How to calculate the exponents in the differential rate law?

GIVEN:
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2 Answers

You use the formulas from the integrated rate laws.

Explanation:

Shen you have a list of concentrations as a function of time, you want to plot something that gives a straight line.

The equation for a straight line is:

#color(blue)(bar(ul(|color(white)(a/a) y = mx + b color(white)(a/a)|)))" "#

We can write the integrated rate laws for zero-, first, and second-order reactions all as straight-line functions.

#ulbb("Order"color(white)(mll)y = color(white)(l)mcolor(white)(l)x + b)#
#color(white)(mll)0 color(white)(mmml)c = color(white)(l)"-"kcolor(white)(ll)t + c_0#
#color(white)(mll)1color(white)(mm)lnc = color(white)(l)"-"kcolor(white)(ll)t + lnc_0#
#color(white)(mll)2color(white)(mmm)1/c = color(white)(ll )kcolor(white)(ll)t + 1/c_0#

Thus,

  • For a 0-order reaction, a plot of #c# vs #t# should give a straight line with slope #"-"k#
  • For a 1st-order reaction, a plot of #ln(c)# vs #t# should give a straight line with
    slope #"-"k#
  • For a 2nd-order reaction, a plot of #1/c# vs #t# should give a straight line with slope #k#

So, we plot each of these functions and see which gives the best straight line.

I used Excel to convert your data into the following table.

Data

Then I plotted each of the three functions.

Zero-Order Plot

Zero-order

The graph appears to be a curve.

First-Order Plot

First-Order

The graph appears to be a straight line.

Second-Order Plot

Second-Order

The graph appears to be a curve.

Conclusion

Your data give the best fit to a first-order rate law of the form

#rate = k["SO"_2"Cl"_2]^1#

Nov 23, 2017

Jose suggested a different way to approach this, so I figured I'd share it.


The starting rate law is:

#r(t) = k["SO"_2"Cl"_2]^m#

Jose suggested that one could take the #ln# of both sides like this:

#ln r(t) = mln["SO"_2"Cl"_2] + lnk#

But this requires that we generate a new column of data:

#r(t) = -(Delta["SO"_2"Cl"_2])/(Deltat)#

and for #n# data points, we generate #n-1# new data points to use here. We use the midpoint concentrations of #"SO"_2"Cl"_2# (call them #["SO"_2"Cl"_2]_"mp"#) and generate each #r(t)# as described above.

#ul(ln[r(t) xx 10^4] ("M/s")" "" "" "ln["SO"_2"Cl"_2]_"mp" ("M")" ")#

#ln1.24" "" "" "" "" "" "" "" "" "ln0.0938#
#ln1.08" "" "" "" "" "" "" "" "" "ln0.0822#
#ln0.95" "" "" "" "" "" "" "" "" "ln0.0721#
#ln0.83" "" "" "" "" "" "" "" "" "ln0.0632#
#ln0.73" "" "" "" "" "" "" "" "" "ln0.0554#
#ln0.64" "" "" "" "" "" "" "" "" "ln0.0485#

You get the idea. I'll leave out the remaining #5# data points because six is quite sufficient to establish a trend.

Plotting #lnr(t)# vs. #ln["SO"_2"Cl"_2]#, the y-intercept should be #lnk#, and the slope should be whatever the order of reaction is.

The regression equation was found as follows using my trusty TI-83 calculator.

  1. #"STAT"# #-># #"Edit"#, enter all the data in there. Have the #x# variable in #"L1"# and #y# variable in #"L2"#.
  2. #"STAT" -> "CALC" -> "LinReg"("ax+b")#

This gave:

#lnr(t) = 1.000765854 cdot ln["SO"_2"Cl"_2]_"mp" - 6.630082087#

This implies that the order of reaction is approx. #color(blue)(1)# and that

#lnk = -6.630082087#

and

#color(blue)(k) = e^(-6.630082087) = color(blue)("0.0013 s"^(-1))#

So this method works too. However, you can see it is a big hassle, when you have this many data points, to rewrite all of them into #lnr(t)# and #ln["SO"_2"Cl"_2]_"mp"#. I would say Ernest's method is faster.