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Chapter 14: Problem 22

The rate of disappearance of HCl was measured for the following reaction: $$ \mathrm{CH}_{3} \mathrm{OH}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{Cl}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ The following data were collected: $$ \begin{array}{rl} \hline \text { Time (min) } & \text { [HCI] (M) } \\ \hline 0.0 & 1.85 \\ 54.0 & 1.58 \\ 107.0 & 1.36 \\ 215.0 & 1.02 \\ 430.0 & 0.580 \\ \hline \end{array} $$ (a) Calculate the average rate of reaction, in \(M / \mathrm{s}\), for the time interval between each measurement. (b) Calculate the average rate of reaction for the entire time for the data from \(t=0.0 \mathrm{~min}\) to \(t=430.0 \mathrm{~min} .\) (c) Graph [HCl] versus time and determine the instantaneous rates in \(M / \min\) and \(M / s\) at \(t=75.0 \mathrm{~min}\) and \(t=250\) min.

Short Answer

Expert verified
(a) The average rates for the time intervals are as follows: - $0.0$ to $54.0$ min: \(r_{avg} = -8.5 \times 10^{-4} \ M/min\) or \(-1.42 \times 10^{-5} \ M/s\) - $54.0$ to $107.0$ min: \(r_{avg} = -4.26 \times 10^{-4} \ M/min\) or \(-7.1 \times 10^{-6} \ M/s\) - $107.0$ to $215.0$ min: \(r_{avg} = -2.0 \times 10^{-4} \ M/min\) or \(-3.33 \times 10^{-6} \ M/s\) - $215.0$ to $430.0$ min: \(r_{avg} = -1.03 \times 10^{-4} \ M/min\) or \(-1.72 \times 10^{-6} \ M/s\) (b) The average rate for the entire time \(t=0.0\) to \(t=430.0\) min is \(r_{avg(total)} = -2.93 \times 10^{-4} \ M/min\) or \(-4.88 \times 10^{-6} \ M/s\). (c) The average rate between \(t=54.0\) and \(t=215.0 \mathrm{~min}\) (\(-2.0 \times 10^{-4}\) M/min) is greater than the average rate between \(t=107.0\) and \(t=430.0 \mathrm{~min}\) (\(-1.03 \times 10^{-4}\) M/min). (d) Using the graph of concentration versus time, the instantaneous rates are approximately: - At \(t=75.0 \min\): \(r_{inst} = -6.5 \times 10^{-4} \ M/min\) or \(-1.08 \times 10^{-5} \ M/s\) - At \(t=250 \min\): \(r_{inst} = -1.5 \times 10^{-4} \ M/min\) or \(-2.5 \times 10^{-6} \ M/s\)

Step by step solution

01

Calculate the change in concentration and time intervals

Since we are asked to find the average rate of reaction for different time intervals, we should first determine the change in concentration of HCl and the time intervals between each measurement. We will find the differences in concentration between consecutive measurements, and the corresponding time intervals between them.
02

Calculate average rates

To calculate the average rate of reaction for each time interval, we will use the following formula: \(r_{avg} = \frac{\Delta[\mathrm{HCl}]}{\Delta t}\), where \(\Delta[\mathrm{HCl}]\) represents the change in concentration of HCl, and \(\Delta t\) represents the time interval between the consecutive measurements. We must calculate this for each time interval. #Phase 2: Average Rate for the Entire Duration#
03

Calculate the average rate for the entire time

To determine the average rate of reaction for the entire time, we will calculate the difference in concentration of HCl between \(t=0\) and \(t=430\) minutes and divide it by the total time interval. The average rate can be calculated as follows: \(r_{avg(total)} = \frac{\Delta[\mathrm{HCl}]}{\Delta t}\) #Phase 3: Comparing Average Rates#
04

Compare the average rates

To find which average rate is greater, between \(t=54.0\) and \(t=215.0 \mathrm{~min}\), or between \(t=107.0\) and \(t=430.0 \mathrm{~min}\), we will use the calculated average rates from Step 2 and compare them. #Phase 4: Instantaneous Rates Calculation#
05

Graph concentration versus time

In order to determine the instantaneous rates at \(t=75.0 \min\) and \(t=250 \min\), we must first graph the concentration of HCl versus time. Make sure to plot the points with given concentrations and connect them with a line.
06

Determine the instantaneous rates

With the graph, we can determine the instantaneous rates for\(t=75.0 \min\) and \(t=250 \min\) by calculating the slope of the tangent line at each of these points. To do this visually, we can draw the tangent line at each point and then determine its slope, which represents the instantaneous rate at that point, by dividing the change in concentration on the graph by the corresponding change in time represented by the tangent line. Remember to convert the instantaneous rates to \(M/ \mathrm{s}\) by dividing the obtained rates by 60 (since there are 60 seconds in a minute).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Average Rate of Reaction
Understanding the average rate of reaction is crucial for studying the kinetics of a chemical process. In essence, this rate measures how quickly reactants are converted into products over a specified time period. The average rate is calculated by dividing the change in concentration of a reactant by the total time taken for that change to occur.

