Unit 16 – Equlibrium 16.1 How Chemical Reactions Occur 16.2 Conditions That Affect Reaction Rates 16.3 The Equilibrium Condition 16.4 Chemical Equilibrium 16.5 The Equilibrium Constant 16.6 Heterogeneous Equilibria 16.7 LeChatelier’s Principle 16.8 Applications Involving the Equilibrium Constant 16.9 Solubility Equilibria Text Pages

Unit 16 – Equilibrium Upon completion of this unit, you should be able to do the following: Understand the collision model of how chemical reactions occur Understand the activation energy. Understand how a catalyst speeds up a reaction. Learn how equilibrium is established Learn about the characteristics of chemical equilibrium Understand the law of chemical equilibrium and learn how to calculate values for the equilibrium constant Understand the role that liquids and solids play in contructing the equilibrium expression. Learn to predict the changes that occur when a system at equilibrium is disturbed Learn to calculate equilibrium concentrations from equilibrium constants Learn to calculate the solubility product of a salt, given its solubility and vice versa 2

Chemical Reactions The collision model is based upon the idea that chemical reactions occur during molecular collisions. Some collisions are violent enough to break bonds, allowing the reactants to rearrange to form the products. The collision model explains why a reaction proceeds faster if the concentrations of the reacting molecules is increased. Higher concentrations lead to more collisions and therefore more reactions The collision model also explains why reactions go faster at higher temperatures. 3

Chemical Reactions Not all collisions possess enough energy to break bonds. The minimum amount of energy required for a reaction to occur is called the activation energy, E a. If a collision has more energy than E a, a reaction will occur. As temperature increases, the speed of the molecules increases and the collisions have more energy. 4

Chemical Reactions Figure 16.3 – When molecules collide, a certain minimum energy called the activation energy (E a ) is needed for a reaction to occur. 5

Chemical Reactions A catalyst is a substance that speeds up a reaction without being consumed. A catalyst lowers the activation energy of a reaction. 6

Equilibrium Chemists define equilibrium as the exact balancing of two processes, one of which is the opposite of the other. In unit 13 we discussed vapor pressure in terms of rates of evaporation and condensation. Equilibrium occurs when the rate of evaporation equals the rate of condensation. 7

Equilibrium While many chemical reactions run to completion, where all of the limiting reactant is consumed, there are some reactions that reach a state of equilibrium. The reaction of nitrogen dioxide, a reddish brown gas, to form dinitrogen tetroxide, a colorless gas, is an example. NO 2 (g) + NO 2 (g) → N 2 O 4 (g) If pure NO 2 is placed in a sealed vessel at 25 o C, the initial dark brown color will decrease as some of the gas is converted to N 2 O 4. However, even over a long period of time, the contents of the vessel do not become colorless. This indicates that the reaction stops short of completion and an equilibrium is reached, where the concentrations of reactants and products remain constant. 8

Equilibrium The situation is similar to that of vapor pressure. When the NO 2 is first placed in the vessel, there is no N 2 O 4. As collisions occur between NO 2 molecules, N 2 O 4 is formed. Some of the N 2 O 4 decomposes back into NO 2. Both a forward and reverse reaction occur. NO 2 (g) + NO 2 (g) → N 2 O 4 (g) N 2 O 4 (g) → NO 2 (g) + NO 2 (g) This reaction can also be shown as 2NO 2 (g) N 2 O 4 (g) 9

Dynamic Equilibrium Because no changes occur in the concentrations of reactants or products, it may appear that everything has stopped. However, equilibrium is dynamic. There is activity at the molecular level. Consider the reaction When CO and H 2 O are mixed, they react to form H 2 and CO 2 The concentration of the reactants decreases while the concentration of the products increases. As the concentration of the reactants decreases, the forward rate of reaction decreases. As the concentration of products increases, the reverse rate of reaction increases. 10

Dynamic Equilibrium When the forward and reverse rates of reaction are the same, equilibrium is reached. 11

