This exercise illustrates the following important statement on p. 135 (Levenspiel, CRE^3):
"For ... any n-th order reaction, n>0 ... the reactors should be ... ordered so as to keep the concentration of reactant as high as possible if the rate-concentration curve is concave (n>1), and as low as possible if the curve is convex (n<1)."
It is a follow-up on Problem #4 on Exam #1.

dimTime = 2.0;   
cA0 = 4;  
n1 = 0.5;
n2 = 1.0;
n3 = 1.5;

Let's first re-analyze the unique n=1.0 case (Exam #1). First PFR/MFR:

c1FirstPFR1st = 
      Solve[dimTime == Log[cA0/cA], cA] // N
cA1FirstPFR1st = cA /. First[c1FirstPFR1st]
xA1FirstPFR1st = 1 - cA1FirstPFR1st/cA0

c2FirstPFR1st = 
      Solve[dimTime == (cA1FirstPFR1st - cA2)/cA2, 
        cA2]
cA2FirstPFR1st = cA2 /. First[c2FirstPFR1st]
xAfFirstPFR1st = 1 - cA2FirstPFR1st/cA0

xA1FirstPFR1st
xAfFirstPFR1st

This is n=1.0 case, MFR/PFR:

c1FirstMFR1st = 
      Solve[dimTime == (cA0 - cA)/cA, cA] // 
        N
cA1FirstMFR1st = cA /. First[c1FirstMFR1st]
xA1FirstMFR1st = 1 - cA1FirstMFR1st/cA0

c2FirstMFR1st = 
      Solve[dimTime == Log[cA1FirstMFR1st/cA2], 
        cA2]
cA2FirstMFR1st = cA2 /. First[c2FirstMFR1st]
xAfFirstMFR1st = 1 - cA2FirstMFR1st/cA0

xA1FirstMFR1st
xAfFirstMFR1st

Let's analyze now the n=0.5 case (convex rA vs. xA curve). First PFR/MFR:

c1HalfPFR1st = 
      Solve[(n1 - 1)*(cA0^(n1 - 1))*
            dimTime == ((cA/cA0)^(1 - n1)) - 1, 
        cA]
cA1HalfPFR1st = cA /. First[c1HalfPFR1st]
xA1HalfPFR1st = 1 - cA1HalfPFR1st/cA0

c2HalfPFR1st = 
      Solve[dimTime == (cA1HalfPFR1st - cA2)/(cA2^n1), 
        cA2]
cA2HalfPFR1st = cA2 /. First[c2HalfPFR1st]
xAfHalfPFR1st = 1 - cA2HalfPFR1st/cA0

xA1HalfPFR1st
xAfHalfPFR1st

This is n=0.5 case, MFR/PFR:

c1HalfMFR1st = 
      Solve[dimTime == (cA0 - cA)/cA^n1, 
        cA]
cA1HalfMFR1st = cA /. First[c1HalfMFR1st]
xA1HalfMFR1st = 1 - cA1HalfMFR1st/cA0

c2HalfMFR1st = 
      Solve[(n1 - 1)*cA1HalfMFR1st^(n1 - 1)*
            dimTime == (cA2/cA1HalfMFR1st)^(1 - n1) - 1, 
        cA2]
cA2HalfMFR1st = cA2 /. First[c2HalfMFR1st]
xAfHalfMFR1st = 1 - cA2HalfMFR1st/cA0

xA1HalfMFR1st
xAfHalfMFR1st

Let's now analyze the n=1.5 case (concave rA vs. xA curve). First PFR/MFR:

c1ThreeHalvesPFR1st = 
      Solve[(n3 - 1)*cA0^(n3 - 1)*
            dimTime == (cA/cA0)^(1 - n3) - 1, 
        cA]
cA1ThreeHalvesPFR1st = 
      cA /. First[c1ThreeHalvesPFR1st]
xA1ThreeHalvesPFR1st = 1 - cA1ThreeHalvesPFR1st/cA0

c2ThreeHalvesPFR1st = 
      Solve[dimTime == (cA1ThreeHalvesPFR1st - cA2)/cA2^n3, 
        cA2]
cA2ThreeHalvesPFR1st = 
      cA2 /. First[c2ThreeHalvesPFR1st]
xAfThreeHalvesPFR1st = 1 - cA2ThreeHalvesPFR1st/cA0

xA1ThreeHalvesPFR1st
xAfThreeHalvesPFR1st

Now MFR/PFR, n=1.5:

c1ThreeHalvesMFR1st = 
      Solve[dimTime == (cA0 - cA)/cA^n3, 
        cA]
cA1ThreeHalvesMFR1st = 
      cA /. First[c1ThreeHalvesMFR1st]
xA1ThreeHalvesMFR1st = 1 - cA1ThreeHalvesMFR1st/cA0

c2ThreeHalvesMFR1st = 
      Solve[(1.5 - 1)*cA1ThreeHalvesMFR1st^(1.5 - 1)*
            dimTime == (cA2/cA1ThreeHalvesMFR1st)^(1 - 1.5) - 
            1, cA2]
cA2ThreeHalvesMFR1st = 
      cA2 /. First[c2ThreeHalvesMFR1st]
xAfThreeHalvesMFR1st = 1 - cA2ThreeHalvesMFR1st/cA0

xA1ThreeHalvesMFR1st
xAfThreeHalvesMFR1st

A graph of -1/rA vs. xA can explain all this in an "illuminating" way.

k = 20;
Plot[{k*cA^n1, k*cA^n2, k*cA^n3}, {cA, 0, 10^3}, 
        PlotRange -> {0, 5000}, 
        PlotStyle -> {Hue[0.01], Hue[0.3], Hue[0.6]}];

As hue # increases, go from red to green to blue (RGB)... Which curve is above and which is below the 1st order straight line depends, of course, on whether cA is <1 or >1. But n>1 is concave (w/r to x-axis) and n<1 is convex (w/r to x-axis).

