Chemical Equilibrium Chapter 17

Equilibrium vs. Kinetics Kinetics:speed of a reaction or process how fast? Equilibrium:extent of reaction or process how much?

Chemical Equilibrium Reactant and product concentrations remain constant Molecular level:rapid activity (dynamic) Macroscopic level:unchanging At equilibrium:rate forward = rate reverse Does not limit time

Law of Mass Action If:Then:And: K = equilibrium constant m, n=coefficients in balanced equation

Law of Mass Action For: Equilibrium expression:

Value of K Favors K < 0.01 Reactants K > 100 Products 0.01 < K < 100 Neither

Direction of Reaction and Q For: Equilibrium expression: Reaction Quotient:

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

Practice 1.Writing expression for K 2. Q vs. K and reaction direction 3.K for a multistep process 4.K for reaction “multiples”

Write K The decomposition of dinitrogen pentoxide: N 2 O 5 (g) NO 2 (g) + O 2 (g) The combustion of propane gas: C 3 H 8 (g) + O 2 (g) CO 2 (g) + H 2 O(g)

Write K The decomposition of dinitrogen pentoxide: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) The combustion of propane gas: C 3 H 8 (g) + 5 O 2 (g) 3 CO 2 (g) + 4 H 2 O(g)

Write K 4 NH 3(g) + O 2(g) 4 NO (g) + 6 H 2 O (g) 2 NH 3(g) + 5/2 O 2(g) 2 NO (g) + 3 H 2 O

K vs. Q For the reaction: N 2 O 4 (g) 2NO 2 (g) K c = 0.21 at C. At a point during the reaction, [N 2 O 4 ] = 0.12M and [NO 2 ] = 0.55M. (a) Find Q. Is the reaction at equilibrium? (b) If not, in which direction is it progressing?

K vs. Q N 2 O 4 (g) 2NO 2 (g)K c = 0.21 at C. At a point, [N 2 O 4 ] = 0.12M and [NO 2 ] = 0.55M. (a) Find Q. Is the reaction at equilibrium? (b) If not, in which direction is it progressing?

K for multistep reactions Nitrogen dioxide, a toxic pollutant that contributes to photochemical smog, can develop in combustion engines from N 2 and O 2. (1) N 2 + O 2 2NO K c1 = 4.3 x (2) 2NO + O 2 2NO 2 K c2 = 6.4 x 10 9 (a) Show that Q c for the overall reaction is the same as the product of Q c s of the individual reactions. (b) Calculate K c for the overall reaction.

K for multistep reactions (1) N 2 + O 2 2NO K c1 = 4.3 x (2)2NO + O 2 2NO 2 K c2 = 6.4 x 10 9 N O 2 2 NO 2 (a) Q c (b) K c,overall

K for multistep reactions For the following (1) Br 2 2 Br (2) Br + H 2 HBr + H (3) H + Br HBr (a) Write the overall balanced reaction. (b) Write out the individual expressions for Q c and show that their product is equivalent to the overall Q c.

Multiples of K For the ammonia reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) K c is 2.4x10 -3 at 1000K. Find K for the following: (a) 1/3 N 2 + H 2 2/3 NH 3 (b) NH 3 1/2 N 2 + 3/2 H 2

Multiples of K N 2 (g) + 3H 2 (g) 2NH 3 (g), K c = 2.4x10 -3 (a) 1/3 N 2 + H 2 2/3 NH 3 (b) NH 3 1/2 N 2 + 3/2 H 2

Multiples N 2(g) + O 2(g) 2 NO (g) K c = 1 x Write the expression for Q and determine its value for ½ N 2(g) + ½ O 2(g) 2 NO (g) H 2(g) + Cl 2(g) 2 HCl (g) K c = 7.6 x 10 8 Write the expression for Q and determine its value for 2/3 HCl (g) 1/3 H 2(g) + 1/3 Cl 2(g)

Heterogeneous Equilibrium PURE solids and liquids do not appear in expression for K (or Q).

K c vs. K p For:

K p and K c For the ammonia reaction, N 2 (g) + 3H 2 (g) 2NH 3 (g), K c = 2.4x10 -3 Find K p at 1000 K.

K p and K c N 2 (g) + 3H 2 (g) 2NH 3 (g), K c = 2.4x10 -3 Find K p at 1000 K.

K p and K c For the following reaction, PCl 3(g) + Cl 2(g) PCl 5(g), K c = 1.67 at 500 K Find K p at 500 K.

K vs. Q For the reaction: CH 4(g) + Cl 2 CH 3 Cl (g) + HCl K p = 1.6x10 4 at 1500 K. At a point during the reaction, P CH4 = 0.13 atm, P Cl2 = atm, P CH3Cl = 0.24 atm, and P HCl = 0.47 atm. (a) Find Q. Is the reaction at equilibrium? (b) If not, in which direction is it progressing?

Problems 1.Given equilibrium concentrations or pressures, find K or Q. 2. Given K and initial conditions (conc’s or P’s), find equilibrium quantities (conc’s or P’s).

Le Châtelier’s Principle... if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

Le Châtelier’s Principle 1.Concentration 2.Temperature 3.Pressure 4.Volume 5.Catalysts*