# How to calculate the exponents in the differential rate law?

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GIVEN:

GIVEN:

##### 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.

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.

Then I plotted each of the three functions.

**Zero-Order Plot**

The graph appears to be a curve.

**First-Order Plot**

The graph appears to be a straight line.

**Second-Order Plot**

The graph appears to be a curve.

**Conclusion**

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

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 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

#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

Plotting

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

#"STAT"# #-># #"Edit"# , enter all the data in there. Have the#x# variable in#"L1"# and#y# variable in#"L2"# .#"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.

#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