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FOUNTAINHEAD PRESS
1. Reaction Rates. For each of the reactions E1 through E5, calculate the absorbance the
absorbance from % transmittance using Eq. 8 and plot the absorbance of the solution vs.
time. You may either plot the data on the graph paper provided, or you can enter the data
into a spreadsheet (e.g., Excel).
The rate of a given reaction is calculated as the initial slope of the plot of absorbance vs
time. If you view the plot, the slope may decrease over time, so it is important to calculate
the slope in the initial portion of the graph (i.e., first 100 to 200 seconds) in a region where
the plot appears to be linear. If you are using a spreadsheet program, you can take advantage
of the “add trend line” option to calculate the best-fit straight line, which will include a
calculation of the slope.
Record the reaction rates for each reaction mixture on the Calculation Sheet.
2. Reaction order. Since the CV concentration was monitored spectrophotometrically, the
absorbance of the reaction solution was directly related to the [CV] by Beer’s Law.
Therefore, we can take advantage of the integrated rate laws to determine the order of the
reaction with respect to CV. Using the absorbance data vs. time for reaction E1, calculate
ln[Abs] and 1/[Abs] for each data point and record these data in Data Table 3.
Using experimental data for reaction mixture E1, prepare two graphs. For the first order
graph, plot ln[Abs] vs. time; for the second order graph, plot 1/[Abs] vs. time. You can plot
the data using the graph paper provided or you can use a spreadsheet program (e.g., Excel).
Based on the graph that produces a straight line, record the value of x on the Data Sheet.
To determine the value of z we take advantage of the fact that only one reactant
concentration was changed during the experiment. Note that the [CV] was the same for
Reactions E1--E5, while the [OH
-
] varied. The generic rate law can be written as
Rate = k [CV]
x
[OH
-
]
z
Taking the natural log of both sides of the equation yields:
ln rate = lnk + xln[CV] + z ln[OH
-
] (3)
Since x and [CV] are both constants for E1--E5, we can combine them with ln k to make a
new constant ln k’, which yields:
ln rate = lnk’ + z ln[OH
-
] (10)
This new equation resembles the equation for a straight line, y = mx + b. Plotting ln rate vs.
ln [OH
-
] for experiments E1--E5 yields a straight line with a slope = z.
Using the volumes and concentrations for CV and NaOH solutions, calculate the initial
concentration of CV and OH
-
in each of the reactions E1--E5, and record these
concentrations under Part B3 in Calculations. Using the rates for E1--E5, plot ln rate vs. ln
[OH
-
]. Calculate the slope of this plot and record it as z on your Calculations Sheet. You
should round your results to the nearest integer.