1. Tutorial on Equilibrium ConstantCalculation, Le ChateliersPrinciple, Equilibrium Law. Prepared by Lawrence Kok http://lawrencekok.blogspot.com

2. Equilibrium LawEffect of Concentration on equilibrium constant, Kc Increase in Conc - position of equilibrium shift to right/left - Conc will be Reduced Decrease in Conc position of equilibrium shift to right/left - Conc will be Increased Equilibrium Law applies to system at equilibrium Kc is constant at constant temperature Kc unaffected by Concentration, Pressure and Catalyst Addition of reactants and products will only shift position of equilibrium to right or left changing to a new equilibriumconcentration but Kc remains the same Magnitude of Kc indicate the extend or how far the reaction forms product/reactants but not how fast Kc High High product but rate can be very slow Kc Low Low product but rate can be very fast 3. Equilibrium LawEffect of Concentration on equilibrium constant, Kc Increase in Conc - position of equilibrium shift to right/left - Conc will be Reduced Decrease in Conc position of equilibrium shift to right/left - Conc will be Increased Equilibrium Law applies to system at equilibrium Kc is constant at constant temperature Kc unaffected by Concentration, Pressure and Catalyst Addition of reactants and products will only shift position of equilibrium to right or left changing to a new equilibriumconcentration but Kc remains the same Magnitude of Kc indicate the extend or how far the reaction forms product/reactants but not how fast Kc High High product but rate can be very slow Kc Low Low product but rate can be very fast Reaction between H2 + I2 2HIKc = 46.4 at 730K Kc = [HI]2 = 46.4[H2][I2]Initial ConcH2 + I2 2HIEquilibrium Conc 4. Equilibrium LawEffect of Concentration on equilibrium constant, Kc Increase in Conc - position of equilibrium shift to right/left - Conc will be Reduced Decrease in Conc position of equilibrium shift to right/left - Conc will be Increased Equilibrium Law applies to system at equilibrium Kc is constant at constant temperature Kc unaffected by Concentration, Pressure and Catalyst Addition of reactants and products will only shift position of equilibrium to right or left changing to a new equilibriumconcentration but Kc remains the same Magnitude of Kc indicate the extend or how far the reaction forms product/reactants but not how fast Kc High High product but rate can be very slow Kc Low Low product but rate can be very fast Reaction between H2 + I2 2HIKc = 46.4 at 730K Kc = [HI]2 = 46.4[H2][I2]Initial ConcH2 + I2 2HIEquilibrium ConcReaction contain different Initial Conc of H2 and I2 and HI but at equilibium Kc remains the same regardless of initial concKc = [HI]2 = 46.4[H2][I2] 5. How dynamic equilibrium is achieved in a closed system? N2(g) + 3H2(g) 2NH3 (g)As reaction proceeds As reaction proceedsConc of H2 decrease over time Conc of NH3 increase over timeNH3H2 6. How dynamic equilibrium is achieved in a closed system?N2(g) + 3H2(g) 2NH3 (g) As reaction proceeds As reaction proceeds Conc of H2 decrease over time Conc of NH3 increase over timeNH3H2Rate of forward rxn, Rf decrease over timeRate of reverse rxn, Rr increase over timeNH3 H2 7. How dynamic equilibrium is achieved in a closed system?N2(g) + 3H2(g) 