Equilibrium

Ammonia N 2 + 3H 2 2NH 3

Ammonia 3H 2 (g) + N 2 (g) 2NH 3 (g)

N 2 O 4 (g) 2NO 2 (g)

Chemical Equilibrium reaction that proceeds forward (reactants become products) and reverse (“products” become “reactants”) at the same rate does NOT mean that amounts of reactants and products are equal

Equilibrium Constant an expression and resulting numerical value that describes the position of equilibrium for a given reaction K eq K sp KcKc KpKp KaKa KbKb

Equilibrium Constant K c = 1.9 x 10 19

Equilibrium Constant K c = 4.1 x

Equilibrium Constant K >> 1 K << 1 K ≈ 1 equilib favors products equilib favors reactants equilib midway btw

Equilibrium Constant aW + bX cY + dZ K c = [Y] c [Z] d [W] a [X] b only solutions and gases appear in the equilibrium constant expression, NOT solids and pure liquids.

Equilibrium Constant ClNO 2 (g) + NO(g) NO 2 (g) + ClNO(g) rate forward = rate reverse k f [ClNO 2 ][NO] = k r [NO 2 ][ClNO] k f [NO 2 ][ClNO] k r [ClNO 2 ][NO] = = K eq

Equilibrium Constant Write the equilibrium constant expression for the following reactions: 3H 2 (g) + N 2 (g) 2NH 3 (g) CaCO 3 (s) CaO(s) + CO 2 (g)

H 2 (g) + I 2 (g) 2HI(g) Calculate a numerical value for K eq if the equilibrium concentrations are [H 2 ] = 0.11 M [I 2 ] = 0.11 M [HI] = 0.78 M

CH 3 CO 2 H (aq) + C 2 H 5 O (aq) CH 3 CO 2 C 2 H 5(aq) + OH - (aq) At 25  C, K c = In an equilibrium mixture, the following concentrations were measured: [CH 3 CO 2 H] = M [CH 3 CO 2 C 2 H 5 ] = M [OH - ] = M What was the equilibrium concentration of ethanol?

Comparing Multiple Equilibria reversible reaction written in opposite direction ½ N 2 O 4 NO 2 K forward = [NO 2 ] [N 2 O 4 ] 1/2 K forward = 0.11 NO 2 ½ N 2 O 4 K reverse = K reverse =

Comparing Multiple Equilibria reversible reaction written in opposite direction = equilibrium constant is inverted

Comparing Multiple Equilibria eqn that is a multiple of another equilib rxn ½ N 2 O 4 NO 2 K c = [NO 2 ] [N 2 O 4 ] 1/2 K c = 0.11 N 2 O 4 2NO 2 K c ’= K c ’=

Comparing Multiple Equilibria eqn that is a multiple of another equilib rxn = equilib constant raised to that factor

Comparing Multiple Equilibria equilib rxns added together to get an overall reaction A 2B 2B 3C KcKc KcKc

Comparing Multiple Equilibria equilib rxns added together to get an overall reaction = equilib constant of overall reaction is the product of the individual equilib constants

Do this without your calculator! Given the following equilibrium reactions: C 2 H 2 O 4 ↔ H + + C 2 HO 4 - K c = 6.5 x C 2 HO 4 - ↔ H + + C 2 O 4 2- K c = 6.1 x Calculate K c for the following reaction: C 2 H 2 O 4 ↔ 2H + + C 2 O 4 2-

Reaction Quotient Q written the same as an equilib constant, but NOT at equilib conditions

Reaction Quotient H 2 (g) + I 2 (g) 2HI(g)K c = 60. (at 350 o C) [H 2 ] = [I 2 ] = 0.010M [HI] = 0.050M

Comparing Q and K Q = K Q < K K > Q Q > K K < Q rxn at equilb proceeds toward products proceeds toward reactants

PCl 3 (g) + Cl 2 (g) PCl 5 (g) At a given temperature, the following equilibrium concentrations were observed: [PCl 3 ] eq = M [Cl 2 ] eq = M [PCl 5 ] eq = M Determine the value of the equilibrium constant at this temperature.

At the same temperature, some Cl 2 (g), PCl 3 (g), and PCl 5 (g) are mixed at the following concentrations: [Cl 2 ] i = M [PCl 3 ] i = M [PCl 5 ] i = M In which direction will the reaction proceed to achieve equilibrium?

ICE Tables I nitial (Concentrations) C hange E quilibrium (Concentrations)