Experimental data are listed for the hypothetical reaction $$\mathrm{x} \longrightarrow \mathrm{Y}+\mathrm{Z}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\ {[\mathrm{X}]} & 0.0038 & 0.0028 & 0.0021 & 0.0016 & 0.0012 & 0.00087 \\\\\hline\end{array}$$ (a) Plot these data as in Figure \(11.2 .\) (b) Draw a tangent to the curve to find the instantaneous rate at \(40 \mathrm{~s}\). (c) Find the average rate over the 10 to 50 s interval. (d) Compare the instantaneous rate at \(40 \mathrm{~s}\) with the average rate over the \(40-\mathrm{s}\) interval.

Short Answer

Expert verified
Question: Compare the instantaneous rate at 40 seconds with the average rate over the 40 seconds interval in the given reaction, and discuss their differences in the context of the reaction. Answer: The instantaneous rate at 40 seconds represents the rate of decrease in concentration of reactant X at that specific moment in time. On the other hand, the average rate over the 40 seconds interval represents the overall rate of decrease in concentration of reactant X over the entire interval. The instantaneous rate may differ from the average rate, as the reaction rate could vary at different time points throughout the interval. Comparing these two rates can help us understand changes in reaction rate and kinetic behavior over the given time period.

Step by step solution

01

Prepare coordinate points from the table.

Extract the given data pairs \((t, [\mathrm{X}])\) from the table: - (0, 0.0038) - (10, 0.0028) - (20, 0.0021) - (30, 0.0016) - (40, 0.0012) - (50, 0.00087)
02

Plot the data points on a graph.

Plot the time values on the x-axis and the concentration values on the y-axis. Complete the graph by connecting the data points with a smooth curve. Please refer to Figure 11.2 for the format. #b) Instantaneous rate at 40 s#
03

Draw the tangent to the curve at 40 s.

Locate the point corresponding to 40 s (40, 0.0012) on the graph. Draw a tangent to the curve at this point.
04

Calculate the slope of the tangent.

The slope of the tangent represents the instantaneous rate at 40 s. To find the slope, select two points on the tangent and compute the change in concentration divided by the change in time. $$\text{Slope} = \frac{\Delta [\mathrm{X}]}{\Delta t}$$ #c) Average rate over 10 s to 50 s interval#
05

Calculate the average rate.

The average rate over the 10 s to 50 s interval is given by the change in concentration divided by the change in time during this interval: $$\text{Average Rate} = \frac{[\mathrm{X}_{50\text{s}}]-[\mathrm{X}_{10\text{s}}]}{50\text{s} - 10\text{s}} = \frac{0.00087 - 0.0028}{40\text{s}}$$ #d) Compare rates#
06

Compare the instantaneous rate and the average rate.

Compare the instantaneous rate calculated from the slope of the tangent to the average rate calculated over the 40 s interval. Discuss any differences in the two rates and their interpretation in the context of the reaction.

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Most popular questions from this chapter

Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)

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Experimental data are listed for the hypothetical reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}+\mathrm{D}$$ $$\begin{array}{lcccccc}\hline \text { Time (s) } & 0 & 10 & 20 & 30 & 40 & 50 \\\\{[\mathrm{~A}]} & 0.32 & 0.24 & 0.20 & 0.16 & 0.14 & 0.12 \\ \hline\end{array}$$ (a) Plot these data as in Figure \(11.2\). (b) Draw a tangent to the curve to find the instantaneous rate at \(30 \mathrm{~s}\). (c) Find the average rate over the 10 to \(40 \mathrm{~s}\) interval. (d) Compare the instantaneous rate at \(30 \mathrm{~s}\) with the average rate over the thirty-second interval.

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