IChO 1992 · Theory problem 1

Diatoms, whales and the carbon cycle

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Diatoms, microscopic organisms, are an abundant food source in the oceans producing carbohydrates from carbon dioxide and water by photosynthesis:

Figure 1

1.1 During the first five years of life blue whales gain 75 kg of mass per day by feeding on krill. The whale must consume ten times this mass of krill each day. The krill must consume 10.0 kg of diatoms to produce 1.0 kg of krill. Assuming that the mass gain in the first years of a whale's life is due to the consumption of carbohydrates (\(\ce{C6H12O6}\)), calculate the volume of \(\ce{CO2}\) at STP (0 °C, 101 kPa) that must be used by the diatoms to produce the carbohydrates consumed by a blue whale in its first five years of life.

1.2 There is 0.23 cm³ of dissolved \(\ce{CO2}\) per one litre sea water (at 24 °C and 101 kPa).

i) If diatoms can completely remove carbon dioxide from the water they process, what volume of water would they process to produce the carbohydrates required by a blue whale during the first five years of life?

ii) What fraction of the total volume of the oceans will be needed to supply the carbon dioxide for the first five years of growth of 1000 blue whales? The volume of the oceans is \(1.37\times10^{18}\ \mathrm{m^3}\).

1.3 Three percent of the mass of a \(9.1\times10^4\) kg adult whale is nitrogen. When a \(9.1\times10^4\) kg blue whale dies, what is the maximum mass of \(\ce{NH4+}\) that can become available for other marine organisms?

1.4 Eighteen percent of a \(9.1\times10^4\) kg whale's mass is carbon. Carbon can be returned to the atmosphere as \(\ce{CO2}\) and then removed from the atmosphere by weathering of rocks containing calcium silicate.

Figure 2

What are the maximum number of grams of \(\ce{CaSiO3}\) that can be weathered by the carbon dioxide produced from the decomposition of 1000 blue whales, the number estimated to die annually?

Official English text of the problem.

Answers

  • 1.1: \(\boxed{1.0\times10^7\ \mathrm{m^3}}\) of carbon dioxide at STP, equivalent to \(1.0\times10^{10}\ \mathrm{L}\).
  • 1.2 i): \(\boxed{4.9\times10^{10}\ \mathrm{m^3}}\) of seawater, equivalent to \(4.9\times10^{13}\ \mathrm{L}\).
  • 1.2 ii): \(\boxed{3.5\times10^{-5}}\) of the ocean volume, or approximately \(0.0035\%\).
  • 1.3: \(\boxed{3.5\times10^3\ \mathrm{kg}}\) of ammonium ions, \(\ce{NH4+}\).
  • 1.4: \(\boxed{7.9\times10^{10}\ \mathrm{g}}\) of calcium silicate, \(\ce{CaSiO3}\).

Solution

Take one year as 365 days, neglecting leap days. Use the ideal-gas law with \[ R=8.314\ \mathrm{L\,kPa\,mol^{-1}\,K^{-1}}. \] Intermediate values retain extra digits; final results are rounded to approximately two significant figures.

1.1 — Carbon dioxide required for photosynthesis

The balanced photosynthesis equation is \[ \ce{6CO2 + 6H2O ->[solar energy] C6H12O6 + 6O2}. \]

The species are carbon dioxide (O=C=O), water (O), a hexose carbohydrate such as glucose (open-chain connectivity SMILES O=CC(O)C(O)C(O)C(O)CO), and dioxygen (O=O). The carbohydrate’s stereochemistry is unspecified and does not affect this calculation.

The whale’s daily krill consumption is \[ 10(75\ \mathrm{kg\,d^{-1}})=750\ \mathrm{kg\,d^{-1}}. \] Producing this krill requires \[ 750\ \mathrm{kg\,d^{-1}} \times\frac{10.0\ \mathrm{kg\ diatoms}}{1.0\ \mathrm{kg\ krill}} =7500\ \mathrm{kg\ diatoms\,d^{-1}}. \]

Under the problem’s carbohydrate mass-accounting assumption, the required carbohydrate mass over five years is therefore \[ m(\ce{C6H12O6}) =(7500\ \mathrm{kg\,d^{-1}})(5\times365\ \mathrm{d}) =1.36875\times10^7\ \mathrm{kg}. \]

Using \(M(\ce{C6H12O6})=180\ \mathrm{g\,mol^{-1}}\), and six moles of carbon dioxide per mole of carbohydrate, \[ n(\ce{CO2}) =6\left(\frac{1.36875\times10^{10}\ \mathrm{g}} {180\ \mathrm{g\,mol^{-1}}}\right) =4.5625\times10^8\ \mathrm{mol}. \]

At \(T=273.15\ \mathrm{K}\) and \(P=101\ \mathrm{kPa}\), \[ \begin{aligned} V_0(\ce{CO2}) &=\frac{nRT}{P}\\ &=\frac{(4.5625\times10^8)(8.314)(273.15)}{101}\ \mathrm{L}\\ &=1.02587\times10^{10}\ \mathrm{L}. \end{aligned} \]

