IChO 1983 · Theory problem 5

Ethanol dehydration equilibrium and its pressure dependence

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On passing ethanol over a catalyst at 400 K, a dehydration reaction occurs resulting in the formation of ethylene:

Figure 1

At the above temperature and \(p_0 = 101.325\ \mathrm{kPa}\), the conversion of ethyl alcohol is 90.6 mol %.

5.1 Calculate the equilibrium constant \(K_p\) of the reaction under given conditions.

5.2 Calculate the values of the equilibrium constants \(K_x\) and \(K_c\) at the above temperature.

5.3 Calculate the ethanol conversion at the following pressures:

\(5p_0\), \(10p_0\), \(50p_0\), \(100p_0\), and \(200p_0\).

5.4 Plot the graph for the variation of conversion vs. pressure.

Official English text of the problem.

Answers

The constants below use the unnormalized pressure and concentration quotients defined in the solution.

  • 5.1 \(\boxed{K_p=4.64\times10^2\ \mathrm{kPa}=4.58\ \mathrm{atm}}\).
  • 5.2 At \(p=p_0\), \(\boxed{K_x=4.58}\); at \(400\ \mathrm{K}\), \(\boxed{K_c=0.140\ \mathrm{mol\,L^{-1}}}\).
  • 5.3 The equilibrium conversions at \(5p_0,\ 10p_0,\ 50p_0,\ 100p_0,\ 200p_0\) are, respectively, \(\boxed{69.1\%,\ 56.1\%,\ 29.0\%,\ 20.9\%,\ 15.0\%}\).
  • 5.4 Plot \[ \boxed{\text{Conversion (\%)}= 100\sqrt{\frac{4.58148}{4.58148+p/p_0}}}. \] The curve decreases smoothly and is convex upwards, approaching \(100\%\) as \(p\to0^+\) and \(0\%\) as \(p\to\infty\).

Solution

5.1 — Pressure equilibrium constant

The balanced equilibrium is

\[ \ce{C2H5OH(g) <=> C2H4(g) + H2O(g)}. \]

The species are ethanol (CCO), ethylene, also called ethene (C=C), and water (O).

Assume a pure ethanol feed, ideal-gas behavior, and that the stated pressures are total equilibrium pressures. Let \(\alpha\) be the fraction of the initial ethanol converted. Starting with \(1\ \mathrm{mol}\) of ethanol gives:

Species Equilibrium amount (\(\mathrm{mol}\)) Mole fraction
Ethanol \(1-\alpha\) \(\dfrac{1-\alpha}{1+\alpha}\)
Ethylene \(\alpha\) \(\dfrac{\alpha}{1+\alpha}\)
Water \(\alpha\) \(\dfrac{\alpha}{1+\alpha}\)
Total \(1+\alpha\) \(1\)

Since \(p_i=x_i p\), the pressure quotient is

\[ K_p= \frac{p_{\ce{C2H4}}p_{\ce{H2O}}} {p_{\ce{C2H5OH}}} = \frac{\left(\dfrac{\alpha p}{1+\alpha}\right)^2} {\dfrac{(1-\alpha)p}{1+\alpha}} = \frac{\alpha^2p}{1-\alpha^2}. \]

At \(p=p_0=101.325\ \mathrm{kPa}\), the conversion is \(\alpha_0=0.906\). Thus

\[ K_p= \frac{0.906^2}{1-0.906^2} (101.325\ \mathrm{kPa}) =464.218\ \mathrm{kPa}. \]

Therefore,

\[ \boxed{K_p=464\ \mathrm{kPa}=4.58\ \mathrm{atm}}. \]

5.2 — Mole-fraction and concentration equilibrium constants

The mole-fraction quotient is

\[ K_x= \frac{x_{\ce{C2H4}}x_{\ce{H2O}}} {x_{\ce{C2H5OH}}} = \frac{\alpha^2}{1-\alpha^2}. \]

At the specified reference pressure,

\[ K_x(p_0)= \frac{0.906^2}{1-0.906^2} =4.58148, \]

so

\[ \boxed{K_x(p_0)=4.58}. \]

Here \(K_x\) depends on total pressure: at fixed temperature,

\[ K_x(p)=\frac{K_p}{p}. \]

For the concentration quotient,

\[ K_c= \frac{c_{\ce{C2H4}}c_{\ce{H2O}}} {c_{\ce{C2H5OH}}}. \]

The change in the number of gaseous moles is

\[ \Delta\nu=(1+1)-1=1. \]

Using \(p_i=c_iRT\),

\[ K_p=K_c(RT)^{\Delta\nu}=K_cRT. \]

With \(R=8.31446\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}\),

\[ K_c= \frac{464.218\ \mathrm{kPa}} {(8.31446\ \mathrm{kPa\,L\,mol^{-1}\,K^{-1}}) (400\ \mathrm{K})} =0.139582\ \mathrm{mol\,L^{-1}}. \]

Hence

\[ \boxed{K_c=0.140\ \mathrm{mol\,L^{-1}}}. \]

5.3 — Conversion at the specified pressures

At constant temperature, \(K_p\) remains unchanged. Rearranging

\[ K_p=\frac{\alpha^2p}{1-\alpha^2} \]

gives

\[ K_p=\alpha^2(K_p+p), \qquad \boxed{\alpha=\sqrt{\frac{K_p}{K_p+p}}}. \]

Writing \(r=p/p_0\), and using \(K_p/p_0=4.58148\),

\[ \alpha(r)=\sqrt{\frac{4.58148}{4.58148+r}}. \]

The resulting conversions are:

Pressure Pressure (\(\mathrm{kPa}\)) Ethanol conversion (mol %)
\(5p_0\) \(506.625\) \(69.1\)
\(10p_0\) \(1013.25\) \(56.1\)
\(50p_0\) \(5066.25\) \(29.0\)
\(100p_0\) \(10132.5\) \(20.9\)
\(200p_0\) \(20265\) \(15.0\)

The forward reaction converts one mole of gas into two. Increasing pressure therefore favors ethanol and reduces its equilibrium conversion.

5.4 — Description of the graph

Use total pressure \(p/p_0\) on the horizontal axis, from \(0\) to \(200\), and ethanol conversion in mol % on the vertical axis, from \(0\) to \(100\).

Draw the smooth curve

\[ y=100\sqrt{\frac{4.58148}{4.58148+p/p_0}} \]

through the reference point \((1,90.6)\) and the five calculated points:

\[ (5,69.1),\quad (10,56.1),\quad (50,29.0),\quad (100,20.9),\quad (200,15.0). \]

The curve approaches \((0,100)\) as pressure tends to zero. It falls steeply at low pressure and progressively flattens; it is convex upwards throughout. At indefinitely increasing pressure, conversion approaches zero asymptotically.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.