Question

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \longrightarrow \mathrm{B}(a q)\). A flask is charged with \(0.065 \mathrm{~mol}\) of \(\mathrm{A}\) in a total volume of \(100.0 \mathrm{~mL}\). The following data are collected: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 10 & 20 & 30 & 40 \\ \hline \text { Moles of A } & 0.065 & 0.051 & 0.042 & 0.036 & 0.031 \\ \hline \end{array} $$ (a) Calculate the number of moles of \(\mathrm{B}\) at each time in the table, assuming that there are no molecules of \(\mathrm{B}\) at time zero, and that \(A\) cleanly converts to \(B\) with no intermediates. (b) Calculate the average rate of disappearance of \(\mathrm{A}\) for each 10 -min interval in units of \(M / \mathrm{s}\). (c) Between \(t=10 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(M / s\) ? Assume that the volume of the solution is constant.

Short answer

(a) Moles of B at each time: 0 min: 0, 10 min: 0.014, 20 min: 0.023, 30 min: 0.029, 40 min: 0.034. (b) Average rate of disappearance of A for each 10-min interval: 0-10 min: 2.33 × 10⁻⁴ M/s; 10-20 min: 1.50 × 10⁻⁴ M/s; 20-30 min: 1.00 × 10⁻⁴ M/s; 30-40 min: 8.33 × 10⁻⁵ M/s. (c) The average rate of appearance of B between t=10 min and t=30 min is 1.25 × 10⁻⁴ M/s.

Step-by-step solution

(a) Calculating the number of moles of B at each time

To find the number of moles of B at each time, we can use the stoichiometry of the reaction and the initial and subsequent number of moles of A at each time point: Moles of B = Initial moles of A - Moles of A at time t Now, we can calculate the moles of B for each time point in the table: Time (min) | Moles of A | Moles of B --- | --- | --- 0 | 0.065 | 0 10 | 0.051 | 0.014 20 | 0.042 | 0.023 30 | 0.036 | 0.029 40 | 0.031 | 0.034

(b) Calculating the average rate of disappearance of A for each 10-min interval

To find the average rate of disappearance of A for each 10-min interval, we can compute the change in concentration of A over each 10-min interval and divide by the length of the interval in seconds: Average rate of disappearance of A = (Δ[A] / Δt) First, let's calculate the concentration of A at each time point: Time (min) | Moles of A | [A] (M) --- | --- | --- 0 | 0.065 | 0.650 10 | 0.051 | 0.510 20 | 0.042 | 0.420 30 | 0.036 | 0.360 40 | 0.031 | 0.310 Now, compute the average rate of disappearance of A for each 10-min interval: Interval | Δ[A] (M) | Δt (s) | Rate (M/s) --- | --- | --- | --- 0-10 | 0.140 | 600 | 2.33 × 10⁻⁴ 10-20 | 0.090 | 600 | 1.50 × 10⁻⁴ 20-30 | 0.060 | 600 | 1.00 × 10⁻⁴ 30-40 | 0.050 | 600 | 8.33 × 10⁻⁵

(c) Calculating the average rate of appearance of B between t=10 min and t=30 min

To find the average rate of appearance of B between t=10 min and t=30 min, we can compute the change in concentration of B over this time interval and divide by the length of the interval in seconds: Average rate of appearance of B = (Δ[B] / Δt) First, let's calculate the concentration of B at 10 min and 30 min: Time (min) | Moles of B | [B] (M) --- | --- | --- 10 | 0.014 | 0.140 30 | 0.029 | 0.290 Now, compute the average rate of appearance of B between t=10 min and t=30 min: Interval | Δ[B] (M) | Δt (s) | Rate (M/s) --- | --- | --- | --- 10-30 | 0.150 | 1200 | 1.25 × 10⁻⁴ Thus, the average rate of appearance of B between t=10 min and t=30 min is 1.25 × 10⁻⁴ M/s.

Key concepts

Rate of Reaction

Understanding the rate of a chemical reaction is crucial for anyone studying chemistry. The rate of reaction, simply put, is the speed at which reactants transform into products. It's usually expressed in terms of concentration change per unit time, such as moles per liter per second (M/s). When we look at a reaction such as \(\mathrm{A}(aq) \longrightarrow \mathrm{B}(aq)\), measuring how fast \(\mathrm{A}\) disappears or how quickly \(\mathrm{B}\) appears gives us the reaction rate.

For example, in our exercise, the average rate of disappearance of \(\mathrm{A}\) over each ten-minute interval was calculated by taking the difference in concentration of \(\mathrm{A}\) and dividing by the time interval in seconds. It's important to translate minutes into seconds to maintain proper units for the rate, which should be in M/s. Hence, shorter time intervals can provide finer resolution in rate measurements, which is useful when studying more complex kinetics.

Improvement Advice for Exercises:

• Practical application examples help illustrate how reaction rates are measured and analyzed in real-world contexts like pharmaceuticals and environmental processes.
• Use of interactive simulations could greatly enhance the understanding of how changes in conditions impact the rate of reaction.
• Contrasting the rates of different types of reactions (e.g., instantaneous versus slow reactions) can provide a broader perspective on the concept.

Stoichiometry

Stoichiometry plays a pivotal role in our reaction example. It's the part of chemistry that relates the quantities of reactants and products in a chemical reaction. It is based on the conservation of mass where the total mass of the reactants equals the total mass of the products. In a balanced chemical equation, stoichiometric coefficients tell us the exact proportion of molecules needed for the reaction to occur without any excess of reactants or shortage of products.

In the given reaction, the stoichiometry is simple: one mole of \(\mathrm{A}\) yields one mole of \(\mathrm{B}\). This one-to-one ratio allows us to easily compute the moles of \(\mathrm{B}\) formed at any given time by simply subtracting the moles of \(\mathrm{A}\) remaining from the initial moles of \(\mathrm{A}\). It's essential to grasp this concept, because knowing how to use stoichiometry can predict the amount of product formed or the amount of reactants required for a reaction.

Improvement Advice for Exercises:

• Examples involving real-world chemical production can illustrate the importance of stoichiometry in industry.
• Exercise variations with limiting reactants and excess conditions can deepen comprehension.
• Inclusion of stoichiometry problems involving different states of matter (e.g., gases, using the ideal gas law) can expand practical understanding.

Concentration of Reactants

Concentration of reactants is a central factor in determining the rate of chemical reactions. In the context of our example with \(\mathrm{A}(aq) \longrightarrow \mathrm{B}(aq)\), the concentration is defined as the amount of a substance (in moles) present in a unit volume of solution (here, expressed in molarity, M). Concentration changes as the reaction progresses; reactants are consumed and products are generated.

Calculating the concentration involves dividing the moles of \(\mathrm{A}\) or \(\mathrm{B}\) by the volume of the solution, which remains constant in this particular exercise. These concentrations allow us to find the rates of disappearance and appearance of \(\mathrm{A}\) and \(\mathrm{B}\), respectively, which are central to studying chemical kinetics. It's a dynamic parameter that often drives the reactants over the energy barrier to the products, following the collision theory.

Improvement Advice for Exercises:

• Introducing scenarios where concentration changes impact the reaction rate can show the importance of this parameter.
• Students can benefit from practicing how concentration affects equilibrium and reaction rates under various conditions.
• Exercises with concentration-time graphs could help visualize and better understand the relation between concentration and reaction progress.