2NH3 (g) As reaction proceeds As reaction proceeds Conc of H2 decrease over time Conc of NH3 increase over timeNH3H2Rate of forward rxn, Rf decrease over timeRate of reverse rxn, Rr increase over timeNH3 H2 At dynamic equilibrium Forward Rate = Reverse Rate Conc of reactant/product remainconstant 8. How dynamic equilibrium is shifted when H2 is added ? N2(g) + 3H2(g) 2NH3 (g) When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc Rate of forward and reverse will increase New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr More products NH3 and less reactants N2 but Kc value remains unchangedAt equilibriumRate forward kf = Rate reverse krKc = 4.07 9. How dynamic equilibrium is shifted when H2 is added ? N2(g) + 3H2(g) 2NH3 (g) When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc Rate of forward and reverse will increase New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr More products NH3 and less reactants N2 but Kc value remains unchanged H2 addedAt equilibriumEquilibrium shifted to rightRate forward kf = Rate reverse kr Rate forward kf > Rate reverse krKc = 4.07 Qc = 2.24 10. How dynamic equilibrium is shifted when H2 is added ? N2(g) + 3H2(g) 2NH3 (g) When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc Rate of forward and reverse will increase New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr More products NH3 and less reactants N2 but Kc value remains unchanged H2 addedNew equilibrium ConcAt equilibriumEquilibrium shifted to rightAt equilibrium againRate forward kf = Rate reverse kr Rate forward kf > Rate reverse kr Rate forward kf = Rate reverse krKc = 4.07 Qc = 2.24Kc = 4.07 11. How dynamic equilibrium is shifted when H2 is added ? N2(g) + 3H2(g) 2NH3 (g) When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc Rate of forward and reverse will increase New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr More products NH3 and less reactants N2 but Kc value remains unchangedEquilibrium Conc for H2 = 0.82MEquilibrium Conc for N2 = 0.20MEquilibrium Conc for NH3 = 0.67MN2(g) + 3H2(g) 2NH3 (g) Kc = (NH3)2 (N2)(H2)3 Kc = (0.67)2(0.20)(0.82)3 Kc = 4.07 12. How dynamic equilibrium is shifted when H2 is added ? N2(g) + 3H2(g) 2NH3 (g) When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc Rate of forward and reverse will increase New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr More products NH3 and less reactants N2 but Kc value remains unchangedH2 addedEquilibrium Conc for H2 = 0.82M New Conc for H2 = 1.00MEquilibrium Conc for N2 = 0.20M Equilibrium Conc for N2 = 0.20MEquilibrium Conc for NH3 = 0.67MEquilibrium Conc for NH3 = 0.67MN2(g) + 3H2(g) 2NH3 (g) N2(g) + 3H2(g) 2NH3 (g) Kc = (NH3)2 Qc = (NH3)2 (N2)(H2)3 (N2)(H2)3 Kc = (0.67)2 Qc = (0.67)2(0.20)(0.82)3 (0.20)(1.00)3 Kc = 4.07 Qc = 2.24 13. How dynamic equilibrium is shifted when H2 is added ? N2(g) + 3H2(g) 2NH3 (g) When more H2 added, position equilibrium will shift to right, to reduce the H2 conc and increase the NH3 conc Rate of forward and reverse will increase New equilibrium concentration will be achieved when Rate of forward kf = Rate of reverse kr More products NH3 and less reactants N2 but Kc value remains unchangedH2 addedNew equilibrium ConcEquilibrium Conc for H2 = 0.82M New Conc for H2 = 1.00MNew Equilibrium Conc for H2 = 0.90MEquilibrium Conc for N2 = 0.20M Equilibrium Conc for N2 = 0.20MNew Equilibrium Conc for N2 = 0.19MEquilibrium Conc for NH3 = 0.67MEquilibrium Conc for NH3 = 0.67M New Equilibrium Conc for NH3 = 0.75MN2(g) + 3H2(g) 2NH3 (g) N2(g) + 3H2(g) 2NH3 (g) N2(g) + 3H2(g) 2NH3 (g) Kc = (NH3)2 Qc = (NH3)2 Kc = (NH3)2 (N2)(H2)3 (N2)(H2)3 (N2)(H2)3 Kc = (0.67)2 Qc = (0.67)2 Kc = (0.75)2(0.20)(0.82)3 (0.20)(1.00)3(0.19)(0.90)3 Kc = 4.07 Qc = 2.24Qc < Kc Kc = 4.07 2.24 < 4.07 Kc must remain constant Shift to the right Increase products and decrease reactants New equilibrium Conc is achieved Qc = Kc again 14. Equilibrium LawWhen a reversible reaction achieved dynamic equilibrium - aA + bB cC and dD Equilibrium constant Kc = ratio of molar conc of product (raised to power of their respective stoichiometry coefficient)to molar conc of reactants (raised to power of their respective stoichiometry coefficient) Kc is constant at constant temperatureaA + bB cC and dD Kc = (C)c(D)d(A)a(B)bAt equilibrium:N2O4 (g) 2NO2 (g)N2O4 (g) 2NO2 (g) Rf (rate forward) = Rr (rate reverse)Kc = k f / k rRf (rate forward) = Rr (rate reverse) kf [N2O4] = kr [NO2]2kf [N2O4] = kr [NO2]2 Kc = (NO2)2 kf/kr = (NO2)2(N2O4) (N2O4) Kc is a ratio of rate constant and is temperature dependent Kc is a ratio of products conc to reactants conc Magnitude of Kc indicate how far/extend of the reaction proceeds towards a product at a given temperatureSmall Kc :N2(g) + O2(g) 2NO(g) Kc = 1 x 10 -30 small Kc = (NO)2(N2)1(O2)1Kc small low product , more reactants , close to no reaction at allLarge Kc :2COg) + O2 2CO2(g) Kc = 2.2 x 10 22 high Kc = (CO2)2(CO)2(O2)1 Kc large high product , small reactants , close to completionIntermediate Kc :2HI(g) H2(g) + I2(g) Kc = 0.02 Kc = (H2)(I2) (HI)2 Kc intermediate significant amount of reactants and products 15. Equilibrium LawA reaction quotient Q is a ratio made up of product conc multiplied together and divided by reactant conc witheach term raised to the power of its stoichiometric coefficientaA + bB cC and dD Initial Conc Equilibrium Conc cd Qc = (C) (D) Qc = Kc at equilibriumKc = (C)c(D)d(A)a(B)b (A)a(B)b If overall reaction is the sum of two or more reactions, the overall reaction quotient.Kc is the product of reaction quotients for single reactions. A reaction quotient Kc for a forward reaction is the reciprocal of reaction quotient for reverse reaction.Kc forward = 1/Kc reverse If all coefficient are multiplied by some factor, that factor becomes the exponent for relating the reaction quotients 16. Equilibrium Law A reaction quotient Q is a ratio made up of product conc multiplied together and divided by reactant conc with each term raised to the power of its stoichiometric coefficient aA + bB cC and dD Initial Conc Equilibrium Conc cdQc = (C) (D) Qc = Kc at equilibrium Kc = (C)c(D)d (A)a(B)b(A)a(B)bIf overall reaction is the sum of two or more reactions, the overall reaction quotient. Kc is the product of reaction quotients for single reactions.N2(g) + O2(g) 2NO Kc1 = 4.3 x 10 -25 N2(g) + O2(g) 2NOKc1 = 4.3 x 10 -252NO(g) + O2(g) 2NO2(g)Kc2 = 6.4 x 10 9 2NO(g) + O2(g) 2NO2(g) Kc2 = 6.4 x 10 9 N2(g) + 2O2(g) 2NO2(g) Kc (overall) = Kc1 x Kc2Kc (overall) = 4.3 x 10 -25 x 6.4 x 10 9 