Thus \[ \boxed{V_0(\ce{CO2})\approx1.0\times10^7\ \mathrm{m^3}}. \]

1.2 i) — Seawater processed

The dissolved carbon dioxide content is expressed as a gas volume at \(24\,^\circ\mathrm{C}\), so the required carbon dioxide volume must first be converted to that temperature. At unchanged pressure, \[ \frac{V_{24}}{V_0}=\frac{297.15}{273.15}. \] Hence \[ V_{24} =(1.02587\times10^{10}\ \mathrm{L}) \frac{297.15}{273.15} =1.11601\times10^{10}\ \mathrm{L} =1.11601\times10^{13}\ \mathrm{cm^3}. \]

Each litre of seawater supplies \(0.23\ \mathrm{cm^3}\) of carbon dioxide at these conditions. Complete extraction therefore requires \[ \begin{aligned} V_{\text{seawater}} &=\frac{1.11601\times10^{13}\ \mathrm{cm^3}} {0.23\ \mathrm{cm^3\,L^{-1}}}\\ &=4.85221\times10^{13}\ \mathrm{L}\\ &=4.85221\times10^{10}\ \mathrm{m^3}. \end{aligned} \]

Thus \[ \boxed{V_{\text{seawater}}\approx4.9\times10^{10}\ \mathrm{m^3}}. \]

1.2 ii) — Fraction of the oceans for 1000 whales

Using the unrounded water volume from part i), \[ f=\frac{1000(4.85221\times10^{10}\ \mathrm{m^3})} {1.37\times10^{18}\ \mathrm{m^3}} =3.54176\times10^{-5}. \]

Therefore \[ \boxed{f\approx3.5\times10^{-5}}, \] equivalent to approximately \(0.0035\%\) of the ocean volume.

1.3 — Maximum ammonium mass

The ammonium ion is \(\ce{NH4+}\), SMILES [NH4+].

The whale contains \[ m(\ce{N})=0.03(9.1\times10^4\ \mathrm{kg}) =2.73\times10^3\ \mathrm{kg} \] of nitrogen.

The maximum occurs when all this nitrogen becomes ammonium. Each ammonium ion contains one nitrogen atom, so \[ n(\ce{NH4+})=n(\ce{N}). \] Using \(M(\ce{N})=14\ \mathrm{g\,mol^{-1}}\) and \(M(\ce{NH4+})=18\ \mathrm{g\,mol^{-1}}\), \[ m(\ce{NH4+}) =(2.73\times10^3\ \mathrm{kg})\frac{18}{14} =3.510\times10^3\ \mathrm{kg}. \]

The hydrogen incorporated into ammonium supplies the additional mass. Thus \[ \boxed{m_{\max}(\ce{NH4+})\approx3.5\times10^3\ \mathrm{kg}}. \]

1.4 — Maximum calcium silicate weathered

The balanced weathering reaction is \[ \ce{CaSiO3(s) + 2CO2(g) + 3H2O(l) -> Ca^2+(aq) + 2HCO3^-(aq) + H4SiO4(aq)}. \]

The aqueous products are calcium(II) ions ([Ca+2]), hydrogen carbonate or bicarbonate ions (O=C(O)[O-]), and orthosilicic acid (O[Si](O)(O)O). Calcium silicate has the formal formula-unit SMILES [Ca+2].[O-][Si](=O)[O-]; this represents its composition, not its extended crystal lattice or coordination geometry.

The carbon mass in 1000 whales is \[ \begin{aligned} m(\ce{C}) &=1000(9.1\times10^4\ \mathrm{kg})(0.18)\\ &=1.638\times10^7\ \mathrm{kg} =1.638\times10^{10}\ \mathrm{g}. \end{aligned} \]

For the maximum, all whale carbon becomes carbon dioxide. Each mole of carbon supplies one mole of carbon dioxide: \[ n(\ce{CO2}) =\frac{1.638\times10^{10}\ \mathrm{g}} {12\ \mathrm{g\,mol^{-1}}} =1.365\times10^9\ \mathrm{mol}. \]

The weathering equation consumes two moles of carbon dioxide per mole of calcium silicate: \[ n(\ce{CaSiO3}) =\frac{1.365\times10^9}{2} =6.825\times10^8\ \mathrm{mol}. \]

With \[ M(\ce{CaSiO3})=40+28+3(16)=116\ \mathrm{g\,mol^{-1}}, \] the maximum mass weathered is \[ \begin{aligned} m(\ce{CaSiO3}) &=(6.825\times10^8\ \mathrm{mol})(116\ \mathrm{g\,mol^{-1}})\\ &=7.917\times10^{10}\ \mathrm{g}. \end{aligned} \]

Therefore \[ \boxed{m_{\max}(\ce{CaSiO3})\approx7.9\times10^{10}\ \mathrm{g}}. \]

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.