13 Dec. 2010 Take Out Homework: Lab Notebook and Kinetics AP Questions. Objective: SWBAT Do now:
Agenda • Do now • Kinetics Problem Set Solutions • Introduction to Equilibrium • Demo Homework: p. #
Chemical Equilibrium Chapter 14
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 and • the concentrations of the reactants and products remain constant • However, there is a lot of activity at the molecular level! • Reactants continue to form products, while products continue to yield reactants!
H2O (l) H2O (g) N2O4(g) 2NO2(g) Physical vs. Chemical Equilibrium Physical equilibrium NO2 Chemical equilibrium
equilibrium equilibrium equilibrium N2O4(g) 2NO2(g) Start with NO2 Start with N2O4 Start with NO2 & N2O4
N2O4(g) 2NO2(g) K = aA + bB cC + dD [NO2]2 [C]c[D]d [N2O4] K = [A]a[B]b Equilibrium Constant = 4.63 x 10-3 Law of Mass Action
Law of Mass Action • For a reversible reaction at equilibrium and a constant temperature, a certain ratio of reactant and product concentration has a constant value, K (the equilibrium constant) • concentrations may vary • but as long as temperature stays the same and the reaction is at equilibrium, K will not change.
aA + bB cC + dD [C]c[D]d K = [A]a[B]b Equilibrium Will K >> 1 Lie to the right Favor products K << 1 Lie to the left Favor reactants
N2O4(g) 2NO2(g) Kc = Kp = In most cases Kc Kp P 2 [NO2]2 NO2 [N2O4] aA (g) + bB (g)cC (g) + dD (g) P N2O4 Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. Kp = Kc(RT)Dn Dn = moles of gaseous products – moles of gaseous reactants = (c + d) – (a + b)
CH3COOH (aq) + H2O (l) CH3COO-(aq) + H3O+ (aq) ′ Kc = ′ Kc [H2O] [CH3COO-][H3O+] [CH3COO-][H3O+] = [CH3COOH][H2O] [CH3COOH] Homogeneous Equilibrium [H2O] = constant Kc = General practice not to include units for the equilibrium constant.
Example 1 Write expressions for Kc, and KP if applicable, for the following reversible reactions at equilibrium: • HF(aq) + H2O(l) H3O+(aq) + F-(aq) • 2NO(g) + O2(g) 2NO2(g) • CH3COOH(aq) + C2H5OH(aq) CH3COOC2H5(aq) + H2O(l) Hints: KP applies only to gaseous reactions; the concentration of the solvent (usually water) does not appear in the equilibrium constant expression.
Problem 1 Write Kc and KP for the decomposition of dinitrogen pentoxide: 2N2O5(g) 4NO2(g) + O2(g)
Example 2 The following equilibrium process has been studied at 230oC: 2NO(g) + O2(g) 2NO2(g) In one experiment, the concentrations of the reacting species at equilibrium are found to be [NO]=0.0542 M, [O2]=0.127 M and [NO2]=15.5 M. Calculate the equilibrium constant (Kc) of the reaction at this temperature.
Problem 2 Carbonyl chloride (COCl2), also called phosgene, was used in WWI as a poisonous gas. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form carbonyl chloride CO(g) + Cl2(g) COCl2(g) at 74oC are [CO]=1.2x10-2 M, [Cl2]=0.054 M and [COCl2]=0.14 M. Calculate the equilibrium constant (Kc).
Example 3 • At the equilibrium constant KP for the decomposition of phosphorus pentachloride to phosphorus trichloride and molecular chlorine PCl5(g) PCl3(g) + Cl2(g) is found to be 1.05 at 250oC. If the equilibrium partial pressures of PCl5 and PCl3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl2 at 250oC?
Problem 3 The equilibrium constant KP for the reaction 2NO2(g) 2NO(g) + O2(g) is 158 at 1000 K. Calculate PO2 if PNO2 = 0.400 atm and PNO = 0.270 atm.
