units of M/s. Thus, the units of k can be found by rearranging the equation: \[k = \frac{Rate}{[\mathrm{NO}]^2[\mathrm{O}_{2}]}\] Determine the units of k using the known units for Rate and concentrations.

5. Calculate the Rate Change for an Increased Concentration of NO (d)

: We are asked what would happen to the rate if the concentration of NO were increased by a factor of 1.8. To determine this, we will analyze the rate equation with the new concentration of NO: \[New\ Rate = k[\mathrm{NO}]'^2[\mathrm{O}_{2}]\] Where \([\mathrm{NO}]' = 1.8 \times [\mathrm{NO}]\) The change in the rate can be expressed as: \[Rate \, Change = \frac{New\ Rate}{Old\ Rate}\] Plug in the initial and new concentrations of NO and solve for the Rate Change.

Key concepts

Understanding Rate Law

In chemical reactions, a rate law is a mathematical expression that shows how the rate depends on the concentration of reactants. For the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\), the rate law is given by \(\text{Rate} = k[\mathrm{NO}]^2[\mathrm{O}_{2}]\). This tells us that the rate of reaction is directly proportional to the concentration of \(\mathrm{NO}\) squared and \(\mathrm{O}_{2}\) to the first power.

The rate law helps us understand how different concentrations affect the speed of the reaction. This is crucial for predicting how quickly a reaction proceeds under various conditions, which is valuable in everything from chemical manufacturing to environmental science.

The Role of the Rate Constant

The rate constant, denoted as \(k\), is a number that provides insight into how fast or slow a reaction occurs. In the context of the rate law, it's necessary for calculating the reaction rate given specific concentrations of reactants.

• The value of \(k\) is determined experimentally. In this exercise, the value was found by substituting the known rate and concentrations into the rate law.
• The units of \(k\) are crucial as they help maintain the consistency of the mathematical expression in the rate law. For our reaction, since the overall order is three, the units of \(k\) would be \(\mathrm{M}^{-2}\mathrm{s}^{-1}\).

Understanding \(k\) is essential because it offers insights into the reaction mechanism and helps compare the reactivity of different reactions.

Order of Reaction Explained

The order of a reaction represents the power to which the concentration of a reactant is raised in the rate law. It provides an insight into how sensitive the reaction rate is to changes in concentration.

For the reaction \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\), the order is second with respect to \(\mathrm{NO}\) and first with respect to \(\mathrm{O}_{2}\). This means the rate of reaction changes as the concentration of \(\mathrm{NO}\) is squared and \(\mathrm{O}_{2}\) is taken directly.

• If the concentration of \(\mathrm{NO}\) changes by a factor, the rate changes by the square of that factor.
• This understanding helps predict how alterations in concentrations impact the overall rate.

Learning about reaction order is key for controlling and optimizing chemical reactions in various applications.

Stoichiometry and Its Importance

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. It tells us how much of each substance is consumed or produced.

In our reaction, \(2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\), the stoichiometric coefficients show that two molecules of \(\mathrm{NO}\) react with one molecule of \(\mathrm{O}_{2}\) to produce two molecules of \(\mathrm{NO}_2\). This relationship helps us use the rate of disappearance of \(\mathrm{NO}\) to find the rate of disappearance of \(\mathrm{O}_{2}\).

• By understanding stoichiometry, one can interrelate the amounts of reactants and products formed.
• This concept is important for calculating yields and scaling reactions, especially in industrial and laboratory settings.

Mastery of stoichiometry allows chemists to ensure proper proportions in chemical reactions, optimizing both efficiency and cost.