Virtual Experiment: Part 2

[Table of Contents]

[Introduction]

To begin our experiment we must first collect some data. Below are the initial conditions of our experiment.

Initial concentration of NO: 8.071 x 10-5 moles/L
Initial concentration of O2: 1.550 x 10-3 moles/L

We mix these materials and then monitor the loss of NO at 21oC by IR spectroscopy. Nitric oxide (NO) absorbs strongly in the IR at approximately 1900 cm-1. Take several scans below and monitor the loss of NO as a function of time. Stop taking scans when it appears that the loss of NO has stopped. Using a pencil and paper, write down the approximate intensity of the NO signal at about 1900 cm-1.

Take an IR Scan

When you have completed taking scans and recording data, click below to continue the experiment.

Continue

 

 

 

 

 

 

 

 

If you properly recorded your data, you should have a table that looks something like the following.

Table 1
Time (seconds)
Absorbance (@ 1854 cm-1)
Absorbance (@ 1857 cm-1)
Absorbance (@ 1860 cm-1)
Absorbance (@ 1864 cm-1)
Absorbance (@ 1900 cm-1)
Absorbance (@ 1904 cm-1)
520 0.0120 0.0113 0.00918 0.00867 0.0124 0.0149
586 0.0114 0.0106 0.00856 0.00818 0.0117 0.0141
828 0.00992 0.00930 0.00759 0.00713 0.0103 0.0122
908 0.00909 0.00838 0.00677 0.00653 0.00939 0.0110
1036 0.00826 0.00763 0.00619 0.00588 0.00847 0.0100
1118 0.00785 0.00711 0.00582 0.00562 0.00792 0.00943

While now you have a set of absorbance data, we still need to convert this information into units of concentration. We do this using Beer's law. As you may recall, Beer's law has the formula,

Absorbance = ABC
A = Extinction coefficient
B = Path length
C = Concentration of absorbing species

Now, while it is possible to determine the concentration of NO in our sample by measuring the absorbance at a single wavelength, we can obtain a more accurate measurement by looking at the integrated area under the absorption peaks.

Figure 1: Illustration of absorption spectra for NO in the IR. The base line of the spectra is the portion of the spectra where the sample is not absorbing light. Ideally this should be zero, but often times your instruments can pick up some noise causing baseline to either increase or decrease. In the figure above notice how the baseline is zero at 1800 cm-1, but then it is nonzero at 1930 cm-1. This increase in the baseline is known as drift.
A good "back of the envelope" way to determine the integrated area under the absorption peaks is to merely sum the absorption peaks. We then have to correct this sum for any drift in our baseline measurement. You will be asking yourself, "what is a baseline and why do I have to correct it". To explain this, take a look at the figure to the right. The figure shows the IR absorption spectra of NO. Each peak represents an absorption of light by the NO molecule. The portion of the spectrum where the molecule doesn't absorb light is known as the baseline. The area under the curve can be approximated by measuring the height of the peak, assuming that the baseline stays at zero. But if you look at Figure 1 you will see that the baseline increases as the wavenumber increases. So to properly approximate the area under a peak you take the height of the peak and then subtract off the baseline. In this experiment we will measure the absorbance for three different wavenumbers and then approximate the integrated area of the absorption spectrum by using the formula below.






Integrated area = (Sum of the six absorption measurements in table 1) - 2(sum of the three baseline measurements in table 2) Now, using the measurements of the baseline absorbance below, determine the integrated area for the data provided in Table 1 at each time.





Table 2
Time (seconds) Baseline absorbance (@ 1980 cm-1) Baseline absorbance (@ 1985 cm-1) Baseline absorbance (@ 1978 cm-1)
520 0.000429 0.000471 0.000497
586 0.000559 0.000531 0.000523
828 0.000602 0.000692 0.000636
908 0.000734 0.000792 0.000844
1036 0.000684 0.000727 0.000785
1118 0.000694 0.000696 0.000719

 

When you have calculated the integrated absorbance values for each time interval listed in Table 1 click below to continue the experiment.

Continue.

