IChO 1989 · Theory problem 2

Synthesis gas: compression work and the CO conversion equilibrium

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A mixture of gases containing mainly carbon monoxide and hydrogen is produced by the reaction of alkanes with steam:

Figure 1

2.1 Using equations (1) and (2) write down an overall reaction (3) so that the net enthalpy change is zero.

2.2 The synthesis of methanol from carbon monoxide and hydrogen is carried out either

a) in two steps, where the starting mixture corresponding to equation (3) is compressed from \(0.1\times10^6\ \mathrm{Pa}\) to \(3\times10^6\ \mathrm{Pa}\), and the mixture of products thereof compressed again from \(3\times10^6\ \mathrm{Pa}\) to \(6\times10^6\ \mathrm{Pa}\)

or

b) in one step, where the mixture of products corresponding to equation (3) is compressed from \(0.1\times10^6\ \mathrm{Pa}\) to \(6\times10^6\ \mathrm{Pa}\).

Calculate the work of compression, \(W_a\), according to the two step reaction for \(100\ \mathrm{cm^3}\) of starting mixture and calculate the difference in the work of compression between the reactions 1 and 2.

Assume for calculations a complete reaction at constant pressure. Temperature remains constant at \(500\ \mathrm{K}\), ideal gas behaviour is assumed.

To produce hydrogen for the synthesis of ammonia, a mixture of \(40.0\ \mathrm{mol}\) CO and \(40.0\ \mathrm{mol}\) of hydrogen, \(18.0\ \mathrm{mol}\) of carbon dioxide and \(2.0\ \mathrm{mol}\) of nitrogen are in contact with \(200.0\ \mathrm{mol}\) of steam in a reactor where the conversion equilibrium is established.

Figure 2

2.3 Calculate the number of moles of each gas leaving the reactor.

Official English text of the problem.

Answers

  • 2.1 — Using the enthalpies exactly as printed: \[ \boxed{\ce{5CH4 + 3O2 -> 5CO + 9H2 + H2O}},\qquad \Delta H=0. \] This is \(6(1)-(2)\), requiring reversal of reaction (2). No nonzero combination of both reactions in the forward direction has zero enthalpy with the printed positive values.

If equation (1) was intended to have \(\Delta H=-36\ \mathrm{kJ\,mol^{-1}}\): \[ \boxed{\ce{7CH4 + 3O2 + H2O -> 7CO + 15H2}},\qquad \Delta H=0. \]

  • 2.2 — Reversible isothermal compression, with work positive when supplied to the gas:
Enthalpy interpretation \(W_a\) \(W_b\) Saving, \(W_b-W_a\)
Values as printed \(\boxed{47.0\ \mathrm{J}}\) \(76.8\ \mathrm{J}\) \(\boxed{29.8\ \mathrm{J}}\)
If \(\Delta H_1=-36\ \mathrm{kJ\,mol^{-1}}\) \(\boxed{47.9\ \mathrm{J}}\) \(81.9\ \mathrm{J}\) \(\boxed{34.0\ \mathrm{J}}\)
  • 2.3 — A numerical composition cannot be determined from the data given: an equilibrium constant or equivalent datum is missing. If \(x\ \mathrm{mol}\) undergoes forward conversion, \[ \boxed{ \begin{aligned} n_{\ce{CO}}&=(40.0-x)\ \mathrm{mol},\\ n_{\ce{H2}}&=(40.0+x)\ \mathrm{mol},\\ n_{\ce{CO2}}&=(18.0+x)\ \mathrm{mol},\\ n_{\ce{N2}}&=2.0\ \mathrm{mol},\\ n_{\ce{H2O}}&=(200.0-x)\ \mathrm{mol}. \end{aligned}} \]

Solution

Molecular names and SMILES used here are: methane C; molecular oxygen O=O; carbon monoxide [C-]#[O+]; hydrogen [H][H]; water, including steam, O; methanol CO; carbon dioxide O=C=O; nitrogen N#N; and ammonia N.

2.1

The supplied reactions are \[ \ce{CH4 + 1/2 O2 -> CO + 2H2}, \qquad \Delta H_1=+36\ \mathrm{kJ\,mol^{-1}}, \] and \[ \ce{CH4 + H2O -> CO + 3H2}, \qquad \Delta H_2=+216\ \mathrm{kJ\,mol^{-1}}. \]

For a linear combination \(a(1)+b(2)\) to have zero enthalpy, \[ 36a+216b=0. \] Thus \(a=-6b\): the two coefficients must have opposite signs.

Choosing \(a=6\) and \(b=-1\), multiply reaction (1) by six and reverse reaction (2): \[ \ce{6CH4 + 3O2 -> 6CO + 12H2}, \] \[ \ce{CO + 3H2 -> CH4 + H2O}. \] Adding and cancelling gives \[ \boxed{\ce{5CH4 + 3O2 -> 5CO + 9H2 + H2O}}. \tag{3, printed data} \] Its enthalpy is \[ \Delta H=6(36)-216=0. \]

If the first enthalpy was intended to be negative, six forward partial-oxidation reactions instead supply the heat for one forward steam-reforming reaction: \[ 6(-36)+216=0. \] Then \(6(1)+(2)\) gives \[ \boxed{\ce{7CH4 + 3O2 + H2O -> 7CO + 15H2}}. \tag{3, conditional correction} \]

2.2

Methanol formation is \[ \ce{CO + 2H2 -> CH3OH}. \] The compression comparison explicitly refers to the reactant and product mixtures of reaction (3). Accordingly:

  • Route (a) compresses the reactants to \(3\times10^6\ \mathrm{Pa}\), completes reaction (3) at that pressure, then compresses its products to \(6\times10^6\ \mathrm{Pa}\).
  • Route (b) completes reaction (3) at the initial pressure, then compresses its products directly to \(6\times10^6\ \mathrm{Pa}\).

