2 Equilibrium is a state in which there are no observable changes as time goes by. Chemical equilibrium is achieved when:the rates of the forward and reverse reactions are equal andthe concentrations of the reactants and products remain constantPhysical equilibriumH2O (l)H2O (g)Chemical equilibriumN2O4 (g)2NO2 (g)A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products.At equilibrium, the conversion is still going on,but the two rates are equal.
3 Equilibrium Will aA + bB cC + dD K = [C]c[D]d [A]a[B]b For a reaction at particular temperature,the equilibrium constant K:Here we first discuss the equilibrium constant of forward Rxn.Equilibrium WillK >> 1Lie to the rightFavor productsK << 1Lie to the leftFavor reactants
4 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. For Gaseous Rxn:N2O4 (g) NO2 (g)Kp =NO2P2N2O4PP= (n/V)RTAt constant temp., the pressure is related to the concentration, mol/L. Kp applies to gaseous reaction.Kc =[NO2]2[N2O4]Kc : concentration in molarity, mol/LIn most casesKc KpaA (g) + bB (g) cC (g) + dD (g)Kp = Kc(RT)DnorKc = Kp(RT)-DnDn = moles of gaseous products – moles of gaseous reactants= (c + d) – (a + b)
5 The equilibrium constant Kp for the reaction is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = atm and PNO = atm?2NO2 (g) NO (g) + O2 (g)2Kp =2PNO POPNOPO2= KpPNOPO2= 158 x (0.400)2/(0.270)2= 347 atm14.2
6 The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.CO (g) + Cl2 (g) COCl2 (g)[COCl2][CO][Cl2]=0.140.012 x 0.054Kc == 220Kp = Kc(RT)DnDn = 1 – 2 = -1R =T = = 347 KKp = 220 x ( x 347)-1 = 7.714.2
7 Homogeneous Equilibrium For Liquid Rxn:CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)[CH3COO-][H3O+][CH3COOH][H2O]Kc =‘[H2O] = constant[CH3COO-][H3O+][CH3COOH]Kc =The equilibrium constant of solvent (usually water) does not appear in equilibrium constant expression.14.2
9 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.CaCO3 (s) CaO (s) + CO2 (g)Kc =‘[CaO][CO2][CaCO3][CaCO3] = constant[CaO] = constantKc = [CO2]Kp = PCO2PCO2does not depend on the amount of CaCO3 or CaOThe concentrations of pure solids, pure liquids and solvent do not appear in the equilibrium constant expressions.
11 Consider the following equilibrium at 295 K: The partial pressure of each gas is atm. Calculate Kp and Kc for the reaction?NH4HS (s) NH3 (g) + H2S (g)Kp = PNH3H2SP= x =Kp = Kc(RT)DnKc = Kp(RT)-DnDn = 2 – 0 = 2T = 295 KKc = x ( x 295)-2 = 1.20 x 10-414.2
12 Multiple equilibrium ‘ ‘ ‘ ‘ ‘ Kc = [C][D] [A][B] Kc = [E][F] [C][D] A + B C + DKc‘C + D E + F[E][F][A][B]Kc =A + B E + FKcKc =Kc‘xIf a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions.14.2
13 ‘ Equilibrium Constant for a reversible reaction A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products.Forward RxnReverse RxnN2O4 (g) NO2 (g)2NO2 (g) N2O4 (g)= 4.63 x 10-3K =[NO2]2[N2O4]K =[N2O4][NO2]2‘=1K= 216When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.14.2
14 Chemical Kinetics and Chemical Equilibrium Elementary reaction:ratef = kf [A][B]2A + 2B AB2kfkrrater = kr [AB2]Equilibriumratef = raterkf [A][B]2 = kr [AB2]kfkr[AB2][A][B]2=Kc =14.3
15 What does equilibrium constant tell us? Kc helps us to predict the direction in which a reaction mixture will proceed to achieve equilibrium and to calculate the concentrations of reactants and products once the equilibrium has been achieved.
16 Predict the reaction direction The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.IFQc > Kc system proceeds from right to left to reach equilibriumQc = Kc the system is at equilibriumQc < Kc system proceeds from left to right to reach equilibrium14.4
18 Calculating Equilibrium Concentrations The equilibrium constant (Kc) for the reactionIs 1.1 x If the initial concentrations are [Br2] = M and [Br] = M, calculate the concentrations of these species at equilibrium.Br2 (g) Br (g)Let x be the change in concentration of Br2Br2 (g) Br (g)Initial (M)0.0630.012Change (M)-x+2xEquilibrium (M)xx[Br]2[Br2]Kc =Kc =( x)2x= 1.1 x 10-3Solve for x
19 Kc =( x)2x= 1.1 x 10-34x x = – x4x x = 0-b ± b2 – 4ac2ax =ax2 + bx + c =0x =x =Br2 (g) Br (g)Initial (M)Change (M)Equilibrium (M)0.0630.012-x+2xxxAt equilibrium, [Br] = x = Mor MAt equilibrium, [Br2] = – x = M14.4
21 Le Châtelier’s Principle If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.Changes in ConcentrationRemoveAddAddRemoveaA + bB cC + dDChangeShifts the EquilibriumIncrease concentration of product(s)leftDecrease concentration of product(s)rightIncrease concentration of reactant(s)rightDecrease concentration of reactant(s)left
22 Le Châtelier’s Principle Changes in Volume and PressureA (g) + B (g) C (g)ChangeShifts the EquilibriumIncrease pressureSide with fewest moles of gasDecrease pressureSide with most moles of gasIncrease volumeSide with most moles of gasDecrease volumeSide with fewest moles of gasChange pressure will not affect the concentration of reacting species in condensed phase, e.g. liquid and solid.In general, the increase of pressure favors the net reaction that decreases the total number of moles of gases and decrease of pressure favors the net reaction that increases the total number of moles of gases
24 Le Châtelier’s Principle Changes in TemperatureChangeExothermic RxEndothermic RxIncrease temperatureK decreasesK increasesDecrease temperatureK increasesK decreasesIncrease temperature favors an endothermic reaction, and the temperature decrease favors an exothermic reaction.14.5
25 Le Châtelier’s Principle Adding a CatalystCatalyst lowers Ea for both forward and reverse reactions.does not change equilibrium constant Kdoes not shift the position of an equilibrium systemsystem will reach equilibrium sooneruncatalyzedcatalyzed14.5
26 Le Châtelier’s Principle Change EquilibriumConstantChangeShift EquilibriumConcentrationyesnoPressureyesnoVolumeyesnoTemperatureyesyesCatalystnono14.5
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