A reaction quotient Kc for a forward reaction is the reciprocal of reaction quotient for reverse reaction.Kc forward = 1/Kc reverse 2HI H2 + I2 H2 + I2 2HI Kc forward = 1/Kc reverse Kc1 = (H2) (I2) Kc =(HI)2(HI)2(H2) (I2) Kc1 = 1/Kc Kc = 50 = 1/50 = 0.02If all coefficient are multiplied by some factor, that factor becomes the exponent for relating the reaction quotientsSO2 + 1/2O2 SO3 X factor of 1/22SO2 + O2 2SO3 X factor of 24SO2 + 2O2 ( 4SO3Kc1 = (SO3) Kc = (SO3)2 Kc2 = (SO3)4 (SO2)(O2)1/2(SO2)2(O2)(SO2)4(O2)2Kc1 = (SO3)2 x exponent Kc = 200Kc2 = (SO3)2 x exponent 2 (SO2)2(O2)(SO2)2(O2)Kc1 = Kc x exponent Kc2 = Kc x exponent 2Kc1 = (200)1/2Kc2 = (200)2 17. Equilibrium LawRelationship between Kc and Kp for gaseous system For gaseous partial pressure , Kp can be used PV = nRTConc = n/V P = (n/V) R T P = Conc R TN2O4 (g) 2NO2 (g) p = Conc RTEx 1 2SO2 + O2 2SO3 (pNO2) = Conc (NO2)RT (pN2O4) = Conc(N2O4)RT Partial pressure for SO2 = 0.05atm (pNO2) = (NO2)RT(pN2O4) = (N2O4) RTPartial pressure for O2 = 0.025atmPartial pressure for SO3 = 1.00atmKp =p(SO3)2 p(SO2)2p(O2) =(1.0)2 Kp = (pNO2)2(0.05)2(0.025) (pN2O4) = 1.6 x104 = (NO2)2(RT)2 (N2O4)(RT) Kp = (NO2)2RT Kc = (NO2)2(N2O4) (N2O4) Ex 22SO2 + O2 2SO3 Kp = 11.57atm-1 at 873K, R = 0.0821dm3 atmRelationship between Kp and Kc Kp = Kc(RT)n Kp = Kc(RT)n11.57 = Kc (0.0821 x 873)-1 n = Total number of moles of product Total number of reactantsKc = 11.57 x (0.0821 x 873)= 829.3 dm3mol -1 Relationship between Kp and Kc N2 + 3H2 2NH3Kp = Kc (RT)nKp = Kc(RT)2-4Kp = Kc(RT)-2 18. Equilibrium LawReaction CO + Br 2 COBr2. Equilibrium conc of the gas are shown belowCalculate Kc. (Volume vessel is 1dm3 )Equilibrium Conc : [CO] = 8.78 x 10-3 M, [Br2] = 4.90 x 10-3M, [COBr2] = 3.40 x 10-3MEx 1CO + Br2 COBr2 Equilibrium/Conc8.78 x 10-3 4.90 x 10-3 3.40 x 10-3Equilibrium Conc = Moles /VolumeReaction PCI5 PCI3 + CI2. Calculate equilibrium conc for CI2, Kc = 0.19M. (Volume vessel is 1dm3 )Equilibrium Conc : [PCI5] = 0.20M, [PCI3]= 0.01MEx 2PCI5PCI3 +CI2Equilibrium/Conc0.20 0.01xEquilibrium Conc = Moles /Volume Reaction N2O4 2NO2. Calculate equilibrium conc for N2O4, Kc = 1.06 x 10-5M. (Volume vessel is 1dm3 ) Equilibrium Conc : [NO2] = 1.85 x 10-3 MEx 3N2O4 2NO2Equilibrium/Concx1.85 x 10-3 Equilibrium Conc = Moles /Volume 19. Equilibrium LawReaction CO + Br 2 COBr2. Equilibrium conc of the gas are shown belowCalculate Kc. (Volume vessel is 1dm3 )Equilibrium Conc : [CO] = 8.78 x 10-3 M, [Br2] = 4.90 x 10-3M, [COBr2] = 3.40 x 10-3MEx 1CO + Br2 COBr2 Equilibrium/Conc8.78 x 10-3 4.90 x 10-3 3.40 x 10-3Equilibrium Conc = Moles /VolumeKc = (COBr2 )(Br2)(CO) = (3.40 x 10-3 )(8.78 x 10-3)(4.90 x 10-3)Kc = 79 dm3 mol-1Reaction PCI5 PCI3 + CI2. Calculate equilibrium conc for CI2, Kc = 0.19M. (Volume vessel is 1dm3 )Equilibrium Conc : [PCI5] = 0.20M, [PCI3]= 0.01MEx 2PCI5PCI3 +CI2Equilibrium/Conc0.20 0.01xEquilibrium Conc = Moles /VolumeKc=(PCI3)(CI2)(PCI5)0.19 = (0.01) (x)(0.20)x= 3.8M Reaction N2O4 2NO2. Calculate equilibrium conc for N2O4, Kc = 1.06 x 10-5M. (Volume vessel is 1dm3 ) Equilibrium Conc : [NO2] = 1.85 x 10-3 MEx 3N2O4 2NO2Equilibrium/Concx1.85 x 10-3 Equilibrium Conc = Moles /Volume Kc =(NO2)2(N2O4) 1.06 x10-5 = (1.85 x10-3)2(x) x = 0.323M 20. Equilibrium Law (Using ICE Method) 1 