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# Plotting data for a first-order reaction

Example of graphing first-order rate data to see a linear relationship, and calculating rate constant k from the slope.

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• I got 7.66x10E -4, what did I do wrong?
I used -2.54+2.31 all over 300 seconds.
• The points provided won't produce a perfectly straight line. So the slope between points 1 and 2 may be slightly different between that of 2 and 3, etc. A best fit line is put in place and the slope of the best fit line is what would be used to determine K.
• wouldn't it be easier to take the concentration at 900s minus the initial concentration and divide by 900? Same as rise over run and you don't need to worry about the sloppy graph.
• It is if you know that the data is perfectly linear. This will happen with low level book problems however when you start doing calculations based on lab work there will be a degree of variance in your results. Even if you do flawless work there is systematic and instrument error to take into account. When this is the case you can't just take a shortcut and get accurate results. Even though you are able to do so at this point it is best to practice making calculations based on all the data points so you will be able to do the work later on.
• i did not very well grasp how the slope was calculated..are there any basics for this i should understand first. if yes from where please?
• i'm calculating the slope manually and I keep getting 7.67 x 10 ^-4; although i got the 6.7x10-4 via graphing in calculator as well. can you show me how to find slope manually (using the rise/run formula)
• you can get a pretty close estimate by using the data in the table:
change in [A] / change in time = R
(0.079 - 0.099) / (300s) = R
and then use R / [0.099] = k = 6.734 x 10^-4
• is there no other way to prove if a given reaction is first order or not and to find the rate constant other than plotting a graph?
• Wouldn't a concentration of cyclopropane vs. time graph be an exponential curve?
(1 vote)
• Yes. If we let [A] = the concentration of cyclopropane, the integrated rate law is
[A] = [A]₀e^(- k t)
A plot of [A] vs. t is a curve that starts at [A]₀ and gradually approaches the horizontal axis asymptotically as t increases.
The shape of the curve is called an exponential decay curve.

If you take the natural logarithm of each side of the equation, you get
ln[A] = ln[A]₀ - kt
If you plot ln[A] vs t, you get a straight line with slope = -k and y-intercept =ln[A]₀.
• is the equation thingy only valid for the first-order reaction or is it valid for any