IChO 1990 · Theory problem 7

Hydrogenation of benzene in continuous reactors

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Hydrogenation of benzene to cyclohexane is performed in continuous reactors. The reaction is as follows:

Figure 1

Two basic types of continuous reactors are examined: a continuous plug flow reactor and a continuous stirred reactor.

The continuous plug flow reactor (Fig. 1)

Figure 2

Fig. 1

The reagents are introduced and mixed at the inlet of the reactor (flow F1 and F2). Each slice of the reaction medium (marked zone in the diagram) moves along inside the plug flow reactor without mixing with the others and exits from the reactor (flow F3). When the flow has stabilized, concentrations and conditions are time-independent but dependent on the location in the reactor.

The continuous stirred reactor (Fig. 2)

Figure 3

Fig. 2

The reagents are introduced into the reactor (flows F1 and F2). Inside the reactor they are stirred perfectly and instantaneously. When the flow is stabilized, concentrations and conditions are time-independent and identical at every point of the reactor.

The differential yield of hydrogenation (\(Y\)) is given by \[ Y=-\frac{\mathrm{d}([\ce{C6H12}])}{\mathrm{d}([\ce{C6H6}])}. \]

The proportion of benzene already hydrogenated is given by \[ p=\frac{[\ce{C6H12}]}{[\ce{C6H12}]+[\ce{C6H6}]}. \]

The relationship between \(p\) and \(Y\) is shown in Fig. 3.

Figure 4

Fig. 3

The aim is to hydrogenate almost all the benzene, i.e. \(0.9\le p\le1\) with the best mean yield \[ Y=-\frac{\Delta[\ce{C6H12}]}{\Delta[\ce{C6H6}]}. \]

7.1 What is the value of \(p\) in a continuous stirred reactor leading to the maximal mean yield of hydrogenation? What is the maximal value of \(Y\)?

7.2 For a continuous plug flow reactor, give a rough value of \(Y\) for \(0.9\le p\le1\):

\(Y=0.97,\ 0.98,\ 0.99\) or \(1.00\) ?

7.3 Represent by grey shading in the graph (Fig. 3) the quantities of benzene lost in installation P (plug flow reactor) and installation MP (stirred reactor + plug flow reactor).

7.4 Give the amount of annual benzene lost in installation P and MP assuming an annual consumption of 100 000 tons of benzene.

Official English text of the problem.

Answers

  • 7.1: \(p\approx0.95\); maximal mean yield \(Y_{\max}\approx0.99\), or 99%.
  • 7.2: \(Y\approx0.98\), or 98%.
  • 7.3: For P, shade between \(Y=1\) and the plotted curve over \(0\le p\le1\). For MP, shade between \(Y=1\) and \(Y=0.99\) over \(0\le p\le0.95\), then between \(Y=1\) and the descending plotted curve over \(0.95\le p\le1\).
  • 7.4: Annual benzene losses are approximately \(2.0\times10^3\) tons for P and \(1.0\times10^3\) tons for MP.

Solution

The desired reaction is \[ \ce{C6H6 + 3H2 <=> C6H12} \qquad\text{(catalyst)}. \]

The species are benzene c1ccccc1, hydrogen [H][H], and cyclohexane C1CCCCC1. One mole of benzene gives one mole of cyclohexane in the desired reaction. Thus a differential yield below unity means that some consumed benzene does not appear as cyclohexane.

We write \(\overline Y\) for the mean yield, distinguishing it from the differential yield \(Y(p)\).

7.1

In a perfectly stirred reactor, the composition is uniform. All reaction therefore occurs at the same \(p\), with the same differential yield. Consequently, \[ \overline Y_{\mathrm{stirred}}=Y(p). \]

The maximum of the plotted curve occurs at approximately \[ \boxed{p=0.95,\qquad \overline Y_{\max}=0.99.} \]

7.2

In a plug flow reactor, each slice of material passes successively through compositions ranging from the inlet value \(p=0\) to the outlet value. Its mean yield therefore averages the differential yields encountered along that path.

More precisely, if \(X\) is the fraction of inlet benzene consumed, \[ \overline Y_{\mathrm P} =\frac{1}{X_f}\int_0^{X_f}Y\bigl(p(X)\bigr)\,\mathrm dX. \]

Because the yields are close to unity, \(p\approx X\) is adequate for the graphical estimate requested. Over the principal rising section, the average yield is approximately \[ \overline Y_{\mathrm P} \approx\frac{0.97+0.99}{2} =0.98. \]

Moving the outlet within \(0.9\le p\le1\) changes this only slightly. Hence the appropriate choice is \[ \boxed{\overline Y_{\mathrm P}\approx0.98.} \]

7.3

The quantity \(1-Y\) represents the fraction of an increment of consumed benzene that fails to give cyclohexane. Using \(p\approx X\), its integral is the graphical estimate of the benzene loss.

For the comparison, take both installations to finish at \(p=1\), with the stirred stage of MP operated at its optimum \(p=0.95\).

  • Installation P: Shade the entire region below the horizontal line \(Y=1\) and above the original yield curve, from \(p=0\) to \(p=1\). Its lower boundary rises from \((0,0.97)\) to \((0.95,0.99)\), then falls to \((1,0.985)\).
  • Installation MP: The stirred stage performs its reaction at \(Y=0.99\), so shade the rectangle bounded by \(p=0\), \(p=0.95\), \(Y=0.99\), and \(Y=1\). For the subsequent plug flow stage, add the region between \(Y=1\) and the descending curve from \(p=0.95\) to \(p=1\).

Equivalently, the MP shading is a strip of height \(0.01\) across the full interval \(0\le p\le1\), together with a small triangle beneath it over \(0.95\le p\le1\).

7.4

Reading the graph gives \[ Y(0)\approx0.970,\qquad Y(0.95)\approx0.990,\qquad Y(1)\approx0.985. \]

For P, the shaded area consists of two trapezoids: \[ \begin{aligned} A_{\mathrm P} &=\frac{(1-0.970)+(1-0.990)}{2}(0.95)\\ &\quad+\frac{(1-0.990)+(1-0.985)}{2}(0.05)\\ &=0.019+0.000625\\ &=0.019625. \end{aligned} \]

The graphical estimate of the annual loss is therefore \[ m_{\mathrm{loss,P}} \approx100\,000\ \mathrm{t\,yr^{-1}}\times0.019625 =1962.5\ \mathrm{t\,yr^{-1}}. \]

At the accuracy warranted by the graph, \[ \boxed{m_{\mathrm{loss,P}}\approx2.0\times10^3\ \mathrm{t\,yr^{-1}}.} \]

For MP, the shaded area is a rectangle followed by the same final trapezoid: \[ \begin{aligned} A_{\mathrm{MP}} &=(1-0.990)(0.95) +\frac{(1-0.990)+(1-0.985)}{2}(0.05)\\ &=0.0095+0.000625\\ &=0.010125. \end{aligned} \]

Thus \[ m_{\mathrm{loss,MP}} \approx100\,000\ \mathrm{t\,yr^{-1}}\times0.010125 =1012.5\ \mathrm{t\,yr^{-1}}, \] or appropriately rounded, \[ \boxed{m_{\mathrm{loss,MP}}\approx1.0\times10^3\ \mathrm{t\,yr^{-1}}.} \]

These are masses of benzene lost, so no benzene-to-cyclohexane molar-mass conversion is required.

The shaded-area method is a small-loss approximation: strictly, \(p\) excludes material lost to other products and therefore differs slightly from the actual benzene conversion \(X\). This distinction does not change the quoted rounded answers.

The solution was checked against the official solution.