IChO 2002 · Theory problem 4

Production of Methanol

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Methanol (\(\ce{CH3OH}\)) is a chemical that is used for the production of additives in gasoline and many common plastics. A factory, producing methanol, is based on the reaction:

Figure 1

Hydrogen and \(\ce{CO}\) are obtained by the reaction:

Figure 2

The three units of the factory, namely, the “reformer” for the hydrogen / carbon monoxide production, the “methanol reactor” and a “separator” to separate methanol from \(\ce{CO}\) and \(\ce{H2}\), are schematically shown in Figure 1. Four positions are indicated by \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\).

Figure 3

Figure 1

The flow of methanol at position \(\gamma\) is \(n[\ce{CH3OH},\gamma]=1000\ \mathrm{mol\,s^{-1}}\). The factory is so designed that 2/3 of the \(\ce{CO}\) is converted to methanol. Excess \(\ce{CO}\) and \(\ce{H2}\) at position \(\delta\) are used to heat the first reactor. Assume that the reformer reaction goes to completion.

4.1 Calculate the flow of \(\ce{CO}\) and \(\ce{H2}\) at position \(\beta\).

4.2 Calculate the flow of \(\ce{CO}\) and \(\ce{H2}\) at position \(\gamma\).

4.3 Calculate the flows of \(\ce{CH4}\) and \(\ce{H2O}\) needed at position \(\alpha\).

4.4 At point \(\gamma\) all species are gases. Calculate the partial pressures in MPa for \(\ce{CO}\), \(\ce{H2}\) and \(\ce{CH3OH}\) at position \(\gamma\) using the equation:

\[ p_i=p\frac{n_i}{n_{\mathrm{tot}}} \]

wherein \(n_i\) is the flow and \(p_i\) the partial pressure of the compound \(i\), \(n_{\mathrm{tot}}\) is the total flow at the position considered, and \(p\) the total pressure in the system. (\(p=10\ \mathrm{MPa}\))

When the methanol reactor is large enough the reaction goes to equilibrium. The partial pressures at point \(\gamma\) obey the equation:

\[ K_p=\frac{p_{\ce{CH3OH}}p_0^2}{p_{\ce{CO}}p_{\ce{H2}}^2} \]

wherein \(p_0\) is a constant (0.1 MPa) and \(K_p\) is a function of temperature as is shown in Figure 2 (the vertical scale is logarithmic).

Figure 4

Figure 2

4.5 Calculate \(K_p\) and indicate at which temperature \(T\) the reaction must be operated to achieve this equilibrium.

Official English text of the problem.

Answers

Molecular names and SMILES used below: carbon monoxide, \(\ce{CO}\), [C-]#[O+]; hydrogen, \(\ce{H2}\), [H][H]; methane, \(\ce{CH4}\), C; water, \(\ce{H2O}\), O; methanol, \(\ce{CH3OH}\), CO.

  • 4.1: \(n[\ce{CO},\beta]=1.50\times10^3\ \mathrm{mol\,s^{-1}}\); \(n[\ce{H2},\beta]=4.50\times10^3\ \mathrm{mol\,s^{-1}}\).
  • 4.2: \(n[\ce{CO},\gamma]=5.00\times10^2\ \mathrm{mol\,s^{-1}}\); \(n[\ce{H2},\gamma]=2.50\times10^3\ \mathrm{mol\,s^{-1}}\).
  • 4.3: \(n[\ce{CH4},\alpha]=n[\ce{H2O},\alpha]=1.50\times10^3\ \mathrm{mol\,s^{-1}}\).
  • 4.4: At \(\gamma\), \(p_{\ce{CO}}=1.25\ \mathrm{MPa}\), \(p_{\ce{H2}}=6.25\ \mathrm{MPa}\), and \(p_{\ce{CH3OH}}=2.50\ \mathrm{MPa}\).
  • 4.5: \(K_p=5.12\times10^{-4}\), dimensionless; \(T\approx640\ \mathrm{K}\), read from Figure 2.

