lecture 15 natural sulfur, acid rain Rainout We mentioned a few of things that may rainout: 1.CH 3 OOH (CH 4 oxidation, low NO x ) 2.H 2 O 2 (CO oxidation, low NO x ) 3.HNO 3 (CH 4 and RH oxidation, high NO x ) If hydrocarbons convert to acids during oxidation and dissolve in water  acid rain.

lecture 15 natural sulfur, acid rain History of Acid Rain In the 19th century, Robert Angus Smith discovered high levels of acidity in rain falling over industrial regions. In 1950’s and 1960’s, biologists noticed a decline of fish populations in lakes of southern Norway and North America. Later found that acid rain also affects vegetation, materials, structures.

lecture 15 natural sulfur, acid rain Acid Rain Over North America and Norway

lecture 15 natural sulfur, acid rain Two Main Sources of Acid Rain 1.SO 2 from industry oxidized to H 2 SO 4 2.NO, NO 2 from automobiles oxidized to HNO 3

lecture 15 natural sulfur, acid rain Aqueous Phase Chemical Equilibrium What happens to a species that dissolves in water? partly dissociates into ions (bonds toward water ions stronger than to its own atoms) What happens to water in the process? partially ionizes These are reversible reactions that reach equilibrium rapidly: At equilibrium: K eq : equilibrium constant M = mol L -1

lecture 15 natural sulfur, acid rain Concentration of H 2 O(aq) [H 2 O] is very large: 55.5 M  virtually constant Incorporate [H 2 O] into K eq : K eq w = [H  ][OH  ] = 1.0 x M 2 at 298 K Concentration of ions in pure water: Each water molecule that dissociates produces 1 H  and 1 OH  : Concentration of ions is much smaller than [H 2 O]  water has small conductivity.

lecture 15 natural sulfur, acid rain pH of Pure Water Definition of pH: Pure water at 298 K: acidic: pH < 7 alkaline: pH > 7 neutral: pH = 7

lecture 15 natural sulfur, acid rain pH of Clean Rainwater Clean rainwater is not pure water. It equilibrates with CO 2 : hydrolysis ionization – bicarbonate ion further ionization – carbonate ion

lecture 15 natural sulfur, acid rain Equilibrium Between Gas and Aqueous Phase Equilibrium for first reaction: [H 2 CO 3 (aq)]: aqueous phase concentration in equilibrium with gas phase p CO2 : partial pressure of gas phase species (atm) 1 atm = 760 mm Hg = 760 Torr = mbar = x 10 5 Pa (N m -2 ) K H : Henry’s law constant (for dilute solutions) Soluble gases have large K H.

lecture 15 natural sulfur, acid rain Henry’s Law Constants for Atmospheric Gases

lecture 15 natural sulfur, acid rain What Does [H 2 CO 3 (aq)] Depend On? Does [H 2 CO 3 (aq)] depend on amount of liquid water available? No. Does [H 2 CO 3 (aq)] depend on size of droplet? No. Does [H 2 CO 3 (aq)] depend on temperature? Yes. van’t Hoff equation: similar to Clausius-Clapeyron equation:  H: reaction enthalpy at constant T and P or heat of dissolution L: heat of vaporization K H increases as T decreases  gas more soluble at lower T (less energetic molecules on surface, less evaporation, more stays in solution).

lecture 15 natural sulfur, acid rain Heat of Dissolution for Atmospheric Gases

lecture 15 natural sulfur, acid rain CO 2 /H 2 O System Reactions in CO 2 /H 2 O system:

lecture 15 natural sulfur, acid rain CO 2 /H 2 O System cont. Total dissolved CO 2 :

lecture 15 natural sulfur, acid rain Effective Henry’s Law Constant effective Henry’s Law constant for CO 2 : Is greater than or less than ? Always greater than  Total amount of CO 2 dissolved always exceeds that predicted by Henry’s Law for CO 2 alone (although not by much).

lecture 15 natural sulfur, acid rain Effective Henry’s Law Constant cont. What does depend on? T, pH of solution ([H  ]) As pH increases ([H  ] decreases), does increase or decrease? increase As pH increases, does increase or decrease? increase

lecture 15 natural sulfur, acid rain Effective Henry’s Law Constant of CO 2 as a Function of pH

lecture 15 natural sulfur, acid rain Calculate the pH of CO 2 /H 2 O System – Approximate Method 1.Since K H * is small (compared to 10 7 /10 8 ), assume p CO2 ≈ constant. 2.Since K eq w so small, assume it does not contribute to [H  ]. 3.Since K eq 3 so small, assume [CO 3 2  ] ≈ 0, then every molecule of H 2 CO 3 that dissociates produces 1H  and 1 HCO 3  : [H  ] ≈ [HCO 3  ] pH of pure rainwater

lecture 15 natural sulfur, acid rain Calculate the pH of CO 2 /H 2 O System – More Exact Method 1.Since K H * is small, still assume p CO2 ≈ constant. 2.electroneutrality: concentrations of ions will adjust so that solution is electrically neutral: (Each CO 3 2  ion contributes charge of 2 . Total negative charge is concentration of ions x 2.)

