Chapter 12 - Chemical Equilibrium Chapter 12 Chemical...

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Dr. Manuel P. Soriaga Department of Chemistry Chapter 12 Chemical Equilibrium Chemical Equilibrium Consider the reaction of SO 2 with O 2 to form SO 3 at 1500 K. Initially, only forward reaction occurs: As the product SO 3 accumulates, collisions between them induces a reverse reaction: The more the SO 3 produced, the more significant the reverse reaction: Eventually, the rate of the reverse reaction equals that of the forward reaction (chemical equilibrium): 2SO 2(g) + O 2(g) 2SO 3(g) 2SO 2(g) + O 2(g) 2SO 3(g) 2SO 2(g) + O 2(g) 2SO 3(g) 2SO 2(g) + O 2(g) 2SO 3(g) Establishment of Equilibrium 2SO 2(g) + O 2(g) 2SO 3(g) Concentrations (M) Time t eq 0.400 0.100 [SO 2 ] [SO 3 ] [O 2 ] 0.344 0.056 0.172 A [SO 2 ] t eq 0.400 0.100 Time [SO 3 ] [O 2 ] 0.344 0.172 0.056 B [ ] Concentration at equilibrium Concentration Ratios at Equilibrium A + B C Rate Law for forward reaction: r f = k f C A C B Rate Law for reverse reaction: r b = k b C C At equilibrium: r f = r b r f = k f C A C B = k b C C k f /k b K eq = [C]/[A][B] [ ] equilibrium concentration The Equilibrium Constant For the general reversible reaction: aA + bB cC + dD The equilibrium constant K c at a given temperature is written as: where [ ] equilibrium concentration K c = C c D d A a B b Homogeneous and Heterogeneous Equilibria Homogeneous: single-phase reaction 2H 2 S (g) + 3O 2(g) 2H 2 O (g) + 2SO 2(g) Heterogeneous: multi-phase reaction 2HgO (s) 2Hg (l) + O 2(g) K c = H 2 O 2 SO 2 2 H 2 S 2 O 2 3 K c = O 2 Concentration terms for liquids and solids do not appear in the K c expression
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Gas-phase Homogeneous Equilibria Equilibrium constant can be written as K p : P partial pressure = (n/V)RT n/V concentration Example: 2H 2 S (g) + 3O 2(g) 2H 2 S (g) +2SO 2(g) K p = p H 2 O 2 p SO 2 2 p H 2 S 2 p O 2 3 K p = K c RT Δ n Δ n ( Σ stoich coeff prod ) - ( Σ stoich coeff react ) Properties of K c Dimensionless quantity: concentration terms are normalized with respect to standard states Reaction-specific: each reaction has its own K c value Temperature dependent Dependent only on equilibrium concentrations, not of initial concentrations Its value indicates the extent of the reaction: K c >> 1: [Products] >> [Reactants] K c << 1: [Products] << [Reactants] K c = 1: [Products] ~ [Reactants] The value of K c depends on the direction of the balanced equation for the reaction: If the reaction is reversed, the new K c is the reciprocal of the original K c . 2HBr (g) + Cl 2(g) 2HCl (g) + Br 2(g) Variations in K c K c = HCl 2 Br 2 HBr 2 Cl 2 = 4 x 10 4 K c ' = HBr 2 Cl 2 HCl 2 Br 2 = 1 K c = 2.5 x 10 -5 2HCl (g) + Br 2(g) 2HBr (g) + Cl 2(g) If the reaction is multiplied by a factor n, the new K c is obtained by raising the original K c to the power n: N 2(g) + 3H 2(g) 2NH 3(g) K 1 = 3.5 x 10 8 1 / 2 N 2(g) + 3 / 2 H 2(g) NH 3(g) K 2 = (K 1 ) 1/2 = (3.5 x 10
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This note was uploaded on 02/23/2011 for the course CHEM 101 taught by Professor Williamson during the Spring '08 term at Texas A&M.

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Chapter 12 - Chemical Equilibrium Chapter 12 Chemical...

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