 

 

 

 

 

 

 

 

Now that you know the integrated absorbance of NO at each time interval, you need to calculate the concentration of NO that this integrated signal represents. Fortunately for you, we simplified your life by providing a simple Beer's law calculator for NO that you can use to calculate the concentration of NO for you absorbance measurements. Using this calculator, convert integrated areas from the table into units of concentration

 

Measured Absorbance:

  Concentration of NO: moles/L

                   

 

Now that you have converted your absorbance measurements into concentration values, you can plot the concentration of NO as a function of time. From this plot you should be barely able to determine the reaction order in NO (two of the choices are very close) and the pseudo second order (hint) rate constant for the reaction.

Use the buttons above to plot a more extensive data set for this experiment in different ways. The equation shown in the  figure represent the equation that is best fit to the plotted data.

Now you should know what the order of NO is in this reaction and pseudo second order rate constant for the reaction. Make sure to write that down these numbers because you might need them later. Normally at this point you would have to do another experiment to determine the order of O2 in the reaction. Fortunately for you, a colleague has done that experiment for you and told you that this reaction is first order in oxygen.

Now you have one last experiment - determine the rate constant for this reaction. You have to measure the pseudo second order rate constant for this reaction at different concentrations of O2. The data for this experiment is provided for you below.

Table 3
[NO] (moles/liter) [O2] (moles/liter) C (L*moles-1*sec.-1)
0.0000411 0.00106 13.5
0.0000411 0.00108 15.4
0.0000625 0.00153 22.9
0.0000833 0.00155 23.6
0.0000411 0.00159 23.7
0.000167 0.00161 23.9
0.000104 0.00161 25.5
0.000125 0.00168 27.0
0.000146 0.00174 28.5
0.0000411 0.00212 29.0
0.0000410 0.00265 36.5
0.0000411 0.00318 44.9
0.0000411 0.00371 52.3
0.0000411 0.00424 53.3
0.0000411 0.00424 61.6
0.0000411 0.00467 62.0
0.0000411 0.00478 65.3
0.0000411 0.00531 69.8
0.0000411 0.00564 73.5

Now all that remains is for you to plot this data and then to determine the rate constant k from plotting the observed pseudo second order rate constant, C.

Now that you have determined the rate constant and the order of each reactant in the reaction, you can test your theory using the kinetics simulator.

Problems to be Answered

1) How do the rates of disappearance of oxygen and nitric oxide differ in the experiment?

2) The literature value for the rate constant for the NO + O2 reaction at 25oC is 2 x 10-38 cm6/molecule2 s. How does your rate constant compare with that number? Hint, in converting units there are 1000 cm3/L and 6.02 x 1023 molecules/mole. In making your comparison you should recognize that your experiment monitored the rate of disappearance of NO. By convention, rate constants are normalized by the stoichiometric coefficient in the net reaction.

3) This experiment required the measurement of very low concentrations of NO. What is your estimate about the reproducibility of the rate measurement and the reliability of the determined rate constant?

4) Ozone, O3, reacts with NO in a second order process (rate = k[NO][O3]) with a rate constant at 25oC of 1.6 x 10-14 cm3/molecule s. What would be the ratio of the rates of reaction of 0.00005 M NO emitted from the exhaust pipe of a car with oxygen in air and with the small amount of ozone (~ 80 ppb) in ambient rural air.

5) The integrated rate law can be used, as in the simulator below, to plot


complete concentration vs. time curves. This simulator uses your rate constant and assumes that the oxygen concentration is in excess as compared to the nitric oxide concentration. Why is this latter assumption necessary? Estimate from the concentration vs. time curves the reaction half life expected for oxidation of NO in the atmosphere if the NO concentration were 800 ppm as in exhaust gas, 100 ppb as found in a polluted urban atmosphere, and 0.04 ppb as found in a remote marine environment.

6) The integrated rate law can be used, as in the simulator below, to plot


complete concentration vs. time curves for the NO - ozone reaction. Estimate from the concentration vs. time curves the reaction half life expected for oxidation of NO in the atmosphere if the NO concentration were 100 ppb and the ozone concentration were 120 ppb, as found in a polluted urban atmosphere. How does this compare with the corresponding value for the NO - oxygen reaction in problem 5? Note that the temperature must be input in this simulator.

[Table of Contents] [Introduction]