Chemistry 4 Lecture 4 The Chemistry of Climate Change: CO Equilbria i NC State University The short term carbon cycle Photosynthesis removes carbon atoms from the atmosphere and turns them into food for green plants. This process requires energy derived from the sun s radiation. At night, when plants turn off their photosynthesis and undergo respiration, carbon is replaced back into the air. Also, when the plant dies and begins to decay, much of the trapped carbon within the plant is released back into the atmosphere. There is a small percentage that remains as fixed carbon. Even on a time scale of years this cycle is mostly reversible meaning that at steady state there is little carbon fixed by these processes. They are important however and if the biomass on earth decreases (e.g. deforestation) the net capture of CO will decrease. Carbon Cycle Calcium carbonate in sedimentary rock Roughly 75% of the carbon injected into the atmosphere by non-organic means (usually volcanoes) finds its way into deposits of calcium carbonate (limestone). These deposits make up the largest reservoir in the carbon cycle. Limestone is formed from bicarbonate (HCO-) ions weathered and dissolved in the ocean. The ions, along with the skeletal remains of marine life accumulate on the ocean floor. Limestone formation involves a series of chemical reactions, all of which have a net effect of removing carbon dioxide from the atmosphere. Weathering of limestone deposits by rain tends to return carbon atoms to short-term reservoirs, thereby replenishing the concentration of atmospheric carbon dioxide. General Considerations Fixation of CO in organic form requires an energy source. In biology that energy source is the sun s radiation. The Carbonate-Silicate Geochemical Cycle CaCO SiO CaSiO CO The formation of CaCO is thermodynamically favorable. However the process is slow. The calcium content of seawater is about 80 mg/l. The solubility product of calcium carbonate is K sp = [Ca ][CO ] = 5 x 10-9 at 98 K. The concentration of CO depends on ph because of the two acid equilibria. CO H O HCO - H CO H 1
The Balance between the Atmosphere and the Ocean The most common method of CO sequestration Sources of CO CO Atmosphere Concentration 1945 70ppm 005 50ppm 050 550ppm Henry's Law Chemical Equilibria and the Solubility of Carbon Dioxide Calcium Ocean CO How much of the carbon ends up on the ocean floor? 1.9 Tons of CaCO Biological Process A Chemist s view: the ideal ocean We can define an ideal or inorganic ocean as follows: a body of water in contact with gas- -eous carbon dioxide and containing dissolved strong electrolytes, dissolved carbon dioxide, bicarbonate ions and carbonate ions, all at equilibrium. Note that this ocean does not include any biological activity so it can serve only to provide the inorganic background for the real ocean. Procedure for treating coupled equilibria First step: List the species We consider the process of carbon dioxide in the atmosphere in equilibrium with aqueous carbon dioxide, which can form carbonic acid. Carbonic acid has two dissociations. First, we create a list of chemical species present in the system at equilibrium. We will use: H, CO, H O(l), CO, CO (g), HCO Step : create a Species-by-Element Matrix Create a matrix with species along the top and elements along the side: H CO HO CO CO (g) HCO H 1 0 0 0 1 O 0 1 Each entry in the matrix corresponds to the elemental composition of each species. The original organization is arbitrary, but in the end you will want to have a diagonal unit matrix (row reduction). Row Reduction Procedure The row reduction process can be described as follows: 1. Any row can be multiplied by any number (positive, negative, integral, a fraction). Any row can be added to any other (before or after multiplication - so if you multiply by -1 this means you can add or subtract any row from any other.). You can repeat these operations as many times as you wish. 4. You can interchange any two rows (the species heading the row moves with the row).
