Anyone who has never made a mistake has never tried anything new. - Albert Einstein – Instruction does not prevent wasted time or mistakes; and mistakes themselves are often the best teachers of all. - James Anthony Froude -

CHEMICAL EQUILIBRIUM IN SCIENCE, YOU OFTEN RUN INTO SITUATIONS WHERE CHANGE IS OCCURING IN OPPOSITE DIRECTIONS AT THE SAME RATE. AS A RESULT, THE SYSTEM ITSELF EXHIBITS NO NET CHANGE. SOME EXAMPLES OF THIS ARE 1.THE VAPOR PRESSURE OF A LIQUID 2.THE SOLUBILITY OF SOLID IN A SATURATED SOLUTION 3.AVERAGE KINETIC ENERGY OF ATOMS/MOLECULES IN A SYSTEM AT CONSTANT TEMPERATURE. THE SAME TYPE OF SITUATION CAN OCCUR WITH A CHEMICAL REACTION. THESE ARE ALL EXAMPLES OF DYNAMIC EQUILIBRIUM – CHANGE IS OCCURING IN BOTH DIRECTIONS AT THE SAME RATE, SO THAT THERE IS NO OBSERVABLE NET CHANGE.

SO FAR, IN OUR DISCUSSIONS WE HAVE ASSUMED THAT THE REACTIONS THAT WE HAVE TALKED ABOUT GO TO COMPLETION. THIS IS NOT ALWAYS THE CASE. MANY EXAMPLES EXIST WHERE THE REACTION IS REVERSIBLE. LET’S CONSIDER A HYPOTHETICAL REACTION A + B = C + D IF YOU START WITH JUST THE REACTANTS, THE RATE OF THE FORWARD REACTION WOULD PROBABLY BE HIGH. AS THE REACTION PROCEEDS, THE CONCENTRATIONS OF A AND B ARE REDUCED, SO THE FORWARD REACTION SLOWS DOWN.

NOW, LET’S ASSUME THAT THE THE REVERSE REACTION IS NOT ONLY POSSIBLE, BUT PROBABLE. AS THE REACTION PROCEEDS, THE CONCENTRATIONS OF C AND D START TO INCREASE, SO THE REVERSE REACTION STARTS, AND ITS RATE INCREASES AS THE CONCENTRATIONS OF C AND D INCREASE. EVENTUALLY, A POINT WOULD BE REACHED WHERE THE RATES WOULD BE EQUAL, AND THE REACTION WOULD BE IN EQUILIBRIUM. THIS IS ILLUSTRATED IN THE DIAGRAM ON THE NEXT SLIDE.

OUR EQUATION FOR THE REACTION IS A + B = C + D THE RATE EXPRESSIION FOR THE FORWARD REACTIION IS Rate f = k f [A][B] THE RATE EXPRESSION FOR THE REVERSE REACTION IS Rate r = k r [C][D] AT EQUILIBRIUM Rate f = Rate r SO k f [A][B] = k r [C][D] and dividing across k f /k r = [C][D]/[A][B] or K EQ = [C][D]/[A][B]

THIS IS SOMETIMES REFERRED TO AS THE LAW OF MASS ACTION, AND IT EXPRESSES THE RELATIONSHIP BETWEEN THE CONCENTRATIONS OF THE REACTANTS AND PRODUCTS AT EQUILIBRIUM. THE EQUILIBRIUM CONSTANT IS EQUAL TO THE PRODUCT OF THE CONCENTRATIONS OF THE PRODUCTS DIVIDED BY THE PRODUCT OF THE CONCENTRATIONS OF THE REACTANTS, EACH RAISED TO THE POWER EQUAL TO ITS COEFFICIENT IN THE BALANCED CHEMICAL EQUATION. THE EQUILIBRIUM CONSTANT DEPENDS ONLY ON THE STOICHIOMETRY OF THE REACTION AND NOT ON ITS MECHANISM. EQUILIBRIUM CAN BE APPROACHED FROM EITHER DIRECTION.