For example, with the reaction involving the disappearance of HCl, the average rate between two time points can be calculated using the formula:
\r(r_{avg} = \frac{\Delta[\mathrm{HCl}]}{\Delta t})

Here, \(\Delta[\mathrm{HCl}]\) is the change in concentration of HCl and \(\Delta t\) is the time interval. Notably, the rates for each interval can vary, reflecting how the reaction might speed up or slow down over its course. This is especially visible when comparing the average rates over different time intervals, showing the dynamic nature of chemical reactions.
Concentration Change Over Time
The concept of concentration change over time lies at the heart of reaction kinetics and provides insights into the reaction's mechanism. A graph of HCl concentration versus time for the given reaction would typically show a decreasing curve, indicating the consumption of HCl as the reaction progresses. Understanding this concept helps predict how much reactant will be present at any given moment in the future, which is essential for planning in industrial applications and lab experiments.

The significance of plotting this graph becomes evident when assessing the overall rate, i.e., the average rate over the entire reaction course. For the given exercise, this would involve finding the total change in HCl concentration from the beginning to the end and dividing by the total time elapsed. Moreover, observing the shape of the curve can hint at whether the reaction rate changes steadily or if there are phases where the reaction speeds up or slows significantly.
Instantaneous Rate of Reaction
Diving deeper into reaction kinetics, the instantaneous rate of reaction tells us the rate at a particular moment, akin to taking a snapshot of the reaction at a specific time. Unlike the average rate which considers a time interval, the instantaneous rate examines the change at a specific point and requires the slope of the tangent to the concentration-time graph at that point.

Using the provided data, students can visualize this concept by drawing the concentration-time graph and then the tangent lines at selected times, such as 75.0 min and 250 min. The slope of these tangents gives the instantaneous rate at those exact moments. It's important to convert these rates to units of \(M/s\) when required, which elucidates how quickly conditions are changing at that instant 鈥� a vital parameter for controlling chemical processes.

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Most popular questions from this chapter

(a) Two reactions have identical values for \(E_{a} .\) Does this ensure that they will have the same rate constant if run at the same temperature? Explain. (b) Two similar reactions have the same rate constant at \(25^{\circ} \mathrm{C}\), but at \(35^{\circ} \mathrm{C}\) one of the reactions has a larger rate constant than the other. Account for these observations.

Urea \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right)\) is the end product in protein metabolism in animals. The decomposition of urea in \(0.1 \mathrm{M} \mathrm{HCl}\) occurs according to the reaction $$ \begin{aligned} \mathrm{NH}_{2} \mathrm{CONH}_{2}(a q)+\mathrm{H}^{+}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \mathrm{NH}_{4}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \end{aligned} $$ The reaction is first order in urea and first order overall. When \(\left[\mathrm{NH}_{2} \mathrm{CONH}_{2}\right]=0.200 \mathrm{M},\) the rate at \(61.05^{\circ} \mathrm{C}\) is \(8.56 \times 10^{-5} \mathrm{M} / \mathrm{s}\). (a) What is the rate constant, \(k ?\) (b) What is the concentration of urea in this solution after \(4.00 \times 10^{3} \mathrm{~s}\) if the starting concentration is \(0.500 \mathrm{M}\) ? (c) What is the half-life for this reaction at \(61.05^{\circ} \mathrm{C}\) ?

A colored dye compound decomposes to give a colorless product. The original dye absorbs at \(608 \mathrm{nm}\) and has an extinction coefficient of \(4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}\) at that wavelength. You perform the decomposition reaction in a \(1-\mathrm{cm}\) cuvette in a spectrometer and obtain the following data: $$ \begin{array}{rl} \hline \text { Time (min) } & \text { Absorbance at } 608 \mathrm{nm} \\ \hline 0 & 1.254 \\ 30 & 0.941 \\ 60 & 0.752 \\ 90 & 0.672 \\ 120 & 0.545 \end{array} $$ From these data, determine the rate law for the reaction "dye \(\longrightarrow\) product" and determine the rate constant.

For each of the following gas-phase reactions, write the rate expression in terms of the appearance of each product and disappearance of each reactant: (a) \(2 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (c) \(2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{H}_{4}(g)\)

The following mechanism has been proposed for the gasphase reaction of \(\mathrm{H}_{2}\) with ICl: $$ \begin{array}{l} \mathrm{H}_{2}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{HI}(g)+\mathrm{HCl}(g) \\ \mathrm{HI}(g)+\mathrm{ICl}(g) \longrightarrow \mathrm{I}_{2}(g)+\mathrm{HCl}(g) \end{array} $$ (a) Write the balanced equation for the overall reaction. (b) Identify any intermediates in the mechanism. (c) If the first step is slow and the second one is fast, which rate law do you expect to be observed for the overall reaction?
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