Question

\(\begin{aligned} \text { Consider the reaction. } \\ & 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned}\) \begin{equation} \begin{array}{l}{\text { a. Express the rate of the reaction in terms of the change in }} \\ {\text { concentration of each of the reactants and products. }} \\ {\text { b. In the first } 15.0 \text { s of the reaction, 0.015 mol of } \mathrm{O}_{2} \text { is produced in a }} \\ {\text { reaction vessel with a volume of } 0.500 \text { L. What is the average rate }} \\ {\text { of the reaction during this time interval? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Predict the rate of change in the concentration of } \mathrm{N}_{2} \mathrm{O} \text { during this }} \\ {\text { time interval. In other words, what is } \Delta\left[\mathrm{N}_{2} \mathrm{O}\right] / \Delta t?}\end{array} \end{equation}

Short answer

a. The rate of the reaction expressed in terms of of each substance is \rate = -\frac{1}{2} \frac{\Delta [ \mathrm{N}_2 \mathrm{O} ]}{\Delta t} = \frac{\Delta [\mathrm{N}_2]}{\Delta t} = \frac{1}{2}\frac{\Delta [\mathrm{O}_2]}{\Delta t}. b. The average rate of reaction for O2 is 0.0020 \text{ mol/L/s}. c. The rate of change in the concentration of \text{N}_2\text{O during this time interval is \text{0.0040 mol/L/s}.

Step-by-step solution

Express the rate of the reaction (Part a)

The rate of the reaction can be expressed in terms of the disappearance of N2O and the appearance of N2 and O2. It is important to note that the stoichiometric coefficients should be taken into account for the reaction. The rate of the reaction (\text{rate}) can thus be expressed as: \( \text{rate} = -\frac{1}{2} \frac{\Delta [ \mathrm{N}_2 \mathrm{O} ]}{\Delta t} = \frac{\Delta [\mathrm{N}_2]}{\Delta t} = \frac{1}{2}\frac{\Delta [\mathrm{O}_2]}{\Delta t} \) where \(\Delta [\ldots]\) represents the change in concentration and \(\Delta t\) represents the change in time. The minus sign indicates the concentration of N2O is decreasing.

Calculate the average rate of the reaction (Part b)

In the first 15.0 s of the reaction, 0.015 mol of O2 is produced in a vessel with a volume of 0.500 L. To find the average rate of production of O2, we need to calculate the change in concentration over the time interval: \( \text{Average rate} = \frac{\Delta [\mathrm{O}_2]}{\Delta t} = \frac{0.015 \text{ mol} / 0.500 \text{ L}}{15.0 \text{ s}} = \frac{0.030 \text{ mol/L}}{15.0 \text{ s}} = 0.0020 \text{ mol/L/s} \) Since the reaction produces O2 at half the rate that N2O is consumed, the average rate at which N2O is consumed is twice that of O2 production rate. Thus, the average rate of the reaction in terms of the disappearance of N2O is \(0.0040 \text{ mol/L/s}\).

Predict the rate of change in N2O concentration (Part c)

Using the relationship established in part a, we can now predict the rate of change in concentration of N2O. Since the appearance of O2 happens at half the rate of the disappearance of N2O, we can apply this ratio to the calculated average rate of O2 to find the rate of N2O. \ ( \Delta[\mathrm{N}_2\mathrm{O}] / \Delta t = 2 \times \text{Average rate of O}_2 = 2 \times 0.0020 \text{ mol/L/s} = 0.0040 \text{ mol/L/s} \) Therefore, the rate of change in concentration of N2O is \(0.0040 \text{ mol/L/s}).

Key concepts

Reaction Rate Calculation

Understanding the calculation of a chemical reaction rate is fundamental in chemistry. It involves measuring how quickly reactants are converted into products. In the given exercise, we consider the reaction of dinitrogen monoxide (N₂O) turning into nitrogen gas (N₂) and oxygen gas (O₂). To calculate the reaction rate, you need to consider the changes in concentrations of reactants or products over time.

The formula provided in the solution,
\( \text{rate} = -\frac{1}{2} \frac{\Delta [\mathrm{N}_2\mathrm{O}]}{\Delta t} = \frac{\Delta [\mathrm{N}_2]}{\Delta t} = \frac{1}{2}\frac{\Delta [\mathrm{O}_2]}{\Delta t} \),
is a clear representation of how the stoichiometry of the reactants and products affects the calculation of the reaction rate. The term \( \Delta t \) refers to the change in time, and \( \Delta [\ldots] \) indicates the change in concentration. Negative signs denote the consumption of reactants, and positive signs indicate the formation of products.

We can also refine our understanding of the rates at which reactants are used and products are formed by considering the moles of substance per unit volume per unit time, typically expressed as mol/L/s. A concrete understanding of these principles is crucial to mastering reaction rate calculations.

Stoichiometry in Reaction Rates

Stoichiometry plays an essential role in reaction rate calculations, as it allows us to relate the quantities of reactants and products in a balanced chemical equation. In the step-by-step solution given for the exercise, stoichiometry is used to determine how the rate at which one reactant is used up or a product is produced compares to others involved in the reaction.

For example, in our current equation, every two moles of N₂O gas yield two moles of N₂ gas and one mole of O₂ gas. This stoichiometric ratio can be used for calculating how a change in the concentration of one substance affects the others. Since the production of O₂ occurs at a stoichiometric rate that is half of the consumption rate of N₂O (due to the coefficients in the balanced equation), the change in concentration of O₂ will occur at half the rate of the change in concentration of N₂O.

It's crucial for students to remember to incorporate these stoichiometric relationships when calculating how the concentration of one species will change relative to another over time, leading to a better understanding of the dynamics of a chemical reaction.

Average Rate of Reaction

The average rate of reaction provides information about the overall speed of the chemical reaction over a specific timeframe. It's defined as the change in concentration of reactants or products divided by the change in time. In the exercise, to determine the average rate, we assess the quantity of oxygen gas produced over 15 seconds and divide by the reaction volume and time.

The calculation, \(\text{Average rate} = \frac{0.030 \text{ mol/L}}{15.0 \text{ s}} = 0.0020 \text{ mol/L/s}\),
shows the speed at which O₂ appears in the reaction vessel. However, to comprehend the reaction as a whole, it's important to also ascertain the average rate at which N₂O is consumed, which in this case is twice the rate of O₂ formation due to stoichiometry, or \(0.0040 \text{ mol/L/s}\).

Consistently following this logic allows students to accurately predict the rates at which various species in a reaction are consumed or produced. When dealing with homework problems, such a structured approach in rate calculations can simplify complex scenarios into understandable concepts, ultimately aiding in grasping the overall kinetics of the reaction.