lecture 15 natural sulfur, acid rain CO 2 + H 2 O = H 2 CO 3 H 2 CO 3 = H + + HCO 3 - = 2H + + CO 3 = K 1 = 4.3x10 -7 K 2 = 5.3x  C = [H + ] = [HCO 3 - ]  : fraction of concentration in the form of ions C: concentration (1-  )C = [H 2 CO 3 ] 1-  : fraction of concentration in the form of acid K 1 = [H + ][HCO 3 - ] / [H 2 CO 3 ] = (  C) 2 / (1 -  )C =  2 C/(1 -  ) Calculate the pH of CO 2 /H 2 O System – Third Method

lecture 15 natural sulfur, acid rain CO 2 solubility H 2 O CO 2 Partial pressure P p is 345 ppm n = VP p /RT n = x 345x10 -6 / (0.082 x 298) = 1.07x10  5 mol In 1 liter C = 1.07 x 10  5 mol/liter 4.3 x 10  7 = 1.07 x 10  5  2 / (1 -  )   =0  = 0.18 [H + ] = 0.18 x 1.07 x 10  5 =1.93x10 -5 M pH =  log [H + ] = 5.7 Calculate the pH of CO 2 /H 2 O System – Third Method cont.

lecture 15 natural sulfur, acid rain SO 2 /H 2 O System Reactions in SO 2 /H 2 O system: bisulfite ion sulfite ion

lecture 15 natural sulfur, acid rain SO 2 /H 2 O System cont. Total dissolved sulfur:

lecture 15 natural sulfur, acid rain S(IV) Sulfur occurs in 5 oxidation states in the atmosphere. Chemical reactivity decreases with sulfur oxidation state. Water solubility increases with sulfur oxidation state. Essentially all dissolved species that come from SO 2 are in oxidation state 4.

lecture 15 natural sulfur, acid rain Sulfur Oxidation States

lecture 15 natural sulfur, acid rain Sulfur Oxidation States cont.

lecture 15 natural sulfur, acid rain Total Dissolved Sulfur Total dissolved sulfur: Effective Henry’s Law constant for SO 2 :

lecture 15 natural sulfur, acid rain Effective Henry’s Law Constant of SO 2 as a Function of pH increases by ~7 orders of magnitude with pH  Acid-base equilibrium pulls more material into solution. (Which material? [H 2 SO 3 (aq)] does not depend on pH.)

lecture 15 natural sulfur, acid rain If Assume Constant p SO2 Open system: unlimited LWC, unlimited SO 2 : [S(IV) Tot (aq)] increases dramatically with pH

lecture 15 natural sulfur, acid rain If Don’t Assume Constant p SO2 Closed system: supply of SO 2 limited  Cannot assume p SO2 ≈ constant to calculate concentrations, but can calculate mole fractions as a function of pH:

lecture 15 natural sulfur, acid rain Mole Fractions cont.

lecture 15 natural sulfur, acid rain S(IV) Mole Fractions as a Function of pH pH < 2: S(IV) mainly in the form of H 2 SO 3 (aq) 3 < pH < 6: S(IV) mainly in the form of HSO 3  pH > 7: S(IV) mainly in the form of SO 3 2 

lecture 15 natural sulfur, acid rain How Does This Affect Aqueous Phase Reactions? Since concentrations depend on pH, reaction rates in solution will depend on pH. Why is this important? We still have not calculated the pH of the sulfur system with varying p SO2 (closed system). So far we have: H 2 SO 3 (aq), HSO 3 ,SO 3 2   We don’t yet have the acid H 2 SO 4.

lecture 15 natural sulfur, acid rain Sulfuric Acid What oxidation state is H 2 SO 4 in? 6 We need to convert S(IV) to S(VI) via aqueous phase reactions. S(IV) reacts with many species in solution: O 3, H 2 O 2, CH 3 OOH, O 2, OH, NO 2, HCHO, Mn, Fe, … Of these, O 3 and H 2 O 2 are the most important for converting S(IV) to S(VI).

lecture 15 natural sulfur, acid rain Aqueous Phase Reaction Rate S(IV) + A(aq)  S(VI) + …rate constant k in M -1 s -1 R = k [S(IV)] [A(aq)] in M s -1 (mol L -1 s -1 )

lecture 15 natural sulfur, acid rain The S(IV)/O 3 (aq) System Reactions in the S(IV)/O 3 (aq) system:

lecture 15 natural sulfur, acid rain Rate Constant of the S(IV)/O 3 (aq) System as a Function of pH An increase in pH results in an increase in equilibrium [HSO 3  ] and [SO 3 2  ]  results in an increase in d[S(IV)]/dt.

