Problem 19 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider the following hypothetical aqueous reaction: \(\mathrm{A}(a q) \rightarrow \mathrm{B}(a q)\). 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 A for each 10 -min interval in units of \(M /\) s. (c) Between \(t=0 \mathrm{~min}\) and \(t=30 \mathrm{~min},\) what is the average rate of appearance of \(\mathrm{B}\) in units of \(\mathrm{M} / \mathrm{s}\) ? Assume that the volume of the solution is constant.

Step-by-Step Solution

Answer

(a) Moles of B at each time: 0 min: 0 mol 10 min: 0.014 mol 20 min: 0.023 mol 30 min: 0.029 mol 40 min: 0.034 mol (b) Average rate of disappearance of A (in M/s) for each 10-min interval: 0-10 min: -2.33 x 10^-5 M/s 10-20 min: -1.5 x 10^-5 M/s 20-30 min: -1.0 x 10^-5 M/s 30-40 min: -8.33 x 10^-6 M/s (c) Average rate of appearance of B between t=0 min and t=30 min (in M/s): 1.61 x 10^-4 M/s

Step 1: (a) Calculate the number of moles of B at each time

Since the reaction is A → B, the number of moles of B at each time can be determined by subtracting the number of moles of A remaining from the initial number of moles of A (0.065 mol). At time 0 min: Moles of B = Initial moles of A - Moles of A = 0.065 - 0.065 = 0 mol At time 10 min: Moles of B = 0.065 - 0.051 = 0.014 mol At time 20 min: Moles of B = 0.065 - 0.042 = 0.023 mol At time 30 min: Moles of B = 0.065 - 0.036 = 0.029 mol At time 40 min: Moles of B = 0.065 - 0.031 = 0.034 mol

Step 2: (b) Calculate the average rate of disappearance of A for each 10-min interval

To find the average rate of disappearance of A, calculate the change in concentration of A (∆[A]) divided by the change in time (∆t) for each time interval. Concentration is obtained by dividing moles by volume (in L). Note: 100.0 mL = 0.100 L At 0-10 min: Average rate = (0.051 - 0.065) mol / (0.100 L * 10 min * (1/60) h) = -0.014 mol / (1/6 h) = -0.084 M/h At 10-20 min: Average rate = (0.042 - 0.051) mol / (0.100 L * 10 min * (1/60) h) = -0.009 mol / (1/6 h) = -0.054 M/h At 20-30 min: Average rate = (0.036 - 0.042) mol / (0.100 L * 10 min * (1/60) h) = -0.006 mol / (1/6 h) = -0.036 M/h At 30-40 min: Average rate = (0.031 - 0.036) mol / (0.100 L * 10 min * (1/60) h) = -0.005 mol / (1/6 h) = -0.030 M/h Convert the average rates from M/h to M/s by multiplying by (1/3600) s/h: At 0-10 min: -0.084 M/h * (1/3600) s/h = -2.33 x 10^-5 M/s At 10-20 min: -0.054 M/h * (1/3600) s/h = -1.5 x 10^-5 M/s At 20-30 min: -0.036 M/h * (1/3600) s/h = -1.0 x 10^-5 M/s At 30-40 min: -0.030 M/h * (1/3600) s/h = -8.33 x 10^-6 M/s

Step 3: (c) Average rate of appearance of B between t=0 min and t=30 min

To find the average rate of appearance of B, calculate the change in concentration of B (∆[B]) divided by the change in time (∆t) for the time interval 0-30 min. ∆[B] = (0.029 mol - 0.000 mol) / 0.100 L = 0.29 M ∆t = 30 min * (1/60) h = 0.5 h Average rate of appearance of B = ∆[B] / ∆t = 0.29 M / 0.5 h = 0.58 M/h Convert the average rate from M/h to M/s by multiplying by (1/3600) s/h: 0.58 M/h * (1/3600) s/h = 1.61 x 10^-4 M/s

Key Concepts

Moles and Molarity Calculations

Moles and molarity are key concepts in chemistry to quantify substances and their concentration in solutions. Here, molarity is defined as moles of solute per liter of solution. For instance, the initial solution has 0.065 mol of substance \( \mathrm{A} \) in 0.100 L, giving a molarity of 0.65 M.
Moles of substance \( \mathrm{B} \) can be calculated by subtracting the moles of \( \mathrm{A} \) remaining from the initial amount. For example, at 10 minutes, this would be 0.065 mol (initial) - 0.051 mol (remaining) = 0.014 mol of \( \mathrm{B} \).
By repeating this for all time intervals, you can track how \( \mathrm{B} \) forms over time, showing direct conversions from \( \mathrm{A} \) due to the simple one-to-one reaction. This approach helps understand how substances react and change concentration over time.

Rate of Reaction

The rate of reaction tells us how quickly reactants turn to products. It's commonly expressed as concentration change per unit time. To find the average rate of disappearance of \( \mathrm{A} \), you calculate the change in molarity of \( \mathrm{A} \) over each time period, then convert to seconds for better precision.
The calculation between 0 and 10 minutes:

• Change in moles = 0.051 mol - 0.065 mol = -0.014 mol
• Change in concentration = \(-0.014\) mol / 0.100 L = \(-0.140\) M
• Rate = \(-0.140\) M / 10 min = \(-0.084\) M/h, converted to \(-2.33 \times 10^{-5}\) M/s

This method gives a clear picture of how fast reactants diminish and products form, crucial for understanding and controlling reactions.

Chemical Equations Balance

Balancing chemical equations ensures that you follow the conservation of mass, stating that matter cannot be created or destroyed. In our reaction \( \mathrm{A} \rightarrow \mathrm{B} \), each molecule of \( \mathrm{A} \) converts into one molecule of \( \mathrm{B} \).
This simple one-to-one balance means for every decrease in \( \mathrm{A} \), we see an increase of \( \mathrm{B} \) by the same amount. It's a perfect demonstration of stoichiometry—a foundational concept in chemistry.
Understanding this balance in reactions allows you to calculate how much of each reactant is needed or how much product can be expected, essential for practical applications like synthesizing compounds or predicting the outcome of chemical processes.

Phase Changes in Reactions

Reactions involving phase changes often involve understanding how the physical state of a substance can impact the reaction. While the task focuses on an aqueous phase, it's crucial to note how different states (solid, liquid, gas) can affect reaction processes.
In the context of our aqueous reaction, both \( \mathrm{A} \) and \( \mathrm{B} \) remain in the same phase throughout, as denoted by \((aq)\). This consistency simplifies calculations and observations as there's no energy absorbed or released due to phase change.
However, when phase changes are involved, energy changes can become significant, influencing reaction rates and equilibrium. An understanding of phase-related behavior enhances predictability in complex reactions.