mol of ethanoic acid added to 1 mol of ethanol and at equilibrium, 0.67mol ester was produced. Calculate Kc . (Reaction volume is 1dm3)Ex 4CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial mole1.0 1.000 Change(1-0.67)(1-0.67) 0.67 0.67 Change = Reacted and Produced Equilibrium/Conc0.33/10.33/1 0.67/1 0.67/1 Equilibrium Conc = Moles /Volume In esterification, 2 mol of ethanoic acid, 3 mol of ethanol and 2 mol H2O were mixed . Calculate the equilibrium concentration if Kc for reaction is 4.0 (Reaction volume is 1dm3 )Ex 5CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial mole 2 3 0 2 Change (2- x)(3-x) x 2+xChange = Reacted and Produced Equilibrium/Conc (2-x)/1 (3-x)/1 x/1(2 + x)/1 Equilibrium Conc = Moles /Volume 21. Equilibrium Law (Using ICE Method) 1 mol of ethanoic acid added to 1 mol of ethanol and at equilibrium, 0.67mol ester was produced. Calculate Kc . (Reaction volume is 1dm3)Ex 4CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial mole1.0 1.000 Change(1-0.67)(1-0.67) 0.67 0.67 Change = Reacted and Produced Equilibrium/Conc0.33/10.33/1 0.67/1 0.67/1 Equilibrium Conc = Moles /Volume Kc = (CH3COOC2H5 ) (H2O )(CH3COOH ) (C2H5OH ) Kc = (0.67) (0.67) (0.33) (0.33) Kc = 4.12 In esterification, 2 mol of ethanoic acid, 3 mol of ethanol and 2 mol H2O were mixed . Calculate the equilibrium concentration if Kc for reaction is 4.0 (Reaction volume is 1dm3 )Ex 5CH3COOH + C2H5OH CH3COOC2H5 + H2O Initial mole 2 3 0 2 Change (2- x)(3-x) x 2+xChange = Reacted and Produced Equilibrium/Conc (2-x)/1 (3-x)/1 x/1(2 + x)/1 Equilibrium Conc = Moles /VolumeKc = (CH3COOC2H5) (H2O)(CH3COOH) (C2H5OH )Kc = (x) (2+x) (2-x) (3-x)Conc CH3COOC2H5 = 1.33M4 = (x) (2+x) Conc H2O = 3.33M(2-x) (3-x) Conc CH3COOH = 0.67Mx = 1.33Conc C2H5OH = 1.67M 22. Equilibrium Law (Using ICE Method)Reaction X + Y Z. Initial amount of X (0.02) and Y (0.01). At equilibrium 9.00 x 10-3 Z was producedCalculate equilibrium constant Kc. (Volume vessel is 600cm3 )Ex 6X+ Y Z Initial mole 0.02 0.01 0 Change (0.02 9.00 x 10-3) (0.01 - 9.00 x 10-3) 9.00 x 10-3Change = Reacted and Produced Equilibrium/Conc 0.011/0.60.001/0.69.00 x 10-3 /0.6 Equilibrium Conc = Moles /VolumeMixture of 1.9 mol H2 and 1.9 mol I2 was added to vessel . At equilibrium concentration 3mol of HI was formedCalculate Kc at 400C. (Volume vessel is 1dm3 )Ex 7 H2+ I22HIInitial mole 1.91.9 0Change (1.9 1.5) (1.9 1.5)3 Change = Reacted and ProducedEquilibrium/Conc 0.4/1 0.4/13/1 Equilibrium Conc = Moles /VolumeReaction SO2CI2 SO2 + CI2 . 0.05 mol of SO2CI2 was used and at equilibrium 0.0345 mol of CI2 was formedCalculate equilibrium constant Kc. (Volume vessel is 2dm3 )Ex 8 SO2CI2 SO2 + CI2Initial mole 0.050 0Change (0.05 0.0345) 0.03450.0345Change = Reacted and ProducedEquilibrium/Conc 0.0155/20.0345/20.0345/2Equilibrium Conc = Moles /Volume 23. Equilibrium Law (Using ICE Method)Reaction X + Y Z. Initial amount of X (0.02) and Y (0.01). At equilibrium 9.00 x 10-3 Z was producedCalculate equilibrium constant Kc. (Volume vessel is 600cm3 )Ex 6X+ Y Z Initial mole 0.02 0.01 0 Change (0.02 9.00 x 10-3) (0.01 - 9.00 x 10-3) 9.00 x 10-3Change = Reacted and Produced Equilibrium/Conc 0.011/0.60.001/0.69.00 x 10-3 /0.6 Equilibrium Conc = Moles /Volume Kc =(Z) (X)(Y) Kc = (0.015)(0.0183) (0.0016) Kc = 491M-1Mixture of 1.9 mol H2 and 1.9 mol I2 was added to vessel . At equilibrium concentration 3mol of HI was formedCalculate Kc at 400C. (Volume vessel is 1dm3 )Ex 7 H2+ I22HIInitial mole 1.91.9 0Change (1.9 1.5) (1.9 1.5)3 Change = Reacted and ProducedEquilibrium/Conc 0.4/1 0.4/13/1 Equilibrium Conc = Moles /Volume Kc = (HI)2(H2)(I2)=(3)2 (0.4)(04) Kc = 56Reaction SO2CI2 SO2 + CI2 . 