Both routes are compared for the same initial \(100\ \mathrm{cm^3}\) reactant charge.

For reversible isothermal compression of an ideal gas, the work supplied is \[ W=-\int_{V_i}^{V_f}p\,dV =nRT\ln\!\left(\frac{V_i}{V_f}\right) =nRT\ln\!\left(\frac{p_f}{p_i}\right). \]

Let \(n_0\) be the initial amount and \(r=n_{\mathrm{products}}/n_0\) the mole-number ratio for complete reaction (3). From the initial state, \[ n_0RT=p_0V_0 =(0.1\times10^6\ \mathrm{Pa})(100\times10^{-6}\ \mathrm{m^3}) =10\ \mathrm{J}. \] Thus the specified temperature need not be substituted separately.

For route (a), \[ \begin{aligned} W_a &=n_0RT\ln\!\left(\frac{3\times10^6}{0.1\times10^6}\right) +rn_0RT\ln\!\left(\frac{6\times10^6}{3\times10^6}\right)\\ &=(10\ \mathrm{J})[\ln30+r\ln2]. \end{aligned} \]

For route (b), \[ W_b=rn_0RT\ln\!\left(\frac{6\times10^6}{0.1\times10^6}\right) =(10\ \mathrm{J})r\ln60. \]

Therefore, \[ \boxed{W_b-W_a=(10\ \mathrm{J})(r-1)\ln30}. \] Route (a) saves work because fewer moles undergo the initial compression.

With the printed enthalpies, reaction (3) converts \(5+3=8\) moles of reactants into \(5+9+1=15\) moles of products, so \[ r=\frac{15}{8}. \] Consequently, \[ W_a=10\left[\ln30+\frac{15}{8}\ln2\right]\ \mathrm{J} =\boxed{47.0\ \mathrm{J}}, \] \[ W_b=10\left(\frac{15}{8}\right)\ln60\ \mathrm{J} =76.8\ \mathrm{J}, \] \[ \boxed{W_b-W_a=29.8\ \mathrm{J}}. \]

With the conditional negative sign for \(\Delta H_1\), reaction (3) converts \(7+3+1=11\) moles into \(7+15=22\) moles, giving \(r=2\). Hence \[ \boxed{W_a=10[\ln30+2\ln2]\ \mathrm{J}=47.9\ \mathrm{J}}, \] \[ W_b=20\ln60\ \mathrm{J}=81.9\ \mathrm{J}, \qquad \boxed{W_b-W_a=34.0\ \mathrm{J}}. \]

These are compressor works; they do not include the expansion work during the constant-pressure chemical reaction.

2.3

The water-gas shift equilibrium is \[ \ce{CO + H2O <=> CO2 + H2}. \]

Let \(x\ \mathrm{mol}\) react in the forward direction. Stoichiometry gives:

Gas Initial amount / mol Change / mol Outlet amount / mol
Carbon monoxide, \(\ce{CO}\) \(40.0\) \(-x\) \(40.0-x\)
Hydrogen, \(\ce{H2}\) \(40.0\) \(+x\) \(40.0+x\)
Carbon dioxide, \(\ce{CO2}\) \(18.0\) \(+x\) \(18.0+x\)
Nitrogen, \(\ce{N2}\) \(2.0\) \(0\) \(2.0\)
Water vapour, \(\ce{H2O}\) \(200.0\) \(-x\) \(200.0-x\)

Nitrogen is unchanged. The total gas amount remains \(300.0\ \mathrm{mol}\), since the reaction has no net change in gaseous mole number.

For ideal gases, the equilibrium condition is \[ K_p =\frac{p_{\ce{CO2}}p_{\ce{H2}}} {p_{\ce{CO}}p_{\ce{H2O}}} =\frac{(18.0+x)(40.0+x)} {(40.0-x)(200.0-x)}. \] The total pressure cancels. Equivalently, \[ (1-K_p)x^2+(58.0+240.0K_p)x+720.0-8000.0K_p=0. \]

A supplied value of \(K_p\) would determine the physically admissible root, with \[ -18.0<x<40.0 \] for a finite positive equilibrium constant. Negative \(x\) would signify net reverse conversion.

However, the problem gives neither \(K_p\) nor equivalent data from which it can be calculated. The initial reaction quotient, \[ Q_{p,0}=\frac{18.0\times40.0}{40.0\times200.0}=0.0900, \] does not establish the equilibrium constant. Therefore, the outlet amounts can only be stated in the parameterized form above. The complete-reaction assumption in 2.2 does not determine this separately specified equilibrium.

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.