CHAPTER 6 : EQUILIBRIA

CHAPTER 6 : EQUILIBRIACHEMICAL EQUILIBRIA

During a game, players enter and leave. Always the same number of players on field.

H2O(l) H2O(g)

Photochromicsunglasses AgCl + light Ag + Cl(transparent) (dark)

Reversible reactiona chemical reaction that can occur in both the forward and the reverse directionsN2(g) + 3H2(g) 2NH3(g)

Chemical equilibriuma state in which the forward and reverse reactions balance each other because they take place at equal ratesdynamic state; no net changeLaw of chemical equilibriumAt a given temperature, a chemical system might reach a state in which a particular ratio of reactant and product concentrations has a constant value.Equilibrium constant expressionratio of molar concentrations of products to reactants; each raised to a power equal to coefficient in balanced equation

aA + bB cC + dD

Equilibrium constant (Keq)numerical value of the ratio of product to reactant concentrationsconstant only at a specific temperatureValue of equilibrium constant (Keq) shows the extent to which reactants are converted into products.Keq > 1: Products are favored at equilibriumKeq < 1: Reactants are favored at equilibriumHomogenous equilibrium applies to reactions in which all reacting species are in the same phase.N2O4 (g) 2NO2 (g)Kc = [NO2]2[N2O4]Kp = NO2P2N2O4PaA (g) + bB (g) cC (g) + dD (g)Kp = Kc(RT)DnDn = moles of gaseous products moles of gaseous reactants= (c + d) (a + b)In most casesKc KpHomogeneous EquilibriumCH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)Kc =[CH3COO-][H3O+][CH3COOH][H2O][H2O] = constantKc = [CH3COO-][H3O+][CH3COOH]= Kc [H2O]General practice not to include units for the equilibrium constant.

14.2Homogeneous equilibriumall reactants and products in same phaseWrite an equilibrium constant expression for:N2(g) + 3H2(g) 2NH3(g)Keq =[NH3]2 ___________[N2] [H2]3

The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.

CO (g) + Cl2 (g) COCl2 (g)Kc = [COCl2][CO][Cl2]=0.140.012 x 0.054= 220Kp = Kc(RT)DnDn = 1 2 = -1R = 0.0821T = 273 + 74 = 347 KKp = 220 x (0.0821 x 347)-1 = 7.7The equilibrium constant Kp for the reaction

is 158 at 1000K. What is the equilibrium pressure of O2 if the PNO = 0.400 atm and PNO = 0.270 atm?

22NO2 (g) 2NO (g) + O2 (g)Kp = 2PNO PO2PNO22PO2= KpPNO22PNO2PO2= 158 x (0.400)2/(0.270)2= 347 atmHeterogeneous equilibriumreactants and products present in more than one physical stateH2O(l) H2O(g)

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] = Kc x[CaCO3][CaO]Kp = PCO2The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.Heterogeneous equilibriumSince concentrations of pure liquids and solids remain constant, these substances are omitted from the equilibrium constant expression.Write an equilibrium constant expression for:H2O(l) H2O(g)Keq =[H2O(g)]

Equilibrium constant expression Keq = . . .Products over reactants raised to power of coefficient; leave out pure solids and liquids.Equilibrium concentrations can vary from trial to trial.Equilibrium position

H2(g) +I2(g) 2HI(g)

constantN2O4 (g) 2NO2 (g)N2O4 (g) 2NO2 (g)= 4.63 x 10-3K = [NO2]2[N2O4]2NO2 (g) N2O4 (g)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.

