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Chemical Kinetics [A][B] [D] [C] [A][B] [B] [A] K R Dt D Dt D K R Dt D Dt D

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Chemical Kinetics [A][B] [D] [C] [A][B] [B] [A] K R Dt D Dt D K R Dt D Dt D Chemical kinetics (Nazaroff & Alvarez-Cohen, Section 3.A.2) This is the answer to what happens when chemical equilibrium is not reached. For example, take the two-way reaction A + B ↔ C + D As A and B come into contact with each other, they start to react with each other, and the reaction rate can be expressed as Rforward = kf [A] [B]. This expression reflects the fact that the more of A and B there is, the more encounters occur, and the more reactions take place. This rate depletes the amounts of both A and B, and generates amounts of C and D: d[A] d[B] R k [A][B] dt dt forward f d[C] d[D] R k [A][B] dt dt forward f But, at the same time, C and D come in contact, too, and carry their own, reverse reaction, at a rate proportional to their amounts Rreverse = kr [C] [D] . This rate depletes the amounts of both C and D, and adds to the amounts of A and B: d[A] d[B] R k [C][D] dt dt reverse r d[C] d[D] R k [C][D] dt dt reverse r Because the two reactions occur simultaneously, we subtract the rate of one from the other: d[A] d[B] R R k [A][B] k [C][D] dt dt forward reverse f r d[C] d[D] R R k [A][B] k [C][D] dt dt forward reverse f r 1 Finally, in a system with open boundaries, we can add the imports and exports: d[A] V Qin [A]in Qout [A]out krV[C][D] kfV[A][B] dt inlets outlets d[B] V Qin [B]in Qout [B]out krV[C][D] kfV[A][B] dt inlets outlets d[C] V Qin [C]in Qout [C]out kfV[A][B] krV[C][D] dt inlets outlets d[D] V Qin [D]in Qout [D]out kfV[A][B] krV[C][D] dt inlets outlets imports exports sources sinks where V is the volume of the system for which the budget is written. Note: The variables here are concentrations ([…]’s), expressed in moles per liter (mol/L or M), not in mass per liter (such as mg/L). The latter would be obtained after invoking the molecular weights of the individual chemicals. Chemical Kinetics – Reaction order (Nazaroff & Alvarez-Cohen, Section 3.A.3) The previous example assumed that it takes two chemicals for each reaction. This is not always the case. For example, when a substance dissolves in water + – H2CO3 ↔ H + HCO3 The forward reaction needs only one concentration Rforward = kf [H2CO3] Nomenclature: d[A] Second-order reaction: k [A][B] dt d[A] First-order reaction: k [A] dt d[A] Zeroth-order reaction: k dt Note: The units of the k coefficients vary accordingly! 2 Chemical Equilibrium (back to beginning of Nazaroff & Alvarez-Cohen, Section 3.A.2) A system is in chemical equilibrium when 1. It does not vary in time (steady state); 2. It is well mixed; 3. There is no net flow of mass, heat or species with the surroundings; 4. The net rate of all chemical reactions is zero. Such a state corresponds to a local minimum in “free energy”: GGRTQ0 ln( ) 0 difference between both sum of free energies of sides of chemical reaction components under standard conditions [C]c [D]d whereQ is called the reaction quotient. [A]a [B]b for the reaction a A+ b B↔ c C+ d D G = 0 implies Q = constant. There is a better way to see this, though… Many chemical reactions in the environment are two-way reactions. Symbolically, A + B ↔ C + D during which A and B react to produce C and D, and at the same time C and D react to produce A and B. The rate of reaction between A and B is Rforward = kf [A] [B] The rate of reaction in the reverse direction is Rreverse = kr [C] [D] Chemical equilibrium occurs when the two rates equal each other so that the rate of depletion of A and B equals the rate of their replenishment: kf [A] [B] = kr [C] [D] That is, when: [C] [D] k f const. [A][B] kr 3 Equilibrium constant from equilibrium the equilibrium equation kf [A] [B] = kr [C] [D] we can write [C][D] k f K [A][B] kr The “constant” K is determined in the laboratory. Like most chemical “constants,” it varies with temperature. Generalization to reaction with arbitrary numbers in the stoichiometry: a A+ b B↔ c C+ d D a b c d Rforward = kf [A] [B] Rreverse = kr [C] [D] a b c d Equilibrium exists when kf [A] [B] = kr [C] [D] c d [C] [D] kf a b K [A] [B] kr Important particular case: Water dissociation (Nazaroff & Alvarez-Cohen, Section 3.C.2) Water always tends to dissociate a little, via the reaction + – H2O ↔ H + OH There is an equilibrium constant for this, and it is: [H ][OH ] 1.831016 moles/L [H2O] Since [H2O] is always the same and equal to 55.4 moles/L (see Lecture 2), we can write: 16 [H ][OH ] (1.8310 mol/L)[H2O] (1.831016 mol/L) (55.4 mol/L) 1.0110-14 (mol/L)2 4 pH of Pure Water In the absence of any other ion (pure water!), to any [H+] ion there is a companion [OH–] ion. Thus, [H ] [OH ] It follows [H ] 1.011014 (mol/L)2 1.0 107 mol/L pH log10[H ] 7.0 Inorganic impurities in water and air First, a distinction needs to be made between organic and inorganic compounds: “organic” compounds = those with carbon, – 2– except for these basic elements: C, CO, CO2, HCO3 and CO3 Acids & Bases: acids make H+ common acids are HCl, HOCl, H2S, H2SO4, H2SO3, HNO3 bases make OH– common bases are NH3, NaOH, CaCO3, Ca(OH)2 – + – + OH takes away H via OH + H → H2O Dissolution in water → ions cations are positive, anions are negative + + 2+ – 2– – 2– most common ions are NH4 , Na , Ca , Cl , SO4 , HCO3 , CO3 5 Electroneutrality Principle (Nazaroff & Alvarez-Cohen, page 45) It can almost always be assumed that the initial water solution is electrically neutral, that is, it has no net electric charge. Then, conservation of electrons demands that this state of neutrality persist over time. Therefore, for every extra electron on an anion in the solution, there must be a missing electron from a cation in the same solution, and zi [Ai] = 0 where zi = charge of ion Ai (+ for cation, – for anion), most often +1, +2, – 1 or – 2 [Ai] = molar concentration of ion Ai . Example: The Carbonate System (Nazaroff & Alvarez-Cohen, section 3.C.4) Carbon dioxide CO2 from the atmosphere partially dissolves in water. In water, it then forms carbonic acid H2CO3. + – In turn, H2CO3 dissolves into H and bicarbonate ion HCO3 , – + 2– and HCO3 into another H and carbonate ion CO3 . Thus, the following reactions take place in water exposed to the atmosphere: CO2 + H2O ↔ H2CO3 + – H2CO3 ↔ H + HCO3 – + 2– HCO3 ↔ H + CO3 + – In addition, water H2O always decomposes slightly into H and OH : + – H2O ↔ H + OH Electroneutrality requires: + – 2– – [H ] = [HCO3 ] + 2[CO3 ] + [OH ] 6 The equilibrium relations are: CO in the water CO in the air [CO ] K P , K 0.034 M/atm 2 2 2 H CO2 H [H2CO3 ] 3 CO2 in the water H2O H2CO3 1.58 10 [CO2 ] [H ][HCO3 ] 4 H2CO3 H HCO3 2.834 10 M [H2CO3 ] 2 2 [H ][CO3 ] 11 HCO3 H CO3 4.68 10 M [HCO3 ] 14 2 H2O H OH [H ][OH ] 10 M Do we have enough equations? – 2– + – Number of unknowns = 6 for [CO2] , [H2CO3] , [HCO3 ] , [CO3 ] , [H ] and [OH ] Number of equations = 6 = 5 (above) + 1 (electroneutrality at bottom of previous slide). So, yes, we have enough equations to solve the problem. 412 For a partial pressure of CO2 in the atmosphere equal to 350 ppm, we have 412 1.40 6 5 [CO2 ] (0.034 M/atm) (350 10 atm) 1.19 10 M From the first chemical equilibrium, we deduce: 1.40 2.21 3 3 5 8 [H2CO3 ] (1.58 10 )[CO2 ] (1.58 10 )(1.19 10 M) 1.88 10 M We solve the remaining simultaneous equations and obtain: 6 -6 [HCO3 ] 2.3110 M 2.50 x 10 M 2 11 -11 [CO3 ] 4.67 10 M 4.67 x 10 M [OH ] 4.33109 M 3.99 x 10-9 M [H ] 2.31106 M 2.51 x 10-6 M 5.60 Thus, the pH of natural (clean) water is pH log10[H ] 5.64 Clean, natural water is slightly acidic. And, it is getting increasingly acidic. There is fear that coral in the sea will not survive. 7 http://coralreef.noaa.gov/education/oa/resources/22-4_kleypas.p df Question: What can be done about coral erosion short of curbing CO2 in the atmosphere? Answer : Use calcium, or simply rely on the calcium that is naturally there. CaCO3 Ca CO3 CO3 H HCO3 HCO3 H H2CO3 H2CO3 H2O CO2 CO2 in water CO2 in air The net effect of limestone is to deliver calcium ions (Ca++) that displace protons (H+), and thus reduce acidity. There is efficiency in this because one Ca++ eliminates two H+’s. 8 But think: - If a body of water has become too acidic because of some acid input, it is better to eliminate that acid source than to remediate the water body.
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