lecture 15 natural sulfur, acid rain Self-Limiting Reaction The strong increase in d[S(IV)]/dt with pH makes the reaction self- limiting. Why? Production of H 2 SO 4 (acid) lowers the pH and slows further reaction.  The reaction of S(IV) with O 3 (aq) is a source of cloud water acidification when pH >~ 4 and an important sink of gas phase SO 2 when pH >~8 (sea spray).

lecture 15 natural sulfur, acid rain The S(IV)/H 2 O 2 (aq) System [H 2 O 2 (aq)] is ~6 orders of magnitude higher than [O 3 (aq)] Reactions in the S(IV)/ H 2 O 2 (aq) system:

lecture 15 natural sulfur, acid rain The S(IV)/H 2 O 2 (aq) System cont. Steady state approximation on SO 2 OOH  :

lecture 15 natural sulfur, acid rain Rate Constant of the S(IV)/H 2 O 2 (aq) System as a Function of pH As pH increases ([H  ] decreases), first (2) first becomes faster (denominator dominates), then (2) becomes slower (numerator dominates).  The reaction of S(IV) with H 2 O 2 (aq) is a source of cloud water acidification when pH <~ 4 and an important sink of gas phase SO 2 when pH <~7 (clouds).

lecture 15 natural sulfur, acid rain pH of the Sulfur System with Varying p SO2 (Closed System) SO 2 (g) becomes depleted  less production of S(IV)  less acidic? O 3 (aq) and H 2 O 2 (aq) become depleted  less S(IV)  S(VI)  less acidic? NH 3 becomes depleted  less neutralization  more acidic? less S(IV)  pH lower (Figure 6.7)  faster S(IV)  S(VI) via H 2 O 2 (aq)  more acidic

lecture 15 natural sulfur, acid rain The Sulfur Cycle

lecture 15 natural sulfur, acid rain Assumptions: 1. All sulfate is in the form of H 2 SO 4 2. All the acid is dissolved and dissociated to ions (2H  and SO 4  ) 3. Liquid water content is 1.0 gr/m 3 (very heavy cloud) Note: Total water content (vapor+liquid) is 30 gr per m 3 at 25 o C and 100% humidity…… For [SO 4  ] = 1.0  g/m 3 (typical for remote pacific area) [H  ] = 2 (moles H  per mole H 2 SO 4 ) x10  6 (g SO 4  per m 3 air) / 1.0 (g H 2 O per m 3 air) / 98 (g H 2 SO 4 per mole H 2 SO 4 ) x 10 3 (g H 2 O per L H 2 O) = 2.04x10  5 M pH =  log[H  ] = 4.7 For a thin cloud (0.1 gr/m 3 ); [H  ] = 2.04x10 -4 M pH = 3.7 pH of Sulfur System as a Function of Cloud Liquid Water Content

lecture 15 natural sulfur, acid rain pH of Sulfur System as a Function of Cloud Liquid Water Content cont. SO 4 = (  g/m 3 ) Cloud Water Content

lecture 15 natural sulfur, acid rain The HNO 3 /H 2 O System Reactions in HNO 3 /H 2 O system: nitrate ion very soluble dissociates quickly

lecture 15 natural sulfur, acid rain The HNO 3 /H 2 O System cont. Total dissolved nitric acid: Effective Henry’s law constant for HNO 3 :

lecture 15 natural sulfur, acid rain If Don’t Assume Constant p HNO3 The Henry’s law constant of HNO 3 is very high.  Cannot assume p HNO3 ≈ constant to calculate concentrations, but can calculate mole fractions as a function of pH:

lecture 15 natural sulfur, acid rain HNO 3 Mole Fractions as a Function of pH Since K eq2 is so high, K eq2 /[H  ] >> 1   Dissolved nitric acid in clouds exists exclusively as nitrate. Aqueous fraction of nitric acid as a function of pH and cloud LWC:

lecture 15 natural sulfur, acid rain H 2 SO 4 (g) and NO X (g) We looked at the aqueous phase equilibrium of SO 2 /H 2 O and HNO 3 /H 2 O. But SO 2 (g)  H 2 SO 4 (g) and NO X (g)  HNO 3 (g). What about solubility of H 2 SO 4 (g) and NO X (g)? They are soluble, but not important contributors to acid rain. Rate of gas phase oxidation of SO 2 (g) to H 2 SO 4 (g) by OH: 0.3%-3% / hr Rate of gas phase oxidation of NO 2 (g) to HNO 3 (g) by OH: 10 x faster.

lecture 15 natural sulfur, acid rain The NH 3 /H 2 O System