This is the final form that we want to obtain! H 1 0 0 1 CO 0 1 0 1 1 1 HO 0 0 1 1 1 0 This matrix is in echelon form. Note that the species have changed in the left hand column. We will go through the steps used in this case. Multiply the carbon row by - H 1 0 0 0 1 C 0-1 0-1 -1-1 O 0 1 Add the C row to the O row. Multiply the O row by -. H 1 0 0 0 1 CO - O 0 1 0 0 1-1 -1 0 H 1 0 0 0 1 CO - O 0 1 0 0-0 Add the O row to the H row. H 1 0 0 0 0 01 CO - HO O 0 0 1-1 -1 0 Step : Row Reduce the Matrix to Echelon Form H 1 0 0 1 CO 0 1 0 1 1 1 HO 0 0 1 1 1 0 The species forming the unit matrix (red square) are the independent species. We can form independent net reactions from the columns (blue rectangles).
The Chemical Significance of the Row Reduction In the original species-by-element matrix each column is a vector in a space in which the elements are basis vectors. Row reduction changes the basis vectors to species in the system at equilibrium i.e. each column expresses the species labeling the column as a combination of species in the system, i.e each nontrivial column is a net reaction. The process of row reduction assures that the column vectors are independent. Each column expresses each species as a combination of other species CO = H CO H O(l) CO(g) = H CO HO(l) HCO = CO H These chemical reactions are a complete set of independent net reactions. As such they provide a basis for a complete consideration of the equilibrium state. They may be added and subtracted with the cancellation of species to obtain alternative sets of independent net reactions, but each set will contain exactly three net reactions and will provide a basis for the same consideration of the equilibrium state. An alternative set of independent net reactions The following are linear combinations from the set found by row-reduction: CO = CO(g) CO HO(l) = H HCO HCO = CO H For each independent net reaction there is an equilibrium constant In the Henry s Law constant the pressure is in bar, activities are taken equal to molalities. K = P / m 0 1 CO CO = = m m / m HCO H CO = m m / m CO H HCO K = 4.5 10 K = 5.6 10 7 11 Add any constraints to the system In the study of the ocean we can take the concentration of hydrogen ions and the partial pressure of carbon dioxide as known for a given state of the atmosphere and the ocean. Based on the known ph and partial pressure of CO 9 [ H ] = 6. 10 P =.8 10 bar CO 4 we can calculate the expected value of other species from the equilibria. We do not need a constraint for H O since it does not appear in the equilibrium constants. Solve for concentrations We therefore have three equations snd three unknowns and we can solve for the molalities of the carbon dioxide bearing species: [CO ] =.80 x 10-4 / 0 = 1.7 x 10-5 [HCO - ] = 4.5 x 10-7 [CO ] / [H ] = 9.07 x 10-4 [CO ] = 5.6 x 10-11 [HCO - ] / [H ] = 8.06 x 10-6 We also know that: [H ] [OH - ] = 10-14 4
Charge balance condition In order to consider other states of the system it is necessary to consider charge conservation. The hydrogen ion concentration is far below the concentrations of the negative ions. There is no charge balance without additional cations. We therefore determine the molality of cations required to balance the charges due to CO, HCO - and H : [cb] = [OH - ] [HCO - ] [CO ] [H ] = 9.x10-4 The majority of these compensating ions are calcium. These ions enter the ocean through riverine fluxes (i.e. weathering). Determining total amounts of CO The total mass of the earth s oceans is: m ocean = 1. x 10 1 kg N atm eq represents the number of moles of CO from the atmosphere that have dissolved in the ocean: N atm eq = (1.7 x 10-5 molal)(1. x 10 1 kg) = 165 1.65 x 10 16 moles The number of moles of CO in the atmosphere is: Force = (Area)(Pressure) = (5.15 x 10 14 m )(10 5 Pa) n atm CO = x CO (Force [N])/ g [m/s ] x (MW [kg/mol]) MW is the mol. weight of the atmosphere ~ 0.09 kg/mol n atm CO = (.8 x 10-4 )(5 x 10 19 )(10 /9)/9.8 = 6.7 x 10 16 kg Or n atm CO/ N atm aq ~ 0.5 The ocean is saturated in CaCO Now to return to the question of the ocean as a significant sink for anthropogenic carbon dioxide. When the species were listed CaCO (s) was not included. In order for it to form the system must be saturated with