This is an attempt to clarify the important distinction between space time and residence time in flow reactors. (See Chapter 5 and Problem 5.1 in Levenspiel.) Let's analyze the defining equation for the mean residence time:

t = (Subscript[C, A0]) (∫_0^((Subscript[X, Af])))\(TraditionalForm\`\(\(dX\_A\/\(\((\(-r\_A\))\) \((1 + \(ε\_A\) X\_A)\)\)\)\(\ \)\)\)

cA0 = 1;
k = 1;
epsA1 = 0   ;
epsA2 = -1   ;
epsA3 = 1   ;

timeConstVol1 = 
      cA0*Integrate[
          1/(k*cA0*(1 - xA)/(1 + epsA1*xA)*(1 + 
                    epsA1*xA)), {xA, 0, 0.7}]
timeContr1 = 
      cA0*Integrate[
          1/(k*cA0*(1 - xA)/(1 + epsA2*xA)*(1 + 
                    epsA2*xA)), {xA, 0, 0.7}]
timeExp1 = 
      cA0*Integrate[
          1/(k*cA0*(1 - xA)/(1 + epsA3*xA)*(1 + 
                    epsA3*xA)), {xA, 0, 0.7}]

So, for 1st order rxns the mean residence time is independent of epsA (because epsA is based on the LINEAR volume change assumption)! Therefore, we cannot tell what will happen to the mean residence time (with respect to space time) when there is an expansion of the reacting mixture (as in Levenspiel's Problem 5.1).  

[See, however, Example 2-3 in Missen et al., 1999!? There the authors show, for a 1st order gas-phase reaction A -> R + S, that the residence time is shorter than the space time:]

t = (Subscript[C, A0]) (∫_0^((Subscript[X, Af])))\(TraditionalForm\`dX\_A\/\(\((\(-r\_A\))\) \((1 + \(ε\_A\) X\_A)\)\)\  = 
        1\/k\ \(∫\_0\%\(X\_Af\)\ dX\_A\/\((1 - X\_A)\)\)\)

τ = \(TraditionalForm\`V\/v\_0\  = \ \(C\_A0\/F\_A0\ \(∫\_0\%\(X\_Af\)\ \(\(F\_A0\) dX\_A\)\/\(\(kC\_A0\) 1 - X\_A\/\(1 + \(ε\_A\) X\_A\)\)\)\  = \ 
          1\/k\ \(∫\_0\%\(X\_Af\)\(\(1 + \(ε\_A\) 
                        X\_A\)\/\(1 - X\_A\)\) dX\_A\)\)\)

xAf = 0.2;
k = 0.254;
epsA = 1;
resTime = 
      1/k*Integrate[1/(1 - xA), {xA, 0, xAf}]
tau = 
      1/k*Integrate[(1 + epsA*xA)/(1 - xA), {xA, 0, xAf}]

Let's see what happens when the reaction is of 2nd order.

timeConstVol2 = 
      cA0*Integrate[
          1/(k*cA0^2*(1 - xA)^2/(1 + epsA1*xA)^2*(1 + 
                    epsA1*xA)), {xA, 0, 0.7}]
timeContr2 = 
      cA0*Integrate[
          1/(k*cA0^2*(1 - xA)^2/(1 + epsA2*xA)^2*(1 + 
                    epsA2*xA)), {xA, 0, 0.7}]
timeExp2 = 
      cA0*Integrate[
          1/(k*cA0^2*(1 - xA)^2/(1 + epsA3*xA)^2*(1 + 
                    epsA3*xA)), {xA, 0, 0.7}]

So, if there is contraction, mean t decreases; and if there is expansion, mean t increases (counterintuitive?). How general is this? Not general at all, unfortunately... The conclusion really depends on the conversion dependence of the reaction rate vs. the conversion dependence of the volume. To illustrate this, let's try n=0.5.

timeConstVolHalf = 
      cA0*Integrate[
          1/(k*cA0^0.5*(1 - xA)^0.5/(1 + epsA1*xA)^0.5*(1 + 
                    epsA1*xA)), {xA, 0, 0.7}]
timeContrHalf = 
      cA0*Integrate[
          1/(k*cA0^0.5*(1 - xA)^0.5/(1 + epsA2*xA)^0.5*(1 + 
                    epsA2*xA)), {xA, 0, 0.7}]
timeExpHalf = 
      cA0*Integrate[
          1/(k*cA0^0.5*(1 - xA)^0.5/(1 + epsA3*xA)^0.5*(1 + 
                    epsA3*xA)), {xA, 0, 0.7}]

Now, it's the other way around!? When there is contraction, mean t increases and when there is expansion, mean t decreases (more intuitively appealing?)
We'll perform a more rigorous analysis of this important issue (residence time distribution analysis) when we discuss Chapter 11.
You can now plot t as a function of reaction order for different epsA values.
At this point, we just want to conclude that VISUALIZING EQUATIONS is an important tool in confirming our intuition or in understanding counterintuitive or difficult-to-grasp concepts.

plotConstVol1 = 
        1/(k*cA0*(1 - xA)/(1 + epsA1*xA)*(1 + 
                  epsA1*xA));
plotContr1 = 
        1/(k*cA0*(1 - xA)/(1 + epsA2*xA)*(1 + 
                  epsA2*xA));
plotExp1 = 
        1/(k*cA0*(1 - xA)/(1 + epsA3*xA)*(1 + 
                  epsA3*xA));
Plot[{plotConstVol1, plotContr1, plotExp1}, {xA, 0, 0.7}, 
        PlotStyle -> {Thickness[0.004], Thickness[0.007], 
            Thickness[0.010]}];

plotConstVol2 = 
        1/(k*cA0^2*(1 - xA)^2/(1 + epsA1*xA)^2*(1 + 
                  epsA1*xA));
plotContr2 = 
        1/(k*cA0^2*(1 - xA)^2/(1 + epsA2*xA)^2*(1 + 
                  epsA2*xA));
plotExp2 = 
        1/(k*cA0^2*(1 - xA)^2/(1 + epsA3*xA)^2*(1 + 
                  epsA3*xA));
Plot[{plotConstVol2, plotContr2, plotExp2}, {xA, 0, 0.7}, 
        PlotStyle -> {Thickness[0.004], Thickness[0.007], 
            Thickness[0.010]}];

plotConstVolHalf = 
        1/(k*cA0^0.5*(1 - xA)^0.5/(1 + epsA1*xA)^0.5*(1 + 
                  epsA1*xA));
plotContrHalf = 
        1/(k*cA0^0.5*(1 - xA)^0.5/(1 + epsA2*xA)^0.5*(1 + 
                  epsA2*xA));
plotExpHalf = 
        1/(k*cA0^0.5*(1 - xA)^0.5/(1 + epsA3*xA)^0.5*(1 + 
                  epsA3*xA));
Plot[{plotConstVolHalf, plotContrHalf, plotExpHalf}, {xA, 0, 0.7},  
        PlotStyle -> {Thickness[0.004], Thickness[0.007], 
            Thickness[0.010]}];