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Chemical Equilibria. Equilibrium. The point at which the rate of a forward reaction = the rate of its reverse reaction. Equilibrium. The concentration of all reactants & products become constant at equilibrium. Equilibrium.
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Equilibrium • The point at which the rate of a forward reaction = the rate of its reverse reaction
Equilibrium • The concentration of all reactants & products become constant at equilibrium
Equilibrium • Because concentrations become constant, equilibrium is sometimes called steady state
Equilibrium • Reactions do not stop at equilibrium, forward & reverse reaction rates become equal
Reaction • aA(aq)+ bB(aq) cC(aq)+ dD(aq) • Ratef = kf[A]a[B]b • Rater = kr[C]c[D]d • At equilibrium, Ratef = Rater • kf[A]a[B]b = kr[C]c[D]d
At equilibrium, Ratef = Rater kf[A]a[B]b = kr[C]c[D]d kf /kr = ([C]c[D]d)/ ( [A]a[B]b) kf /kr = Kc = Keq in terms of concentration Kc = ([C]c[D]d)/ ( [A]a[B]b)
Reaction aA(g)+ bB(g)<-->cC(g)+ dD(g) Ratef = kfPAaPBb Rater = krPCcPDd At equilibrium, Ratef = Rater kfPAaPBb = krPCcPDd
At equilibrium, Ratef = Rater kfPAaPBb = krPCcPDd kf /kr = (PCcPDd)/ ( PAaPBb) kf /kr = Kp = Keq in terms of pressure Kp = (PCcPDd)/ ( PAaPBb)
All Aqueous aA + bB pP+ qQ
Equilibrium Expression ( Products)p (Reactants)r Keq=
Equilibrium Applications • When K >1, [p] > [r] • When K <1, [p] < [r]
Equilibrium Calculations • Kp = Kc(RT)Dngas
Equilibrium Expression • Reactants or products not in the same phase are not included in the equilibrium expression
Equilibrium Expression aA(s)+ bB(aq)<--> cC(aq)+ dD(aq) [C]c [D]d [B]b Keq=
Reaction Mechanism • Sequence of steps that make up the total reaction process
Reaction Mechanism • 1) A + B <---> C Fast • 2) A + C <---> D Fast • 3) B + D <---> H Fast • 4) H + A -----> P Slow
Reaction Mechanism • The rate determining step is the slowest step • H + A ----> P Slow • Rate = k4[H][A]
Reaction Mechanism • Rate = k4[H][A] • Because H is not one of the original reactants, H cannot be used in a rate expression
Reaction Mechanism • 3) B + D <---> H • K3 = [H]/([B][D]) • [H] = K3[B][D]
Reaction Mechanism • [H] = K3[B][D] • Rate = k4[H][A] • Rate = k4K3[B][D][A]
Reaction Mechanism • 2) A + C <---> D • K2 = [D]/([A][C]) • [D] = K2[A][C]
Reaction Mechanism • [D] = K2[A][C] • Rate = k4K3[B][D][A] • Rate = k4K3[B]K2[A][C][A] • Rate = k4K3 K2[B][A]2[C]
Reaction Mechanism • 1) A + B <---> C • K1 = [C]/([A][B]) • [C] = K1[A][B]
Reaction Mechanism • [C] = K1[A][B] • Rate = k4K3 K2[B][A]2[C] • Rate = k4K3 K2[B][A]2K1[A][B] • Rate = k4K3 K2K1 [B]2[A]3 • Rate = K[B]2[A]3
Solve Rate Expression • 1) A + B <---> 2C Fast • 2) A + C <---> D Fast • 3) B + D <---> 2H Fast • 4) 2H + A ----> P Slow
Reaction Mechanism • When one of the intermediates anywhere in a reaction mechanism is altered, all intermediates are affected
Reaction Mechanism • 1) A + B <---> C + D • 2) C + D <---> E + K • 3) E + K <---> H + M • 4) H + M <----> P
Lab Results • % 100 80 60 40 • RT 5.21 8.42 11.9 21.7 • WR 2.75 4.23 7.96 11.2
Applications of Equilibrium Constants where [A], [B], [P], and [Q] are molarities at any time. Q = K only at equilibrium.
NH3 H2 + N2 At a certain temperature at equilibrium Pammonia = 4.0 Atm, Phydrogen = 2.0 Atm, & Pnitrogen = 5.0 Atm. Calculate Keq:
Equilibrium Applications • When K > Q, the reaction goes forward • When K < Q, the reaction goes in reverse
Le Chatelier’s Principle • If stress is applied to a system at equilibrium, the system will readjust to eliminate the stress
LC Eq Effects • A(aq) +2 B(aq) <---> • C(aq) + D(aq) + heat • Write equilibrium exp: • What happens if:
LC Eq Effects • 2 A(aq) + B(s) <---> • C(aq) +2 D(aq) + heat • Write equilibrium exp: What happens if:
LC Eq Effects • 2 A(g) + 2 B(g) <---> • 3 C(g) + 2 D(l) • What happens if:
Drill: Write the equilibrium expression & slove for: N2O4(g) 2 NO2(g)
Equilibrium Applications • DG = DH - TDS • DG = - RTlnKeq
Equilibrium Calculations • aA + bB <--> cC + dD • Stoichiometry is used to calculate the theoretical yield in a one directional rxn
Equilibrium Calculations • aA + bB <--> cC + dD • In equilibrium rxns, no reactant gets used up; so, calculations are different
Equilibrium Calculations • CO + H2O CO2 + H2 • Calculate the partial pressure of each portion at eq.when 100.0 kPa CO & 50.0 kPa H2O are combined: • Kp = 3.4 x 10-2
Equilibrium Calculations • CO + H2O CO2 + H2 • Calculate the partial pressure of each portion when 100 kPa CO & 50 kPa H2O are combined: • Kp = 3.4 x 10-2
Equilibrium Calculations • CO H2O CO2 H2 • 100 -x 50 - x x x • Kp =PCO2PH2 = x2 • PCOPH2O(100-x) (50-x) • Kp = 3.4 x 10-2
Equilibrium Calculations x2 x2 (100 -x)(50 - x) = 5000 -150x + x2 = 3.4 x 10-2 x2 = 170 - 5.1x + 0.034x2 0.966x2 + 5.1x - 170 = 0
Write the Eq Expression A(aq)+ 2 B(aq) C(s)+ 2 D(aq) Calculate Keq if: [A] = 0.30 M [B] = 0.20 M C = 5.0 g [D] = 0.30 M
Write the Eq Expression A(aq)+ 3 B(aq) C(g)+ 2 D(aq) Calculate Keq if: [A] = 0.40 M [B] = 0.20 M C = 75 kPa [D] = 0.40 M
Equilibrium Calculations Xe (g) + F2(g) XeF2(g) Calculate the partial pressure of each portion when 50.0 kPa Xe & 100.0 kPa F2 are combined: Kp = 4.0 x 10-6
Equilibrium Calculations Xe (g) + F2(g) XeF2(g) Calculate the partial pressure of each portion when 40.0 kPa Xe & 80.0 kPa F2 are combined: Kp = 4.0 x 10-6