Example 4 Methanol (CH3OH) is manufactured industrially by the reaction CO(g) + 2H2(g) CH3OH(g) The equilibrium constant Kc for the reaction is 10.5 at 220oC. What is the value of KP at this temperature?
Problem 4 For the reaction N2(g) + 3H2(g) 2NH3(g) KP is 4.3x10-4 at 375oC. Calculate Kc for the reaction.
14 Dec. 2010 • Objective: SWBAT write equilibrium constant expressions for reactions with heterogeneous equilibria, and calculate reaction quotients. • Do now: The equilibrium constant KP for the reaction 2SO3(g) ⇄ 2SO2(g) + O2(g) is 1.8x10-5 at 350oC. What is KC for this reaction?
Agenda • Do now • Homework solutions • Heterogeneous equilibrium constants • Reaction quotients p. 649 #23, 25, 27, 29, 31, 35, 37
CaCO3(s) CaO (s) + CO2(g) ′ ′ Kc = Kc x [CaO][CO2] [CaCO3] [CaCO3] [CaO] Kp = PCO 2 Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. [CaCO3] = constant [CaO] = constant Kc= [CO2] = The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.
CaCO3(s) CaO (s) + CO2(g) PCO PCO 2 2 does not depend on the amount of CaCO3 or CaO = Kp
Example 5 Write the equilibrium constant expression Kc, and KP if applicable, for each of the following heterogeneous sytems: • (NH4)2Se(s) 2NH3(g) + H2Se(g) • AgCl(s) Ag+(aq) + Cl-(aq) • P4(s) + 6Cl2(g) 4PCl3(l)
Problem 5 Write equilibrium constant expressions for Kc and KP for the formation of nickel tetracarbonyl, which is used to separate nickel from other impurities: Ni(s) + 4CO(g) Ni(CO)4(g)
Example 6 Consider the following heterogeneous equilibrium: CaCO3(s) CaO(s) + CO2(g) At 800oC, the pressure of CO2 is 0.236 atm. Calculate a) Kp b) Kc for the reaction at this temperature.
Problem 6 Consider the following equilibrium at 395 K: NH4HS(s) NH3(g) + H2S(g) The partial pressure of each gas is 0.265 atm. Calculate KP and Kc for the reaction.
For which of the following reactions is Kc equal to KP? • 4NH3(g) + O2(g) ⇄ 4NO(g) + 6H2O(g) • 2H2O2(aq) ⇄ 2H2O(l) + O2(g) • PCl3(g) + 3NH3(g) ⇄ 3HCl(g) + P(NH2)3(g)
A + B C + D C + D E + F • What if the product molecules of one reversible reaction are involved in a second reaction as the products?
′ ′ Kc = A + B C + D C + D E + F A + B E + F [C][D] [E][F] [E][F] ′ ′′ Kc x Kc Kc = [C][D] [A][B] [A][B] Kc = ′ ′ Kc Kc = Kc ′′ Kc If 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.
N2O4(g) 2NO2(g) 2NO2(g) N2O4(g) ′ K = K = = 4.63 x 10-3 [NO2]2 [N2O4] [N2O4] [NO2]2 1 K = = 216 When the equation for a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.
Writing Equilibrium Constant Expressions • The concentrations of the reacting species in the condensed phase are expressed in M. In the gaseous phase, the concentrations can be expressed in M or in atm. • The concentrations of pure solids, pure liquids and solvents do not appear in the equilibrium constant expressions. • The equilibrium constant is a dimensionless quantity. • In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. • If a reaction can be expressed as a 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.
