Henry s Law. Henry s Law describes the equilibrium absorption of a gas into a liquid (its solubility):

Transcription

1 M 507 Lecture 9 ext reading Chapter 7 PS #6 due Nov. 7 PS #7 due Dec. 6 Paper due Dec. (Final hursday, Dec 0, -3 PM) oday s topic enry s Law; queous Phase tmospheric Chemistry

2 enry s Law enry s Law describes the equilibrium absorption of a gas into a liquid (its solubility): (g) + O(l) O or (g) (aq) his is an equilibrium expression and we can define an equilibrium constant, in this case called a enry s Law coefficient:. he aqueous concentration ( O or (aq)) is usually expressed in moles/liter or M; and (g) in atmospheres (partial pressure), so = [(aq)]/p or [ O]/p ; units M atm -

3 enry s Law is often written [(aq)] = p enry s Law describes only the physical solubility of the gas and not any chemical changes that may occur after dissolution. (For gases that hydrolyze and dissociate, we will later describe an effective enry s Law coefficient *, which takes the additional dissociated species [i.e. chemical changes] into account.) Examples of enry s Law coefficients: 3

4 QUIE INSOLUBLE RELIVELY MODERELY SOLUBLE VERY SOLUBLE Methanol Methyl peroxide Peroxyacetic acid Formic acid cetic acid 4

5 emperature Dependence of enry s Law emperature Dependence of equilibrium constants is given by the van t off equation: d ln d R Δ is the reaction enthalpy (heat of dissolution) don t confuse this with the enry s Law coefficient. Rearrange the van t off equation: d ln R d hen integrate from to, assuming that Δ is constant over that temperature range ln R 5

6 6 ) ( ) ( ln ) ( ln ) ( ln R )exp ( ) ( R OR ll the Δ s for enry s Law equilibrium reactions listed in able 7.3 are negative; so as temperature increases, solubility decreases. hat is, gases are more soluble at lower temperature. (hink bottled soda.)

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8 Pressure Dependence of enry s Law he pressure dependence of the relation is very small the solubility of gases like N and e increases by ~0% as the pressure increases from atm to 0 atm. We will neglect this small effect. For normal tropospheric pressures, enry s Law coefficients will be considered independent of pressure. his may play a small role in the upper troposphere and stratosphere. 8

9 Liquid Water in the tmosphere he liquid water content of air is abbreviated by LWC, or more simply L, and is variously defined as L = g(liquid water)/m 3 (air) Since liquid water very conveniently has a mass density of very nearly g/cm 3, we can also write L = cm 3 (liquid water)/m 3 (air) 9

10 Liquid Water Content of ir It is also sometimes convenient to express LWC as a dimensionless volume mixing ratio w L (vol water/vol air) = 0-6 L (g/m 3 ) Some typical values for L: Clouds: range from 0.05 to 3 g(water)/m 3 most often 0. to 0.3 g(water)/m 3 Fogs: range from 0.0 to 0.5 g(water)/m 3 0

11 Gas/queous-Phase Distribution Factor useful solubility benchmark for atmospheric applications is the distribution of a soluble species between gas and aqueous phases in a typical cloud (i.e., an environment with measurable liquid water). If the species resides mainly in the gas phase, it is considered somewhere between insoluble and slightly soluble. If it resides almost exclusively in the aqueous phase, it is considered very soluble. In between, species are considered moderately soluble. s we saw from able 7., most species are in the slightly soluble to insoluble range.

12 Distribution Factor (cont.) f moles of moles in solution per liter of in air per liter of of air air p p L(0 / R 6 ) 0 6 RL Rw L Notes: hink moles! whether volume is air or water (Ideal gas law: n/v = air conc. = p/r) R = atm L mol - K - L = LWC in g m -3 or cm 3 m -3 p = moles per liter solution nd 0-6 L = (liters of solution)/(liters of air)

13 Distribution Fraction Expressions GS: X g = /(+f ) - gas fraction QUEOUS: X aq = f /(+f ) liquid fraction X aq 6 0 RL 6 0 RL RwL Rw Figure 7.3 shows X aq as a function of L and : If < 400 M atm -, almost always <% dissolved relatively insoluble 000 < < 00,000 M atm -, moderately soluble If > 500,000 M atm -, very soluble Even relatively insoluble and moderately soluble gases, if they are reactive in aqueous forms, can be and are important (i.e., SO and O 3 ). L 3

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15 Examples O 3 and O Consider a case where a cloud forms in a previously clear air mass containing [O 3 ] = 50 ppb and [ O ] = ppb. he cloud temperature is 0º C, and L = 0.3 g/m 3. Ozone is relatively insoluble with O3 (98) =.x0 - M/atm. From figure 7.3 (and/or a calculation of f ) it is clear that almost all (>> 99.99%) of the O 3 will remain in the gas phase, and that the equilibrium gas concentration will be [O 3 ] (g) = 50 ppb. his means a simple application of enry s Law will yield [O 3 ](aq). 5

16 Ozone example (cont.) Step. Find O3 (0ºC)= O3 (73K). Use van t off rule (on board). O3 (73K) = M/atm Step. pply enry s Law to find [O 3 ](aq) Since the gas phase concentration is virtually unchanged at p O3 =5.0x0-8 atm, [O 3 ](aq) = O3 p O3 = M/atm*(5.0x0-8 atm) =.8 x0-9 M 6

17 ydrogen Peroxide Example For O it is likely that a significant fraction of the initial ppb will dissolve into the cloud water and therefore the equilibrium gas phase concentration will not be ppb. We need this equilibrium p O before we can calculate [ O ](aq). Step. Find O (0ºC)= O (73K). Use van t off and Δ from able 7.3. O (73K) = 9.4x0 5 M/atm his is almost 0x bigger then at 98K! 7

18 O (cont.) Step. Calculate the distribution factor and the equilibrium gas phase concentration. Recall L = 0.3 g/m 3 ; w L = 3x0-7 (li water)/(li air) f O = O (73K) Rw L f O = 6.33 Fraction in gas phase: X g O = /(+f O ) X g O = /7.33 =.36 Equilibrium [ O ](g) =.36 ppb; or p O =.36x0-0 atm 8

19 O (cont.) Step 3. pply enry s Law to find [ O ](aq) [ O ](aq) = O p O = 9.4x0 5 M/atm (.36x0-0 atm) =.8x0-4 M For gases of moderate to high solubility (i.e., or * > ) always check the distribution between gas and aqueous phases before using enry s Law. 9