<< Graphics`Graphics`
<< Statistics`LinearRegression`

This example illustrates the use of the integral method to determine the order of a reaction (with respect to a selected reactant). 
Let's illustrate first the use of the fractional life method (Example 3.1 in Levenspiel):
				\(TraditionalForm\`t\_F\  = \ \((\(\((0.8)\)\^\(1 - n\) - 1\)\/\(k(n - 1)\))\) C\_A0\%\(1 - n\)\)

data = 
        List[{0. , 10. }, {20. , 8. }, {40. , 6. }, {60. , 5. }, {120. , 
            3. }, {180. , 2}, {300. , 1. }];
dataPlot = ListPlot[data, PlotStyle -> PointSize[0.02]];
concVsTime = 
      Regress[data, Table[t^i, {i, 0, 4}], t, 
        RegressionReport -> RSquared]
cVsTime = Fit[data, Table[t^i, {i, 0, 4}], t]
Plot[%, {t, 0, 300}];

Show[%, dataPlot];

y1 = 
      Solve[10.054725737331577 - 0.12803785794645872 t + 
            0.0008911174790256581 (t^2) - 3.1223415528696154*^-6 (t^3) + 
            4.130719386676709*^-9 (t^4) == 8. , t]

y[1] = 18.212581253068382

y21 = 
      Solve[10.054725737331577 - 0.12803785794645872 t + 
            0.0008911174790256581 (t^2) - 3.1223415528696154*^-6 (t^3) + 
            4.130719386676709*^-9 (t^4) == 5. , 
        t]
y22 = 
      Solve[10.054725737331577 - 0.12803785794645872 t + 
            0.0008911174790256581 (t^2) - 3.1223415528696154*^-6 (t^3) + 
            4.130719386676709*^-9 (t^4) == 4. , t]

y[2] = 82.35594471641191 - 59.21654843408716

y31 = 
      Solve[10.054725737331577 - 0.12803785794645872 t + 
            0.0008911174790256581 (t^2) - 3.1223415528696154*^-6 (t^3) + 
            4.130719386676709*^-9 (t^4) == 2. , 
        t]
y32 = 
      Solve[10.054725737331577 - 0.12803785794645872 t + 
            0.0008911174790256581 (t^2) - 3.1223415528696154*^-6 (t^3) + 
            4.130719386676709*^-9 (t^4) == 1.6, t]

y[3] = 208.7382182722186 - 180.4794400347306

data2 = 
      List[{Log[10, 10. ], Log[10, y[1]]}, {Log[10, 5. ], 
          Log[10, y[2]]}, {Log[10, 2. ], Log[10, y[3]]}]
dataPlot2 = 
        ListPlot[data2, PlotStyle -> PointSize[0.02]];
timeVsconcZero = 
      Regress[data2, Table[concZero^i, {i, 0, 1}], concZero, 
        RegressionReport -> RSquared]
tVsTime = Fit[data2, Table[concZero^i, {i, 0, 1}], concZero]
Plot[%, {concZero, 0, 1.1}];

Show[%, dataPlot2];

This result differs slightly from that shown on p. 63. Why?