15 Dec. 2010 • Objective: SWBAT calculate reaction quotient and calculate an equilibrium concentration. • Do now: If the equilibrium constant for the reaction N2O4 (g) ⇄ 2NO2 (g) is 4.63 x 10-3 calculate the equilibrium constant for the reaction 2NO2(g) ⇄ N2O4(g)
Agenda • Do now • Homework solutions • Reaction quotient • Equilibrium concentrations Homework: p. 650 #23, 30, 36, 39
kf kr A + 2B AB2 [AB2] = Kc = kf [A][B]2 kr The Relationship between Chemical Kinetics and Chemical Equilibrium ratef = kf[A][B]2 rater = kr[AB2] Equilibrium ratef = rater kf[A][B]2 = kr[AB2]
What does the Equilibrium Constant tell us? • We can use the equilibrium constant to calculate unknown equilibrium concentrations of products or reactants (at a constant temperature) • We can predict the direction in which a reaction will proceed to achieve equilibrium
What if the reaction is not yet at equilibrium? • Calculate reaction quotient (Qc) by substituting the initial concentrations into the equilibrium constant expression.
The reaction quotient (Qc)is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression. • IF • Qc > Kcsystem proceeds from right to left to reach equilibrium • Qc = Kc the system is at equilibrium • Qc < Kcsystem proceeds from left to right to reach equilibrium
Example 7 At the start of the reaction, there are 0.249 mol N2, 3.21x10-2 mol H2 and 6.42x10-4 mol NH3 in a 3.50 L reaction vessel at 375oC. If the equilibrium constant Kc for the reaction N2(g) + 3H2(g) ⇄ 2NH3(g) is 1.2 at this temperature, decide whether the system is at equilibrium. If it is not, predict which way the net reaction will proceed.
Problem 7 The equilibrium constant (Kc) for the formation of nitrosyl chloride, an orange-yellow compound, from nitric oxide and molecular chlorine 2NO(g) + Cl2(g) ⇄ 2NOCl(g) is 6.5x104 at 35oC. In a certain experiment, 2.0x10-2 mole of NO, 8.3x10-3 mole of Cl2, and 6.8 moles of NOCl are mixed in a 2.0 L flask. In which direction will the system proceed to reach equilibrium?
What do you do with the Kc and KP? • If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations!
Calculating Equilibrium Concentrations • Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. • Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. • Having solved for x, calculate the equilibrium concentrations of all species.
Br2 (g) 2Br (g) Example 8 At 1280oC 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.
At 1280oC 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. [Br]2 [Br2] Kc = Br2 (g) 2Br (g) Br2 (g) 2Br (g) = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x Let x be the change in concentration of Br2 Initial (M) 0.063 0.012 ICE Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x Solve for x
-b ± b2 – 4ac x = 2a Br2 (g) 2Br (g) Initial (M) 0.063 0.012 Change (M) -x +2x Equilibrium (M) 0.063 - x 0.012 + 2x = 1.1 x 10-3 Kc = (0.012 + 2x)2 0.063 - x 4x2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x2 + 0.0491x + 0.0000747 = 0 ax2 + bx + c =0 x = -0.0105 x = -0.00178 At equilibrium, [Br] = 0.012 + 2x = -0.009 M or 0.00844 M At equilibrium, [Br2] = 0.062 – x = 0.0648 M
Example 9 A mixture of 0.500 mol H2 and 0.500 mol I2 was placed in a 1.00 L stainless steel flask at 430oC. The equilibrium constant KC for the reaction H2(g) + I2(g) ⇄ 2HI(g) is 54.3 at this temperature. Calculate the concentrations of H2, I2 and HI at equilibrium.
Example 10 For the reaction below, the initial concentrations of each species are H2 = 0.00623 M, I2 = 0.00414 M and HI = 0.224 M at 430oC. The equilibrium constant KC for the reaction is still 54.3. Calculate the concentrations of these species at equilibrium. H2(g) + I2(g) ⇄ 2HI(g)
Homework • p. 650 #23, 30, 36, 39
Problem 8 Consider the reaction below: H2(g) + I2(g) ⇄ 2HI(g) with an equilibrium constant of 54.3. Starting with a concentration of 0.040 M for HI, calculate the concentrations of HI, H2 and I2 at equilibrium.