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Drill:. List five factors & explain how each affect reaction rates. Drill: Solve Rate Law. A + B C + D fast 4 C + A 2G fast 2 K 4D + B fast G + K 2 Q + 2 W fast Q + W Prod. slow. Chemical Equilibria. Equilibrium.

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**Drill:**• List five factors & explain how each affect reaction rates**Drill: Solve Rate Law**A + B C + D fast 4 C + A 2G fast 2 K 4D + B fast G + K 2 Q + 2 W fast Q + W Prod. slow**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=**AP CHM HW**• Read: Chapter 12 • Work problems: 5, 7, & 12 • Page: 365**CHM II HW**• Read: Chapter 17 • Work problems: 17 & 21 • Page: 745**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**Drill: SO2 + O2 SO3**• Determine the magnitude of the equilibrium constant & the partial pressure of each gas is 0.667 Atm.**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 & solve when PNO2 &**PN2O4 = 50 kPa each: 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**• Set & balance rxn • Assign amounts • Write eq expression • Substitute amounts • Solve for x**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**AP CHM HW**• Read: Chapter 12 • Problems: 37 & 39 • Page: 367

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