Solution

4.1 — Flows entering the methanol reactor

The balanced methanol synthesis reaction is

\[ \ce{CO + 2H2 <=> CH3OH}. \]

Each mole of methanol formed consumes one mole of carbon monoxide. Thus, producing \(1000\ \mathrm{mol\,s^{-1}}\) of methanol consumes \(1000\ \mathrm{mol\,s^{-1}}\) of carbon monoxide. Since this represents \(2/3\) of the incoming carbon monoxide,

\[ \frac{2}{3}\,n[\ce{CO},\beta] =1000\ \mathrm{mol\,s^{-1}}, \]

giving

\[ \boxed{n[\ce{CO},\beta]=1500\ \mathrm{mol\,s^{-1}}}. \]

The balanced reforming reaction, taken to completion as instructed, is

\[ \ce{CH4 + H2O -> CO + 3H2}. \]

It produces three moles of hydrogen for each mole of carbon monoxide. Therefore,

\[ n[\ce{H2},\beta]=3\,n[\ce{CO},\beta] =3(1500), \]

\[ \boxed{n[\ce{H2},\beta]=4500\ \mathrm{mol\,s^{-1}}}. \]

4.2 — Flows leaving the methanol reactor

Subtract the quantities consumed in forming methanol:

\[ n[\ce{CO},\gamma] =(1500-1000)\ \mathrm{mol\,s^{-1}} =\boxed{500\ \mathrm{mol\,s^{-1}}}, \]

\[ n[\ce{H2},\gamma] =(4500-2\times1000)\ \mathrm{mol\,s^{-1}} =\boxed{2500\ \mathrm{mol\,s^{-1}}}. \]

The methanol flow is the given \(1000\ \mathrm{mol\,s^{-1}}\).

4.3 — Required methane and water feeds

The reforming reaction consumes one mole each of methane and water per mole of carbon monoxide produced. Complete reforming must supply \(1500\ \mathrm{mol\,s^{-1}}\) of carbon monoxide, so the required stoichiometric feeds are

\[ \boxed{n[\ce{CH4},\alpha]=1500\ \mathrm{mol\,s^{-1}}}, \]

\[ \boxed{n[\ce{H2O},\alpha]=1500\ \mathrm{mol\,s^{-1}}}. \]

4.4 — Partial pressures at \(\gamma\)

The total gas flow is

\[ n_{\mathrm{tot},\gamma} =(500+2500+1000)\ \mathrm{mol\,s^{-1}} =4000\ \mathrm{mol\,s^{-1}}. \]

Using \(p_i=p\,n_i/n_{\mathrm{tot}}\) at \(p=10\ \mathrm{MPa}\),

\[ p_{\ce{CO}} =(10\ \mathrm{MPa})\frac{500}{4000} =\boxed{1.25\ \mathrm{MPa}}, \]

\[ p_{\ce{H2}} =(10\ \mathrm{MPa})\frac{2500}{4000} =\boxed{6.25\ \mathrm{MPa}}, \]

\[ p_{\ce{CH3OH}} =(10\ \mathrm{MPa})\frac{1000}{4000} =\boxed{2.50\ \mathrm{MPa}}. \]

These partial pressures sum to \(10.00\ \mathrm{MPa}\), as required.

4.5 — Equilibrium constant and temperature

Substitute the partial pressures and \(p_0=0.1\ \mathrm{MPa}\) into the given expression:

\[ \begin{aligned} K_p &=\frac{p_{\ce{CH3OH}}p_0^2} {p_{\ce{CO}}p_{\ce{H2}}^2}\\[2mm] &=\frac{(2.50\ \mathrm{MPa})(0.1\ \mathrm{MPa})^2} {(1.25\ \mathrm{MPa})(6.25\ \mathrm{MPa})^2}\\[2mm] &=\boxed{5.12\times10^{-4}}. \end{aligned} \]

The pressure units cancel, so \(K_p\) is dimensionless.

On the logarithmic vertical axis of Figure 2, this value lies between \(10^{-3}\) and \(10^{-4}\), at

\[ \log_{10}K_p=-3.291. \]

Following this level horizontally to the curve and then vertically down to the temperature axis gives

\[ \boxed{T\approx640\ \mathrm{K}}. \]

The solution was checked against the official solution.