Each set of equilibrium concentrations represent an equilibrium position.A system has only one value for Keq at a specific temperature, however there are unlimited number of equilibrium positions.A system at equilibrium must:take place in closed systemtemperature remain constantall reactants and products are present (both reactions can occur)Write the equilibrium constant expression for

N2(g) + O2(g) 2NO(g)

Calculate the value of Keq if [N2] = 0.20 mol/L,[O2] = 0.15 mol/L, and[NO] = 0.0035 mol/L.Keq = 4.1 x 10-4What does the value of Keq tell you about the equilibrium?The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. Vice versa for KpIFQc > 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 equilibrium

At 12800C the equilibrium constant (Kc) for the reaction

Is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) 2Br (g)Br2 (g) 2Br (g)Let x be the change in concentration of Br2Initial (M)Change (M)Equilibrium (M)0.0630.012+x-2x0.063 + x0.012 - 2x[Br]2[Br2]Kc = Kc = (0.012 - 2x)20.063 + x= 1.1 x 10-3ICESolve for x[Br]i2[Br2]iQc = Qc = (0.012)20.063= 2.29 x 10-3 > KcRx proceeds to the left to achieve eqm; prodt decompose to yield rxtKc = (0.012 - 2x)20.063 + x= 1.1 x 10-3ax2 + bx + c =0-b b2 4ac 2ax = Br2 (g) 2Br (g)Initial (M)Change (M)Equilibrium (M)0.0630.012+x-2x0.063 + x0.012 - 2xx = 0.00178x = 0.0105At equilibrium, [Br] = 0.012 - 2x = -0.009 Mor 0.00844 MAt equilibrium, [Br2] = 0.062 + x = 0.0648 MThank you for listening.Now, its question time!1) What are the factors that play a role in balancing the concentration of stratospheric ozone?a. Formation of stratospheric ozoneb. Depletion of stratospheric ozoneAnswer:2) Which of the following are the reactants in the formation of stratospheric ozone?Note: There is more than one answer.Nitrogen monoxideNitrogen dioxideOxygen free radicalsOxygenOzoneAnswer: B & D3) There are two steps in the formation of stratospheric ozone. What are they? Step 1:Step 2:Photo dissociation of nitrogen dioxideFormation of ozone4) In the formation of stratospheric ozone, Step 1 can only take place when there is ______ light. ( bright / infrared / dark / colourful / ultraviolet ) 5) What is the overall chemical equation showing the formation and depletion of stratospheric ozone in a dynamic equilibrium?NO2(g) + O2(g)NO(g) + O3(g)UV lightThats all for now. Hope youve learned something today. = )

What happens when a system is at equilibrium and you upset the balance?When a system in dynamic equilibrium is to change, the system reacts to remove the change so equilibrium reestablish.Le Chteliers Principle Changes in ConcentrationN2 (g) + 3H2 (g) 2NH3 (g)AddNH3Equilibrium shifts left to remove change

Le Chteliers Principle Changes in Concentration continuedChangeShifts the EquilibriumIncrease concentration of product(s)leftDecrease concentration of product(s)rightDecrease concentration of reactant(s)Increase concentration of reactant(s)rightleftaA + bB cC + dD AddAddRemoveRemove

Le Chteliers Principle Changes in Volume and PressurePCl3 (g) + Cl2 (g) PCl5(g)ChangeShifts the EquilibriumIncrease pressure Side with fewest moles of gasDecrease pressureSide with most moles of gasDecrease volumeIncrease volumeSide with most moles of gasSide with fewest moles of gasLe Chteliers Principle Changes in Volume and PressureH2 (g) + I2 (g) 2Hl(g)ChangeShifts the EquilibriumIncrease pressure No effect on eqm system since both sides contain same no of moleculesDecrease pressureDecrease volumeIncrease volumeLe Chteliers Principle Changes in TemperatureChangeExothermic RxEqm position & eqm constant Endothermic Rx

colderhotterThe Effect Temperature on Equilibrium Consider the following exothermic reaction:N2(g) + 3H2(g) 2NH3(g); DHo = -92 kJ, The forward reaction produces heat => heat is a product.When heat is added to increase temperature, reverse reaction will take place to absorb the heat; If heat is removed to reduce temperature, a net forward reaction will occur to produce heat.Exothermic reactions favor low temperature conditions.The Effect Temperature on Equilibrium Consider the following endothermic reaction: N2O4(g) 2 NO2(g); DHo = +57 kJEndothermic reaction absorbs heat heat is a reactant;If heat is added to increasing the temperature, it will cause a net forward reaction.If heat is removed to reduce the temperature, it will cause a net reverse reaction.Endothermic reactions favor high temperature condition.