Plot[
        1/(k*(cA0*(1 - xA))^n1), {xA, 0, 
          0.99}];
Plot[1/(k*(cA0*(1 - xA))^n2), {xA, 0, 0.99}, 
        PlotRange -> {0, 0.3000}];
Plot[1/(k*(cA0*(1 - xA))^n3), {xA, 0, 0.99}, 
        PlotRange -> {0, 0.3000}];

Let's now find the tau values. First for PFR/MFR, n=0.5:

areaPFR1 = 
      Integrate[
        1/(k*(cA0*(1 - xA))^n1), {xA, 0, 
          xA1HalfPFR1st}]
recipRate1 = 
      1/(k*(cA0*(1 - xA))^n1) /. 
        xA -> xAfHalfPFR1st
areaMFR1 = %*(xAfHalfPFR1st - 
            xA1HalfPFR1st)
areaTotal1 = areaPFR1 + areaMFR1
xA1HalfPFR1st
xAfHalfPFR1st

Next for MFR/PFR, n=0.5:

recipRate2 = 
      1/(k*(cA0*(1 - xA))^n1) /. 
        xA -> xA1HalfMFR1st
areaMFR2 = %*(xA1HalfMFR1st - 0)
areaPFR2 = 
      Integrate[
        1/(k*(cA0*(1 - xA))^n1), {xA, xA1HalfMFR1st, 
          xAfHalfMFR1st}]
areaTotal2 = areaMFR2 + areaPFR2
xA1HalfMFR1st
xAfHalfMFR1st

Next for PFR/MFR, n=1.0:

areaPFR3 = 
      NIntegrate[
        1/(k*(cA0*(1 - xA))^n2), {xA, 0, 
          xA1FirstPFR1st}]
recipRate3 = 
      1/(k*(cA0*(1 - xA))^n2) /. 
        xA -> xAfFirstPFR1st
areaMFR3 = %*(xAfFirstPFR1st - 
            xA1FirstPFR1st)
areaTotal3 = areaPFR3 + areaMFR3
xA1FirstPFR1st
xAfFirstPFR1st

Next for MFR/PFR, n=1.0:

recipRate4 = 
      1/(k*(cA0*(1 - xA))^n2) /. 
        xA -> xA1FirstMFR1st
areaMFR4 = %*(xA1FirstMFR1st - 0)
areaPFR4 = 
      NIntegrate[
        1/(k*(cA0*(1 - xA))^n2), {xA, xA1FirstMFR1st, 
          xAfFirstMFR1st}]
areaTotal4 = areaMFR4 + areaPFR4
xA1FirstMFR1st
xAfFirstMFR1st

Next for PFR/MFR, n=1.5:

areaPFR5 = 
      Integrate[
        1/(k*(cA0*(1 - xA))^n3), {xA, 0,  
          xA1ThreeHalvesPFR1st}]
recipRate5 = 
      1/(k*(cA0*(1 - xA))^n3) /. 
        xA -> xAfThreeHalvesPFR1st
areaMFR5 = %*(xAfThreeHalvesPFR1st - 
            xA1ThreeHalvesPFR1st)
areaTotal5 = areaPFR5 + areaMFR5
xA1ThreeHalvesPFR1st
xAfThreeHalvesPFR1st

And finally for MFR/PFR, n=1.5:

recipRate6 = 
      1/(k*(cA0*(1 - xA))^n3) /. 
        xA -> xA1ThreeHalvesMFR1st
areaMFR6 = %*(xA1ThreeHalvesMFR1st - 0)
areaPFR6 = 
      Integrate[
        1/(k*(cA0*(1 - xA))^n3), {xA, xA1ThreeHalvesMFR1st,  
          xAfThreeHalvesMFR1st}]
areaTotal6 = areaMFR6 + areaPFR6
xA1ThreeHalvesMFR1st
xAfThreeHalvesMFR1st

Plot[{1/(k*cA0^n1*(1 - xA)^n1), 
          1/(k*(cA0*(1 - xA))^n2), 
          1/(k*cA0^1.5*(1 - xA)^n3)}, {xA, 0, 0.95}, 
        PlotRange -> {0, 0.2}, 
        PlotStyle -> {Hue[0.01], Hue[0.3], Hue[0.6]}];

From the shapes of these curves, it is now obvious that for n<1 (uppermost curve; convex rate vs. cA), at low conversions there is not much difference between mixed flow and plug flow: the area under the curve is not much smaller than the area of the corresponding rectangle. So mixed flow (low cA) is OK. And at high conversion, when MFR area would be much larger than PFR area, it is important to have plug flow. In contrast, when n>1 (lowest curve; concave rate vs. cA) it is important to keep cA as high as possible (plug flow).