0.05 mol of SO2CI2 was used and at equilibrium 0.0345 mol of CI2 was formedCalculate equilibrium constant Kc. (Volume vessel is 2dm3 )Ex 8 SO2CI2 SO2 + CI2Initial mole 0.050 0Change (0.05 0.0345) 0.03450.0345Change = Reacted and ProducedEquilibrium/Conc 0.0155/20.0345/20.0345/2Equilibrium Conc = Moles /Volume Kc = (SO2) (CI2)(SO2CI2) Kc = (0.0173) (0.0173)(0.00775) Kc = 0.0386M 24. Equilibrium Law (Using ICE Method)Reaction CO + H2O CO2 + H2. Initial amount of 3 mol CO and 3 mol H2O was used.Calculate the equilibrium concentration for H2 if Kc for reaction is 4.0 (Volume vessel is 1dm3 ) Ex 9 CO +H2O CO2 + H2Initial mole 33 0 0Changes(3- x) (3-x) x x Change = Reacted and ProducedEquilibrium/Conc (3-x)/1(3-x)/1 x/1 x/1 Equilibrium Conc = Moles /VolumeReaction PCI5 PCI3 + CI2. Initial amount of PCI5 is 0.01 mol. Kc = 0.19M. (Volume vessel is 0.5dm3 )Calculate equilibrium conc for PCI5, PCI3 and CI2Ex 10PCI5 PCI3+ CI2Initial mole 0.01 0 0Changes0.01 - x x xChange = Reacted and ProducedEquilibrium/Conc (0.01 x)/0.5 x/0.5 x/0.5Equilibrium Conc = Moles /Volume 25. Equilibrium Law (Using ICE Method)Reaction CO + H2O CO2 + H2. Initial amount of 3 mol CO and 3 mol H2O was used.Calculate the equilibrium concentration for H2 if Kc for reaction is 4.0 (Volume vessel is 1dm3 ) Ex 9 CO+H2O CO2 + H2Initial mole33 0 0Changes (3- x) (3-x) x x Change = Reacted and ProducedEquilibrium/Conc(3-x)/1(3-x)/1 x/1 x/1 Equilibrium Conc = Moles /Volume Kc = (CO2) (H2)(CO) (H2O ) Kc = (x) (x)(3-x) (3-x) 4 = (x)(x)(3-x) (3-x) x = 2MReaction PCI5 PCI3 + CI2. Initial amount of PCI5 is 0.01 mol. Kc = 0.19M. (Volume vessel is 0.5dm3 )Calculate equilibrium conc for PCI5, PCI3 and CI2Ex 10 PCI5 PCI3+ CI2Initial mole0.01 0 0Changes 0.01 - x x xChange = Reacted and ProducedEquilibrium/Conc(0.01 x)/0.5 x/0.5 x/0.5Equilibrium Conc = Moles /Volume Kc = (PCI3)(CI2) (PCI5) 0.19 = (x/0.5)2 (0.01 x)/0.5 x= 0.0091M [PCI5] = (0.01 0.0091)/0.5 = 1.8 x 10-3M [PCI3] = 0.0091/0.5 = 1.8 x 10-2M [CI2] = 0.0091/0.5 = 1.8 x 10-2M 26. Questions on IB Equilibrium LawEx 11Equilibrium formed during this reaction was investigated in two experiment carried out at different temperature. The result are shown in table below2HI H2 + I2 For each experiment, deduce the concentrations of other species present at equilibrium. Calculate the values of Kc for the forward reaction for each experiment.EXPT 12HI H2 + I2 Initial mole 0.0600 Change (0.06 0.02) 0.01 0.01Change = Reacted and Produced Equilibrium/Conc 0.04/10.01/1 0.01/1Equilibrium Conc = Moles /VolumeEXPT 22HI H2+I2 Initial mole 0 0.040.04 Change 0.040.020.02Change = Reacted and Produced Equilibrium/Conc 0.04/10.02/10.02/1Equilibrium Conc = Moles /Volume 27. Questions on IB Equilibrium LawEx 11Equilibrium formed during this reaction was investigated in two experiment carried out at different temperature. The result are shown in