Chapter 17 Equilibrium A reversible reaction is a chemical reaction that can occur in both the forward and reverse directions, such as the formation of ammonia. (double arrow) N 2 (g) + 3H 2 (g) 2NH 3 (g) Chemical equilibrium is a state in which the forward and reverse reactions balance each other because they take place at equal rates.

Forward: N 2 (g) + 3H 2 (g)  2NH 3 (g) Reverse: N 2 (g) + 3H 2 (g)  2NH 3 (g) H2H2 NH 3 N2N2

The law of chemical equilibrium states that at a given temperature, a chemical system might reach a state in which a particular ratio of reactant and product concentrations has a constant value. for aA (g) + bB (g) cC (g) + dD (g) equilibrium constant expression K eq = [C] c [D] d [A] a [B] b

K eq is constant only at a specified temperature. K eq > 1: Products are favored at equilibrium K eq < 1: Reactants are favored at equilibrium H 2 (g) +I 2 (g) 2HI(g) This reaction is a homogeneous equilibrium, which means that all the reactants and products are in the same physical state. K eq = __[HI] 2 __ = 49.7 at 731K [H 2 ] [I 2 ] (no unit)

When the reactants and products are present in more than one physical state, the equilibrium is called a heterogeneous equilibrium. Vapor pressure of ethanol in a closed flask C 2 H 5 OH(l) C 2 H 5 OH(g). K = [C 2 H 5 OH(g)] [C 2 H 5 OH(l)] K[C 2 H 5 OH(l)] = [C 2 H 5 OH(g)] = K eq C 2 H 5 OH(l) constant concentration  Combined with K  K eq = [C 2 H 5 OH(g)]

Solids are also pure substances with unchanging concentrations, so equilibria involving solids are simplified in the same way I 2 (s) I 2 (g) K eq = [I 2 (g)] The equilibrium constant depends only on the concentration of the gaseous iodine in the system Equilibrium characteristics: 1) Closed container (nothing leaving or entering) 2) Temperature must remain constant 3) All reactants and products are present and in constant dynamic motion.

Le Chatelier’s principle (Stress is any kind of change in a system that upsets the equilibrium.) If a system is taken out of equilibrium, the reactants/products will adjust themselves to reach equilibrium again. (proposed in 1888) states that if stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress Factors Affecting Chemical Equilibrium

Adding reactants increases the number of effective collisions between molecules and upsets the equilibrium. The equilibrium shifts to the right to more products Adjusting the concentrations… …of either the reactants or the products puts stress on a system in equilibrium and shifts the equilibrium in the direction that relieves the stress.. When the volume of the reaction vessel is decreased, the equilibrium position shifts towards whichever side has fewer total moles of gases.

Lowering the piston de- creases the volume and increases the pressure. The reaction between CO and H 2 is at equilibrium. As a result, more molecules of the products form. Their formation relieves the stress on the system.

Temperature alters equilibrium position and constant k eq. More heat shifts the equilibrium in the direction in which the heat is used up.

A catalyzed reaction reaches equilibrium more quickly, but with no change in the amount of product formed.

Practice Problems 1. For the reaction: N 2 O 4 (g) 2NO 2 (g) At equilibrium, what is the concentration of N 2 O 4 if the concentration of NO 2 is.03 mol/L? K eq =.20 mol/L Step 1: set up equationStep 2: Solve for [N 2 O 4 ] H 2 (g) + I 2 (g) 2HI (g) 1 mole of H 2 and 1 mole of I 2 are sealed in a 1 L flask mol of HI are formed when equilibrium is reached..22 mol of H 2 and.22 mol of I 2 remain. What is the value of K eq ? Step 1: set up equation Step 2: Calculate K eq by finding the concentrations: [HI], [H 2 ], &[I 2 ] 2.

The Solubility Product Constant K sp Some ionic compounds dissolve readily in water, and some barely dissolve at all. The equilibrium constant expression for the dissolving of a sparingly soluble compound is called the solubility product constant, K sp. can be used to determine the solubility of a sparingly soluble compound

K sp can also be used to predict whether a precipitate will form when any two ionic solutions are mixed. Will a precipitate form in this double-replacement reaction? A precipitate is likely to form only if either product, KCl or Fe 4 (Fe(CN) 6 ) 3 has low solubility. KCl K sp = 21.7 high solubility no precipitate Fe 4 (Fe(CN) 6 ) 3 K sp = 3.3 x low solubility… possible… 1) First, calculate the concentrations of the ions for Fe 4 (Fe(CN) 6 ) 3

Ion concentrations 2) Calculate if the concentrations of Fe 3+ and Fe(CN) 6 4- in the mixed solution exceed the value of Ksp when substituted into the solubility product constant expression. This will not likely give you the solubility product constant. Instead, it provides a number called the ion product (Qsp). 1) Careful: When mixing solutions volumes are combined and concentrations change!!!

3 Possibilities: If Qsp < Ksp the solution is unsaturated and no precipitate will form. If Qsp = Ksp the solution is saturated and no change will occur. If Qsp > Ksp a precipitate will form, reducing the concen-trations of the ions in the solution until the product of their concentrations in the Ksp expression equals the numerical value of Ksp. Q sp is a value that can be compared with K sp. Now you can compare Qsp and Ksp. Qsp = 7.8 x and Ksp = 3.3 x

Why is PbCrO 4 less soluble in aqueous solution of K 2 CrO 4 than in pure water? The solution of K 2 CrO 4 already contains CrO 4 2– ions before any PbCrO 4 dissolves  [CrO 4 2– ] increases in formula The Common Ion Effect A common ion is an ion that is common to two or more ionic compounds. The common ion effect is the lowering of the solubility of a substance because of the presence of a common.