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Chapter 11: Problem 3

Consider the following hypothetical reaction: $$\mathrm{X}(g) \longrightarrow \mathrm{Y}(g)$$ A 200.0-mL flask is filled with \(0.120\) moles of \(\mathrm{X}\). The disappearance of \(\mathrm{X}\) is monitored at timed intervals. Assume that temperature and volume are kept constant. The data obtained are shown in the table below. $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{X} & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline\end{array}$$ (a) Make a similar table for the appearance of Y. (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two twenty-minute intervals. (c) What is the average rate of appearance of \(\mathrm{Y}\) between the 20 - and 60-minute intervals?

Short Answer

Expert verified
In this problem, we were asked to find the appearance of product Y and the average reaction rates of the disappearance of reactant X and appearance of product Y during different time intervals. We used stoichiometry and the concept of reaction rates to create a table for moles of Y at different time intervals. We then calculated the molar concentrations of X and Y and determined the average rates of disappearance and appearance for X and Y, respectively. The average disappearance rates of X were around -0.085/1200 M/s and -0.090/1200 M/s, and the average rate of appearance of Y between the 20- and 60-minute intervals was approximately 0.160/2400 M/s.

Step by step solution

01

Finding the moles of Y at different time intervals

To find the moles of Y at different intervals, we can use the stoichiometry of the reaction. At the start, there are no moles of Y, so we can subtract the moles of X at a given time from the initial moles of X to find the moles of Y at that time. Here's the table of moles of Y at different time intervals: $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of } \mathrm{Y} & 0 & 0.017 & 0.035 & 0.049 & 0.054 \\\ \hline\end{array}$$ Now, let's move on to part (b), which is to calculate the average rate of disappearance of X in M/s for the first two twenty-minute intervals.
02

Finding the molar concentrations of X at different intervals

To find the molar concentrations of X, we divide the moles of X by the volume of the flask (200.0 mL or 0.200 L): $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { [X] } (\mathrm{M}) & 0.600 & 0.515 & 0.425 & 0.355 & 0.330 \\\ \hline\end{array}$$
03

Calculating the average rate of disappearance of X for the first two intervals

To find the average rate of disappearance of X, we'll calculate the change in concentration of X over the time interval for the first two twenty-minute intervals: Interval 1: \((0.515 - 0.600) \,\mathrm{M} / (20\,\text{min} * 60\,\text{s/min}) = -0.085\text{/}1200\,\mathrm{M/s}\) Interval 2: \((0.425 - 0.515) \,\mathrm{M} / (20\,\text{min} * 60\,\text{s/min}) = -0.090\text{/}1200\,\mathrm{M/s}\) Now, we'll move on to part (c), which is to find the average rate of appearance of Y between the 20- and 60- minute intervals.
04

Finding the molar concentrations of Y at different intervals

To find the molar concentrations of Y, we divide the moles of Y by the volume of the flask (200.0 mL or 0.200 L): $$\begin{array}{lccccc}\hline \text { Time }(\min ) & 0 & 20 & 40 & 60 & 80 \\\ \text { [Y] } (\mathrm{M}) & 0 & 0.085 & 0.175 & 0.245 & 0.270 \\\ \hline\end{array}$$
05

Calculating the average rate of appearance of Y between the 20 and 60-minute intervals

To find the average rate of appearance of Y, we'll calculate the change in concentration of Y over the time interval between 20 and 60 minutes: $$\frac{(0.245 - 0.085)\,\mathrm{M}}{(60-20)\,\text{min} * 60\,\text{s/min}} = \frac{0.160}{2400}\,\mathrm{M/s}$$ In conclusion, (a) We created a table for the appearance of Y, with the moles of Y at different time intervals. (b) The average disappearance of X in the first two twenty-minute intervals is around \(-0.085/1200\,\mathrm{M/s}\) and \(-0.090/1200\,\mathrm{M/s}\), respectively. (c) The average rate of appearance of Y between the 20- and 60-minute intervals is approximately \(\frac{0.160}{2400}\,\mathrm{M/s}\).

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