table below2HI H2 + I2 For each experiment, deduce the concentrations of other species present at equilibrium. Calculate the values of Kc for the forward reaction for each experiment.EXPT 12HI H2+ I2 Initial mole 0.060 0 Change (0.06 0.02) 0.010.01 Change = Reacted and Produced Equilibrium/Conc 0.04/10.01/10.01/1 Equilibrium Conc = Moles /VolumeKc =(H2) (I2) (HI)2Kc = (0.01) (0.01) (0.04)2Kc = 6.25 x10-2EXPT 22HI H2+ I2 Initial mole 00.04 0.04 Change 0.04 0.02 0.02Change = Reacted and Produced Equilibrium/Conc 0.04/1 0.02/1 0.02/1Equilibrium Conc = Moles /VolumeKc =(H2)(I2) (HI)2Kc = (0.02) (0.02) (0.04)2Kc = 0.25 28. Equilibrium Law (Using ICE Method) 1 mol of HI was heated at 400C and 0.78 mol of HI remains at equilibrium. Calculate Kc. (Volume vessel is 1dm3 )Ex 122HI H2 +I2 Initial mole 1 0 0 Change (1 -0.22) 0.110.11 Change = Reacted and Produced Equilibrium/Conc 0.78/10.11/10.11/1 Equilibrium Conc = Moles /VolumeMixture of 1 mol H2 and 2 mol I2 was added to a reaction vessel .Calculate the equilibrium concentration for HI, H2 and I2 at 400C. (Volume vessel is 1dm3 )Ex 13H2 + I2 2HI Initial mole 1 2 0 Change (1 x)(2-x) 2x Change = Reacted and Produced Equilibrium/Conc (1-x)/1 (2-x)/1 2x/1 Equilibrium Conc = Moles /Volume 29. Equilibrium Law (Using ICE Method) 1 mol of HI was heated at 400C and 0.78 mol of HI remains at equilibrium. Calculate Kc. (Volume vessel is 1dm3 )Ex 12 2HI H2 + I2 Initial mole1 00 Change(1 -0.22) 0.11 0.11 Change = Reacted and Produced Equilibrium/Conc0.78/10.11/1 0.11/1 Equilibrium Conc = Moles /VolumeKc = (H2)(I2) (HI)2Kc = (0.11) (0.11) (0.78)2Kc = 0.02Mixture of 1 mol H2 and 2 mol I2 was added to a reaction vessel .Calculate the equilibrium concentration for HI, H2 and I2 at 400C. (Volume vessel is 1dm3 )Ex 13 H2 + I2 2HI Initial mole1 2 0 Change(1 x)(2-x) 2xChange = Reacted and Produced Equilibrium/Conc(1-x)/1 (2-x)/1 2x/1Equilibrium Conc = Moles /Volume Kc1 = (HI)2 (H2)(I2) Kc1 = 1/Kc50 = (2x)2 Conc HI = 2x = 1.87M = 1/0.02 (1-x)(2-x)Conc H2 = 1 - x = 0.065M Kc1 = 50x = 0.935M Conc I2 = 2 - x = 1.065M 30. Effect of Concentration, Pressure, Temperature and catalyst on equilibrium systemEx 12NO (g) + O2 2NO2 (g) H = -ve What are the effect of Increasing Conc, Pressure, Temp and Catalyst on the rate, rate constant, Kc and position of equilibriumEx 24NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) H = -ve What are the effect of Increasing Conc, Pressure, Temp and Catalyst on the rate, rate constant, Kc and position of equilibrium 31. Effect of Concentration, Pressure, Temperature and catalyst on equilibrium systemEx 12NO (g) + O2 2NO2 (g) H = -ve What are the effect of Increasing Conc, Pressure, Temp and Catalyst on the rate, rate constant, Kc and position of equilibriumEx 24NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) H = -ve What are the effect of Increasing Conc, Pressure, Temp and Catalyst on the rate, rate constant, Kc and position of equilibrium 32. AcknowledgementsThanks to source of pictures and video used in this presentationThanks to Creative Commons for excellent contribution on licenseshttp://creativecommons.org/licenses/Prepared by Lawrence KokCheck out more video tutorials from my site and hope you enjoy this tutorialhttp://lawrencekok.blogspot.com