Chapter 6. Chemical Equilibrium Chem. 1A van Koppen Quantities in Equilibrium Constant (or "Mass Action") Expressions 1. Gases normally enter as partial pressures in atmospheres. 2. Dissolved species in solution enter as concentrations, in units of molarity (moles/L or M). . 3. The concentration of pure solids and pure liquids do not appear in the equilibrium expression. Neither does the concentration of the solvent taking part in a chemical reaction in dilute solution, i.e. ≤ 1 M. 4. Equilibrium constants do not have units. This is accomplished by dividing the pressures of all gases by a standard reference pressure of 1 atm, and dividing the concentrations of all solutes by a standard reference concentration of 1 mole/L. This gives relative pressures and concentrations which have no units, but have the same numerical values. Reference pressures and concentrations are discussed in your text (pg. 586). Manipulation of Equilibrium Constants 1. When you reverse the reaction, invert K. 2. When you multiply the reaction by a constant n, raise K by the power n. 3. When you add reactions, multiply the Ks. When you subtract reactions, divide Ks. Equilibrium Constant, K, and Reaction Quotient, Q 1. The equilibrium constant, K, is calculated using concentrations (or partial pressures) of the reactants and products at equilibrium. The magnitude of K reflects the position of the equilibrium. A large value of K indicates that mostly products will be present at equilibrium. A small value of K indicates mostly reactants are present at equilibrium. 2. The reaction quotient Q is calculated using instantaneous concentrations of the reactants and products (the initial concentrations or whatever they may be). As the reaction approaches equilibrium, Q approaches K. Q predicts the direction that the reaction will go to reach equilibrium. If Q = K , the system is at equilibrium. If Q > K , too much product is present (relative to reactant) => the reaction proceeds to the left, from products to reactants until Q = K. If Q < K , too much reactant is present (relative to product) => the reaction proceeds to the right, from reactants to products until Q = K. Le Chatelier's Principle When a system in equilibrium is subjected to a stress that affects the equilibrium, the system will react so as to counteract (reduce) the stress. Stresses which affect gas phase equilibria: 2 A (g) B (g) + heat ()2ABpPPK =1. Changing the partial pressure of a reactant or product. For example, increasing the partial pressure of B (increasing PB) yields Q > K and the reaction proceeds from products to reactants until Q = K. Adding a noble gas such as argon does not change the partial pressures of A or B. Since there is no change in the partial pressures of A and B, Q = K, and therefore adding a noble gas does not affect the equilibrium. 2. Changing the total pressure by compressing or expanding the gas (i.e. by changing the volume) will only affect those equilibria where there is a change in the total moles of gas in the reaction. Decreasing the volume will favor the side of the reaction with fewer gas molecules. In the above example, decreasing the volume will shift the equilibrium to the right, from reactants to products. 3. Changing the temperature affects the equilibrium by changing the value of the equilibrium constant. If energy in the form of heat is added to a system at equilibrium, the reaction will shift in the direction that consumes energy. For the above reaction, which is exothermic (energy is released in the form of heat), increasing the temperature will shift the equilibrium to the left, decreasing the value of K. aA + bB cC + dD K = Q = [A]a [B]b[C]c [D]d[C]c [D]d[A]a [B]bProcedure for Solving Equilibrium Problems • Write the balanced equation for the reaction. • Write the equilibrium constant expression. aA + bB cC + dD K = K p= [A]a [B]b[C]c [D]d(PC)c (PD)d(PA)a (PB)b • List the initial conditions. • Determine the direction of the shift to equilibrium. Sometimes the direction is obvious. For example, if you have only reactants to start with, the direction of change must be to the right, to form products. Similarly if you only have products to start with, the direction of change must be to the left, to form reactants. If initially you have some of each of the reactants and some of each of the products, the direction of change may not be obvious. In this case, calculate Q to determine the direction of the shift to equilibrium. [C]c [D]d[A]a [B]b(PC)c (PD)d(PA)a (PB)b Q = or Q = Using the initial concentrations or initial pressures given. NOTE: The initial concentration is the concentration of a given species after solutions are mixed but before any reaction takes place. The initial concentration of a given species is the initial number of moles of the species divided by the total volume of the solution. If Q = K , the system is at equilibrium. If Q > K , too much product is present (relative to reactant) => the reaction proceeds to the left, from products to reactants until Q = K. If Q < K , too much reactant is present (relative to product) => the reaction proceeds to the right, from reactants to products until Q = K. • Make a table showing the reaction, the initial concentration, the change in each of the reactants and products, and the equilibrium concentrations or partial pressures in terms of the change. Because we do not know how much the change will be we use a variable, for example, X. See example 6.3 in section 6.6 and in section 6.7 in your text, pg. 203-207. • Substitute the equilibrium concentrations or partial pressures into the equilibrium constant expression, and solve for the unknown variable, X. • Check your calculated equilibrium concentrations by making sure they give the correct value of