Question

Consider the following hypothetical reaction: $$ \mathrm{X}(g) \longrightarrow \mathrm{Y}(g) $$ A \(200.0-\mathrm{mL}\) 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. $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 20 & 40 & 60 & 80 \\ \text { moles of X } & 0.120 & 0.103 & 0.085 & 0.071 & 0.066 \\ \hline \end{array} $$ (a) Make a similar table for the appearance of \(\mathrm{Y}\). (b) Calculate the average disappearance of \(\mathrm{X}\) in \(\mathrm{M} / \mathrm{s}\) in the first two 20 -minute intervals. (c) What is the average rate of appearance of Y between the 20 - and 60 -minute intervals?

Short answer

Question: Calculate the average rate of appearance of Y between the 20-minute and 60-minute intervals. Answer: The average rate of appearance of Y between the 20-minute and 60-minute intervals is 2.22 × 10^{-5} M/s.

Step-by-step solution

(a) Create a table for the appearance of Y

To create a table for the appearance of Y, we first need to determine the initial moles of Y at time zero. Since there is an equal moles ratio between X and Y in the reaction, the moles of Y at time zero would be 0. In addition, for every mole of X that disappears, one mole of Y appears. So, determine the moles of Y at each time interval by subtracting the moles of X at that time from the initial moles of X (0.120 moles). The table for the appearance of Y would be: $$ \begin{array}{lccccc} \hline \text { Time (min) } & 0 & 20 & 40 & 60 & 80 \\\ \text { moles of Y } & 0 & 0.017 & 0.035 & 0.049 & 0.054 \\\ \hline \end{array} $$

(b) Calculate the average disappearance of X in M/s in the first two 20-minute intervals

To calculate the average disappearance of X, we need to find the decrease in moles of X in the first two intervals (0-20 minutes and 20-40 minutes) and divide each by the total time in seconds (converting 20 minutes to seconds: 20 * 60 = 1200 s). Then, we need to divide these rates by the volume of the reaction mixture, 0.200 L or 200.0 mL, to get the concentration-based rate in M/s. For the first interval (0-20 minutes): $$\text{Average rate} = \frac{(\text{Final moles of X - Initial moles of X}) \cdot\mathrm{1\,L\,soln} } {(\text{Volume of soln}) \cdot (\text{Time interval in seconds})}$$ $$\text{Average rate} = \frac{(0.103 - 0.120) \cdot \mathrm{1\,L\,soln}}{0.200\,\mathrm{ L\, soln} \cdot 1200\,\mathrm{s}} = -2.83 \times 10^{-5}\,\mathrm{M/s}$$ For the second interval (20-40 minutes): $$\text{Average rate} = \frac{(0.085 - 0.103) \cdot \mathrm{1\,L\, soln}}{0.200\,\mathrm{ L\, soln} \cdot 1200\,\mathrm{s}} = -2.50 \times 10^{-5}\,\mathrm{M/s}$$

(c) Calculate the average rate of appearance of Y between the 20-minute and 60-minute intervals

To calculate the average rate of appearance of Y between the 20-minute and 60-minute intervals, we need to find the increase in moles of Y during this time period (from the table in part (a)) and then divide by the total time in seconds (converting 40 minutes to seconds: 40 * 60 = 2400 s). Then, we need to divide this rate by the volume of the reaction mixture, 0.200 L or 200.0 mL, to get the concentration-based rate in M/s. $$\text{Average rate} = \frac{(\text{Final moles of Y - Initial moles of Y}) \cdot\mathrm{1\,L\, soln} } {(\text{Volume of soln}) \cdot (\text{Time interval in seconds})}$$ $$\text{Average rate} = \frac{(0.049 - 0.017) \cdot \mathrm{1\,L\, soln}}{0.200\,\mathrm{ L\, soln} \cdot 2400\,\mathrm{s}} = 2.22 \times 10^{-5}\,\mathrm{M/s}$$