IChO 1987 · Theory problem 1

Carbohydrate oxidation and fermentation in a sewage plant

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Treating waste water in a sewage plant, 45 % of its carbohydrate \(\ce{(CH2O)_n}\) is completely oxidized, 10 % undergoes anaerobic decomposition by fermentation (two components) and the rest remains in the sludge. The total gas formation is \(16\ \mathrm{m^3}\) per day (\(25\ ^\circ\mathrm{C}\), \(100\ \mathrm{kPa}\)).

1.1 What is the amount of carbohydrate remaining in the sludge measured in kg per day?

1.2 Using the heat of combustion of methane (\(-882\ \mathrm{kJ\,mol^{-1}}\)), calculate the amount of energy that can be produced by combustion of the methane formed per day.

1.3 Knowing that the concentration of the carbohydrate in the waste water is \(250\ \mathrm{mg\,dm^{-3}}\), calculate the daily amount of waste water processed in the plant in \(\mathrm{m^3}\) of water per day.

Official English text of the problem.

Answers

To two significant figures:

  • 1.1: \(\boxed{16\ \mathrm{kg\,day^{-1}}}\) of carbohydrate remains in the sludge.
  • 1.2: Methane combustion releases \(\boxed{52\ \mathrm{MJ\,day^{-1}}}\).
  • 1.3: The plant processes \(\boxed{1.4\times10^2\ \mathrm{m^3\,day^{-1}}}\) of waste water.

Solution

The molecular species involved are oxygen O=O, carbon dioxide O=C=O, water O, and methane C. The carbohydrate formula \(\ce{(CH2O)_n}\) specifies an empirical composition, so no unique molecular structure or SMILES can be assigned to it. Here, a “\(\ce{CH2O}\) unit” is a stoichiometric counting unit, not a separate molecule.

1.1 — Carbohydrate remaining in the sludge

Complete oxidation produces carbon dioxide and water:

\[ \ce{(CH2O)_n + n O2 -> n CO2 + n H2O}. \]

Part 1.2 identifies methane as a fermentation product; the other gaseous product is carbon dioxide. The average oxidation state of carbon in the carbohydrate is zero. Formation of methane, with carbon at \(-4\), balances formation of an equal amount of carbon dioxide, with carbon at \(+4\):

\[ \ce{2 (CH2O)_n -> n CH4 + n CO2}. \]

Thus, each mole of carbohydrate composition units produces:

  • By complete oxidation: \(1\ \mathrm{mol}\) of \(\ce{CO2}\).
  • By fermentation: \(0.5\ \mathrm{mol}\) of \(\ce{CH4}\) and \(0.5\ \mathrm{mol}\) of \(\ce{CO2}\).

Let \(F\) be the incoming amount of \(\ce{CH2O}\) units in \(\mathrm{mol\,day^{-1}}\). Counting the stated gas formation as the combined carbon dioxide and methane,

\[ \dot n(\ce{CO2})=(0.45+0.10/2)F=0.50F, \]

\[ \dot n(\ce{CH4})=(0.10/2)F=0.05F, \]

so

\[ \dot n_{\mathrm{gas}}=0.55F. \]

Using the ideal-gas law with \(T=298.15\ \mathrm{K}\),

\[ \dot n_{\mathrm{gas}} =\frac{p\dot V}{RT} =\frac{(1.00\times10^5\ \mathrm{Pa})(16\ \mathrm{m^3\,day^{-1}})} {(8.314\ \mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}})(298.15\ \mathrm{K})} =645.5\ \mathrm{mol\,day^{-1}}. \]

Therefore,

\[ F=\frac{645.5}{0.55} =1.174\times10^3\ \mathrm{mol\,day^{-1}}. \]

The fraction remaining in the sludge is

\[ 1-0.45-0.10=0.45. \]

Using \(M(\ce{CH2O})=30.0\ \mathrm{g\,mol^{-1}}\),

\[ \dot m_{\mathrm{sludge}} =0.45F(30.0\times10^{-3}\ \mathrm{kg\,mol^{-1}}) =15.84\ \mathrm{kg\,day^{-1}}. \]

Hence,

\[ \boxed{\dot m_{\mathrm{sludge}}\approx16\ \mathrm{kg\,day^{-1}}}. \]

1.2 — Energy from methane combustion

The methane production rate is

\[ \dot n(\ce{CH4})=0.05F=58.68\ \mathrm{mol\,day^{-1}}. \]

Its combustion reaction is

\[ \ce{CH4 + 2 O2 -> CO2 + 2 H2O}. \]

Using the given combustion enthalpy, the positive amount of heat released per day is

\[ \dot E =-\dot n(\ce{CH4})\Delta_{\mathrm c}H =(58.68\ \mathrm{mol\,day^{-1}}) (882\ \mathrm{kJ\,mol^{-1}}) =5.175\times10^4\ \mathrm{kJ\,day^{-1}}. \]

Thus,

\[ \boxed{\dot E\approx52\ \mathrm{MJ\,day^{-1}}}. \]

1.3 — Daily volume of waste water processed

The total incoming carbohydrate mass is

\[ \dot m_0=F(30.0\times10^{-3}\ \mathrm{kg\,mol^{-1}}) =35.21\ \mathrm{kg\,day^{-1}}. \]

Convert the given concentration:

\[ 250\ \mathrm{mg\,dm^{-3}} =250\ \mathrm{g\,m^{-3}} =0.250\ \mathrm{kg\,m^{-3}}. \]

The required waste-water flow is therefore

\[ \dot V_{\mathrm{water}} =\frac{\dot m_0}{c} =\frac{35.21\ \mathrm{kg\,day^{-1}}} {0.250\ \mathrm{kg\,m^{-3}}} =140.8\ \mathrm{m^3\,day^{-1}}. \]

Hence,

\[ \boxed{\dot V_{\mathrm{water}}\approx1.4\times10^2\ \mathrm{m^3\,day^{-1}}}. \]

The solution was checked against the official solution.