FOR A GIVEN CHEMICAL REACTION, IF: K EQ >> 1 THEN THE EQUILIBRIUM LIES TO THE RIGHT, AND PRODUCTS ARE FAVORED K EQ << 1THEN THE EQUILIBRIUM LIES TO THE LEFT, AND REACTANTS ARE FAVORED HOMOGENEOUS EQUILIBRIA – ALL REACTANTS AND PRODUCTS ARE IN THE SAME PHASE HETEROGENEOUS EQUILIBRIA – DIFFERENT PHASES INVOLVED IF A PURE SOLID OR A PURE LIQUID IS INVOLVED IN A HETEROGENEOUS EQUILIBRIUM, ITS CONCENTRATION IS NOT INCLUDED IN THE EQUILIBRIUM CONSTANT FOR THE REACTION.

EVEN THOUGH THEY DO NOT APPEAR IN THE EQUILIBRIUM CONSTANT EXPRESSION, THE PURE SOLIDS AND LIQUIDS MUST BE PRESENT AT EQUILIBRIUM. WHEN A SOLVENT IS INVOLVED IN THE EQUILIBRIUM, ITS CONCENTRATION IS EXCLUDED FROM THE EQUILIBRIUM CONSTANT EXPRESSION, AS LONG AS THE CONCENTRATIONS OF THE REACTANTS AND PRODUCTS ARE LOW, SO THAT THE SOLVENT IS ESSENTIALLY A PURE SUBSTANCE. APPLICATIONS: 1.MAGNITUDE OF K EQ PREDICTS IF PRODUCTS OR REACTANTS ARE FAVORED. 2.PREDICTS DIRECTION REACTION WILL PROCEED TO ACHIEVE EQUILIBRIUM 3.K EQ CAN BE USED TO CALCULATE CONCENTRATIONS OF REACTANTS OR PRODUCTS AT EQUILIBRIUM

EXAMPLES (courtesy of Brown and LeMay): Write K eq for the following reactions: 3NO (g) = N 2 O (g) + NO 2 (g) CH 4 (g) + 2H 2 S (g) = CS 2 (g) + 4H 2 (g) Ni(CO) 4 (g) = Ni (s) + 4CO (g) HF (aq) = H + (aq) + F - (aq) 2 Ag (s) + Zn +2 (aq) = 2 Ag + (aq) + Zn (s) NH 3 (aq) + H 2 O (l) = NH + 4 (aq) + OH - (aq)

DO THE FOLLOWING REACTIONS FAVOR PRODUCTS OR REACTANTS? N 2 (g) + O 2 (g) = 2NO(g) K eq = 1.5 x SO 2 (g) + O 2 (g) = 2SO 3 (g) K eq = 2.5 x NO (g) + O 2 (g) = 2NO 2 (g) K eq = 5.0 x HBr (g) = H 2 (g) + Br 2 (g) K eq = 5.8 x NOTE IN REACTIONS INVOLVING GASEOUS SUBSTANCES, THE EQUILIBRIUM CONSTANT IS USUALLY GIVEN IN TERMS OF PARTIAL PRESSURES RATHER THAN CONCENTRATIONS.

METHANOL (CH 3 OH) IS PRODUCED COMMERCIALLY BY THE CATALYZED REACTION OF CARBON MONIXIDE AND HYDROGEN: CO (g) + 2 H 2 (g) = CH 3 OH (g) AN EQUILIBRIUM MIXTURE IN A 1 LITER VESSEL IS FOUND TO CONTAIN MOLE CH 3 OH, MOLE CO, AND MOLE H 2 AT 500 o K. CALCULATE THE K EQ AT THIS TEMPERATURE. NOTE IN REACTIONS INVOLVING GASEOUS SUBSTANCES, THE EQUILIBRIUM CONSTANT IS USUALLY GIVEN IN TERMS OF PARTIAL PRESSURES RATHER THAN CONCENTRATIONS.

1.2 CO2 2 CO + O2 The above reaction was allowed to react in a flask at 1400C. At equilibrium, the following results were found: CO2= M; CO = 0.26 M; and O2= 0.13 M. Calculate the equilibrium constant for this reaction. 2. NH3 + 7 O2 4 NO2 + 6 H2O Ammonia and oxygen were put in a flask and allowed to react at 5000C. Analysis at equilibrium gave the following results: NH3 = M; O2 = 1.95 M; NO2 = 1.05 M; and H2O = M. Calculate the equilibrium constant for the reaction.