Equilibria - 1 EQUILIBRIA I Basis of Equilibrium A Q and...

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Unformatted text preview: 1 EQUILIBRIA I. Basis of Equilibrium. A. Q and equilibrium. 1. Consider the general reaction bB + cC dD + eE a. Α s time elapses, [B] and [C] decrease causing the rate of the forward reaction to decrease. Conversely, as the concentrations of product build up, the rate of the reverse reaction increases. When the forward and reverse reaction rates are equal. When this occurs, equilibrium is established. b. At equilibrium: Rate forward = Rate reverse. The concentrations of reactants and products will settle down to their equilibrium values and will no longer change with time. 2. The extent of a reaction can be followed using the proper reaction quotient , Q For a general reaction of the type, bB + cC dD + eE Q = a D d a E e a B b a C c b. Where the a’s are the activities, which are the effective concentrations. For calculations we make the following approximations for the activities: For gases, substitute their partial pressures in atm . For solutes, substitute their molar concentrations . For pure solids and liquids, substitute unity (1 ). c.Since at equilibrium the concentrations of reactants and products reach constant values, Q will equal a constant, K, called the equilibrium constant . Q eq = K d. For gaseous reactions, a X = P X therefore for bB( g ) + cC( g) dD( g) + eE( g) Q P = P D d P E e P B b P C c Where P’s are the instantaneous partial pressures At equilibrium, Q P = K P K P = P D d P E e P B b P C c Where P’s are the equilibrium partial pressures The symbols Q P and K P indicate that the partial pressures are being used for activities 3. Q and the spontaneous reaction direction. a. If Q < K, the reaction is spontaneous in the forward direction. The reactant concentrations will decrease and products increase. 2 b. If Q > K , the reverse direction is the spontaneous one. The reactant concentrations will increase and products decrease. c. In all cases, the reaction proceeds in one direction or the other until Q = K. 3. Some examples of equilibrium constant expressions: a. Gaseous reactions. N 2 (g) + 3H 2 (g) 2 NH 3 (g) K = K P = P NH 3 2 P N 2 P H 2 3 = 1.2x10 –6 at 700 °C Where the partial pressures are their equilibrium values b. Heterogeneous equilibria. C(s) + H 2 O(g) CO(g) + H 2 (g) K = P CO P H 2 P H 2 O Note that the activity of the pure solid, C(s), equals 1. In dealing with heterogeneous equilibria, the pure liquid or solid does not appear in the K, or Q, expression, as long as some is present, the amount is not important. c. Weak acid dissociation. HNO 2 ( aq ) + H 2 O( l ) H 3 O + ( aq ) + NO 2 – ( aq ) K = K a = [H 3 O + ][NO 2 — ] [HNO 2 ] = 7.2x10 —4 at 25 °C The symbol K a is used for the acid dissociation constant. Note that in aqueous solutions, the activity of water=1....
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This note was uploaded on 04/07/2008 for the course CHEM 1304 taught by Professor Prof.maguire during the Spring '08 term at SMU.

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Equilibria - 1 EQUILIBRIA I Basis of Equilibrium A Q and...

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