Environmental Chemistry CHEM 3650
Utah State University
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This 5 page Class Notes was uploaded by Ansel Ankunding PhD on Wednesday October 28, 2015. The Class Notes belongs to CHEM 3650 at Utah State University taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/230511/chem-3650-utah-state-university in Chemistry and Biochemistry at Utah State University.
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Date Created: 10/28/15
mt mm lt39lgt mqm mm w m mm mm m m arm swm Gupta 5 MAW11m Equlhbna The Equalme Cunmnl Wmemdmm mwmmmmmmmm K H17 mmmwmm WWW 39 m mmwmmm a sm m mm 4mm mwumwmmmm kdhsuumms W Mampulaung Ethmum Mampulaung Equlhhnum Cunmnls Cm 39memv Fm mm mmm ma arm mroA K M r K 39muwmhmmltmmmhmm Him aqzwm quotET m Example 0 For the reaction HZO 9HOH39 KKW lOH39J10gtlt1039 If for NH3aqHZO 9 NH OH39 KNHZ 18gtlt10 3 0 What is the equilibrium constant for NH 9 NH3aq H 1 a H 9H K KW NH OH39 aNHAaq 126 K2 119 1 NHXQHNH3aq K3KW 56x10quot NH3 Thermodynamics The equilibrium constant derives from thermodynamics 0 Whether a reaction is favored or not depends upon two parameters 7 Enthalpy the heat absorbed or released during a reaction 7 Entropy the degree of disorder among the producw and reactants Enthalpy The enthalpy change AH for a reaction is the heat absorbed or released under constant pressure 7 Standard enthalpy change AH is heat absorbed When all species are in standard states HClg e Haq c139 aq AH 77485 kJrnol 25 c 0 A negative sign for AHquot indicates that heat is released during the reaction or is exothermic 0 A positive sign for AHquot indicates that heat is absorbed or is endothermic This means that the solution gets colder Entropy Entropy S ofa species is a measure of is disorder A gas is more disordered than a liquid a liquid more than a solid and ions in solution are more disordered than their salt HClg9 Haq Cl39aq AS 71304 JK rnol 25 C KClslt gt Kaq Cl39aq AS 764 JKrnol 25 C 0 AS9 is the change in entropy for a reaction when all species are in their standard states A positive value means producm are more disordered than reactants Free Energy 0 Since enthalpy AH and entropy AS both exert an in uence on whether a reaction proceeds one must account for both to determine whether the reaction is favored We use the change in Gibbs free energy AG for this AG AH 7 TAS 0 If AGlt0 then the reaction is favored Free Energy Example 0 What is the Gibbs free energy change for the dissociation of HCl when all species are in their standard states HCl 9 HWZQ Cl39 341 HClg 9Haq c1 aq AH 77485 kJrnol 25 c HClg aHWaq Cl39aq AS 71304 JKrnol ZSDC AG AH 7 TAS 7485gtlt103 Jrnol 29815 K1304 JK 3597 kJrnol 0 Since AG lt0 then the reaction is favored How Does Free Energy Relate to Equilibrium The equilibrium constant of a reaction K can be related to free energy by 7A6 K 9 RT where R is the gas constant 8314 Imol0K and T is temperature in K Example 0 What is the equilibrium constant for the dissociation of HCl AG 3597 kJmol Mn 7 735 mm3 Jmol K 9 e 8314JmolquotKquotZ9815K 2 00X106 The equilibrium constant is large so HClg is very soluble in water and nearly completely ionized If AGa lt0 or Kgtl we say that the reaction is spontaneous Le Chatelier s Principle 0 If a system at equilibrium is changed Le Chatlier s principle states that the system will proceed in the direction to offset the change B10 2Cr 4HZOH Br39 Crzoi39 8H 2 r a K Br iicrzo7 i 1X101125uc Bro Cr One set of concentrations that exists in equilibrium is H50 M Cr2O7239410 M Cr300030 M Br3910 M BrO39043 M Le Chatelier s Principle 0 If we increase CrZO7239 to 020 M what direction will the reaction proceed Bro 2Cr 4H20 9 Br39 Crzoi39 8H 7 Le Chatelier says to the le Let s con rm this Q is the reaction quotientthe same as K except When the system is not at equilibrium Br39HCrin39HH F 7 10020508 Q Broc 2 0043000301 2x10 Since Q gt K the reaction must indeed go to the left to reduce the numerator and decrease the denominator Effects on Equilibrium 0 If products are added or reactants are removed the reaction goes left 0 If reactants are added or products are removed the reaction goes right 0 K is also dependent on temperature 7 For an endothermic reaction Alf gt0 K increases if the temperature increases 7 For an exothermic reaction Alf lt0 K decreases if the temperature increases Thermodynamics 0 Remember that equilibrium equations make thermodynamic predictions about how a chemical system will procee 7 We calculate What must happen for the system to reach equilibrium 0 Equilibrium equations cannot say anything about how fast the reaction proceeds The rate or kinetics of the move to equilibrium may be fast or extremely slow Solubility Product 0 The solubility product is the equilibrium constant for a reaction in which a solid dissolves and gives ions in solution ngcl2 s aHgi 2cr KS Hg c1392 12gtlt10quot8 0 When a solution has excess undissolved solid the solution is saturated Solubility Product Example 0 What is the concentration of ng2 in a solution saturated with HgZClz Hg2C12segt Hg 2c1 0 Every molecule of HgZCl2 that dissolves produces 1 ng2 ion in solution and 2 Cl39 so if xHg2 than Cl39 must equal 2x Hg c1392 x2x2 12gtlt10quot8 4x3 12x10quot8 4813 pf 67gtlt10397M K 4 J Example with a 2nd Source of Ions 0 What will the concentration of ng2 in a solution containing 0030 M NaCl saturated with HgZClz Hg2C12segt Hg 2c1 Initial Concentration Conc solid 0 0030 Final Concentration Concf solid x 2x 0030 Hg c1392 x2x00302 12x10quot8 4x3 012x2 00009c12gtlt10quot8 0 So what do you do here Do you want to solve this equation Example 2 Cont 0 We learned in the first example that Cl39 from saturated HgZClzaq was 67x10397 M We ve added Cl39 and based on Le Chatlier s principle we would expect even less C139 to be soluble Let s assume that 2x is small compared to 0030 M Hg c1392 x00302 12gtlt10quot8 00009x 12x10quot8 2 x 13gtlt10quot5 Our assumption was valid 2x 26 x103915 M which is much much smaller than 0030 M When to Approximate Approximations are a legitimate way to solve equilibria problems that would difficult otherwise But you must always confinn your assumptions with the final result 0 lfyou have to solve the equation and can t figure out how to do it trialanderror guessing can often be used TrialandError Guessing 0 For the original question we had Hg c1392 x2x 00302 12gtlt10quot8 4x3 012x2 00009x 12x10quot8 0 A series of guesstimates gives the following x4101 25x10395 Way too big x400001 90x10399 still too big x1x103913 90x103917 too big x1x103915 90x103919 too small close 7 Note ifa speci c level of accuracy is speci ed eg to 1 you will have to make a lot ofguesses Separaqu by Frenp zum