uncatalyzedcatalyzedCatalyst lowers Ea for both forward and reverse reactions.Catalyst does not change equilibrium constant or shift equilibrium. Adding a Catalyst does not change K does not shift the position of an equilibrium system system will reach equilibrium soonerLe Chteliers PrincipleRelationship between Keq & TGiven by Vant Hoff equation; ln K = - H + C RTK =eqm constantH =enthalpy change of rxR = gas constantT = temperature (K)C = constanti) Exothermic rx :ii) Endothermic rxln K1/T (K-1)Grad = -H R ln K1/T (K-1)ooooGrad = -H R * +ve slope * -ve slope Chemical Equilibria in Industrial Processes Production of Sulfuric Acid, H2SO4 in Contact process;S8(s) + 8 O2(g) 8SO2(g)2SO2(g) + O2(g) 2SO3(g); DH = -198 kJSO3(g) + H2SO4(l) H2S2O7(l)H2S2O7(l) + H2O(l) 2H2SO4(l)

The second reaction is exothermic and has high activation energy; reaction is very slow at low temperature,. At high temperature reaction goes faster, but the yield would be very low. An optimum condition is achieved at moderate temperatures and using catalysts to speed up the reaction. Reaction also favors high pressure.Chemical Equilibria in Industrial ProcessesThe production of ammonia by the Haber process:N2(g) + 3H2(g) 2NH3(g); DH = -92 kJThis reaction is exothermic and very slow at low temperature. Increasing the temperature will increase reaction rate, but will lower the yield. An optimum condition is achieved at moderate temperature of 250 to 300oC with catalyst added to increase the reaction rate.Increasing the pressure will favor product formation.Reaction favors low temperature and high pressure conditions.

The production of nitric acid by the Ostwald process:Step 1:4NH3(gas) + 5O2(gas) 4NO (gas) + 6H2O (gas)Step 2:2NO (gas) + O2(gas) 2NO2(gas)Step 3:3NO2(gas) + H2O (liquid) 2HNO3(aqueoussolution) + NO (gas)Step 4:4NO2(gas) + O2(gas) + 2H2O (liquid) 4HNO3(aqueous)

reaction is exothermic...conditions that would favor the forward reaction and shift theequilibriumto the right would be decreasingthetemperature, increase the concentration , and increasing the pressure and volume.Thecatalystthat is used for this reaction is a platinumgauze. It would be heated, however sometimes insubstitute..a copper wire/rod can serve as apropercatalystfor this process

Concentration of Stratospheric OzoneConcentration of Stratospheric OzoneFormation of Ozone in StratosphereDepletion of Ozone in StratosphereFormation of ozone in the stratosphereOzone, O3 is formed from chemical reaction between nitrogen dioxide and oxygen in 2 steps.Step 1Photo dissociation of nitrogen dioxideStep 2Formation of OzoneStep 2: Formation of ozoneThe oxygen free radicals rapidly react with oxygen in the air to form ozone molecules.O (g) + O2(g) O3(g)Combining the equations in Steps 1 & 2, we have:

NO2(g) + O2(g) NO(g) + O3(g)Depletion of ozone in stratosphereOzone reacts with nitrogen monoxide (formed in Step 1 of formation of O3) to produce nitrogen dioxide and oxygen.

O3(g) + NO(g) O2(g) + NO2(g)

Its actually the reverse of the formation of stratospheric ozone!Rate of formation of ozoneThe system is in dynamic equilibrium.Rate of depletion of ozone=NO2(g) + O2(g)NO(g) + O3(g)Concentration of ozone remains constantUV lightEveryday, 3x108 tons of stratospheric ozone formed & destroyed by natural processes.Stratospheric O3 IMPORTANT ; photodissociation rx, prevent harmful uv radiation frm reaching earths surface. (200 310 nm) are absorbed frm the uv raysPhotons in this range ; enough energy to cause skin cancer & damage to living plantsFor every 1% in stratospheric ozone, additional 2% damaging uv rays reach earths surface.Le Chteliers PrincipleChangeShift EquilibriumChange Equilibrium ConstantConcentrationyesnoPressureyesnoVolumeyesnoTemperatureyesyesCatalystnono