respect to its formation. It is reported that the surface of the ocean is. x 10 - molal in Ca. The solubility product for calcium carbonate is 4.9 X 10-9 : [Ca ][CO = ]=K sp in the ocean, the ion product is. x 10 - x 8.0 x 10-6 =1.85 X 10-8, which is greater than the solubility product, i.e for the simple system considered here the solution is supersaturated. Saturation and States of Disequilibrium The concentration of the CaCO is not necessarily in equilibrium. In other words, there can be an excess or deficit of Ca and CO in the solution above the CaCO solid. If there is an excess of ions in solution the solution is supersaturated. If there is a deficit then the solution is subsaturated. K sp = [Ca ][CO ] If the system is not in equilbrium then there will be a driving force ΔG to attain equilibrium: ΔG = ΔG o RTlnQ The three possible states are: ΔG > 0, Q = [Ca ][CO ] > K Supersaturated ΔG = 0, Q = [Ca ][CO ] = K Saturated ΔG < 0, Q = [Ca ][CO ] < K Subsaturated The ocean as a CO sink? So could CaCO (s) be a significant sink for anthropogenic carbon dioxide? It has been observed that there are no large deposits of CaCO (s) on the bottom of the deep ocean, so the answer is probably no. There are several possible explanations for this, and all possibly contribute to failure of the precipitate to form or its redissolution in the deep ocean: 1. CO redissolves in the deep ocean. Ionic activities are reduced (high salt concentration). Surface free energy presents a barrier to precipitation. Pressure-dependent equilibrium There are large pressures (up to nearly 1 kbar) in the deep ocean. One scenario is that calcium carbonate forms at the surface and sinks to depths where it redissolves because of effect of pressure on the dissolution reaction: CaCO Ca CO ΔV = -6 cm /mol [molar volume of reaction] The molar volume water of solvation is less around the ion that around the solid (or the bulk). For example, at 8000 m (P ~ 800 atm): ln(k(p)/k) = - ΔV P/R ln(k(p)/k) = - (0.06 L)(800 atm)/0.0806/98 K =.0 K(800 atm) = e.0 K = e.0 4.9 x 10-9 =.6 x 10-8 Millero in Geochimica et Cosmochimica Acta, 59, 661 (1995) 5
Sedimentation and the Snow Line The formation of calcium carbonate as a sediment in the oceans has been occurring for billions of years. This process leads to the formation of limestone (sedimenary rock). One can imagine CaCO forming white particles and settling to the bottom. This is like snow in the ocean since the particles build up a layer on the ocean floor. However, in the deep ocean the pressure shifts the equilibrium so that this snow melts before it reaches the bottom. Below about 5000 m there is no limestone on the ocean floor. This part of the deep ocean is known as the abyssal plain. Ocean acidification Furthermore, if calcium carbonate were precipitating in an inorganic ocean the carbonate forming reactions (below) would be drawn to the right constantly forming hydrogen ions and decreasing the ph HCO = CO H CO HO = CO H unless there were some mechanism consuming hydrogen ions at the surface. Thus, the ph of the oceans is falling in the short term. The oceans are becoming more acidic. Carbon recycling in the ocean We have seen that the equilibria alone do not explain the large quantity (ca. 40 times the amount in air) of carbon dioxide in the sea. However living organisms in the sea die and fall toward the depths. As they do so they are oxidized and produce carbon dioxide, and their skeletal l remains often contain calcium carbonate. At great depths the dissolved carbon dioxide has, over time, become significantly supersaturated relative to the atmosphere because the upwelling required for equilibration is slow, requiring a buildup of carbon dioxide at depth before a steady state is reached. Ocean Carbon Recycling At depth CaCO(s) CO HO(l) HCO the surface: CO CO(g) HCO CO(g) H CO H CO(g) HO(l) Where does human-made CO go? The ocean takes up only 5% of the carbon dioxide output by humans. As a result the ph of the oceans is decreasing. The ph of the ocean has decreased from 8. to 8.1 since the beginning of the industrial revolution. On a longer time scale CO is fixed in the ocean as CaCO. However, CaCO precipitation is slow and the ocean supersaturated in calcium carbonate. 6