Equilibrium Constant In 1864, two Norwegian chemists, Cato Maximilian Guldberg and Peter Waage, proposed the law of chemical equilibrium as a general description of the equilibrium condition. An equilibrium expression can be written for the reaction aA + bB cC + dD K is called the equilibrium constant. The square brackets indicate species concentrations in (mol/L) at equilibrium. 12 [A] a [B] b [C] c [D] d K =

Equilibrium Constant Example: 2 O 3 3 O 2 Place the concentration of the product in the numerator and the concentration of the reactant in the denominator. Use the coefficients as powers. 13 [A] a [B] b [C] c [D] d K = Coefficient Reactant Product [O 3 ] 2 [O 2 ] 3 K =

Equilibrium Constant Example 16.1, page 491 Write the equilibrium expression for the following a.H 2 (g) + F 2 (g) 2HF (g) b. N 2 (g) + 3H 2 (g) 2NH 3 (g) 14

Equilibrium Constant For a given reaction at a given temperature, the ratio of concentrations of the products to the reactants defined by the equilibrium expression will always equal the same number, the equilibrium constant K. Consider the reaction where 1 mol each of N2 and H2 are sealed in a 1-L vessel at 500 oC and allowed to react. At equilibrium the concentrations are [N2] = 0.921M, [H2] = 0.763M and [NH3] = 0.157M N 2 (g) + 3H 2 (g) 2NH 3 (g) 15 [.921][.763] 3 [.157] 2 K = =.0602 = 6.02 x 10 -2

Equilibrium Constant 16

Equilibrium Constant 17 For 3 experiments, with different initial concentrations, the equilibrium constant is the same. Note that the equilibrium concentrations are not the same, but the equilibrium constant, which depends on the ratio of the concentrations is the same.

Equilibrium Constant Example 16.2, page

Heterogenous Equilibria The previous examples involved equilibrium where all reactants and products were in the gaseous state. These are called homogeneous equilibrium. Heterogeneous equilibrium involves reactants and products in more than one state. As an example, the thermal decomposition of calcium carbonate produces lime (CaO) and carbon dioxide. CaCO 3 (s) CaO (s) + CO 2 (g) 19 [CaCO 3 ] [CaO][CO 2 ] K =

Heterogenous Equilibria However, experimental data show that the position of a heterogeneous equilibrium does not depend upon the concentrations of pure solids or pure liquids. The fundamental reason for this is that the concentrations of pure solid and pure liquids cannot change. CaCO 3 (s) CaO (s) + CO 2 (g) or or [CO 2 ]= K 20 C2C2 C 1 [CO 2 ] K’ = C1C1 C 2 K’ [CO 2 ]=

Heterogenous Equilibria Example 16.3, page 495 Write the expression for K for the following processes. a.Solid phosphorous pentachloride is decomposed to liquid phosphorous trichloride and chlorine gas. b. Deep-blue solid copper (II) sulfate pentahydrate is heated to drive off water vapor and form white solid copper (II) sulfate. 21

Le Chatelier’s Principle When a change is imposed on a system at equilibrium, the position of the equilibrium shifts in a direction that will reduce the effect of the change. 22

Le Chatelier’s Principle Consider a change in concentration for the reaction of nitrogen with hydrogen to form ammonia gas. N 2 (g) + 3H 2 (g) 2NH 3 (g) Suppose there is an equilibrium position of [N 2 ] = 0.399M [H 2 ] = M [NH 3 ] = M What will happen if 1.0 mol/L of N 2 is suddenly injected into the system? 23

Le Chatelier’s Principle N 2 (g) + 3H 2 (g) 2NH 3 (g) As the nitrogen is added, there are more collisions between nitrogen and hydrogen molecules. This increases the rate of the forward reaction and the reaction produces more ammonia. As the concentration of ammonia increases, the reverse reaction also increases. However, the new equilibrium position has more ammonia. We say the equilibrium has shifted to the right. [N 2 ] = 0.399M [N 2 ] = M [H 2 ] = M [H 2 ] = M [NH 3 ] = M [NH 3 ] = 0.304M 24

LeChatelier’s Principle N 2 (g) + 3H 2 (g) 2NH 3 (g) [N 2 ] = 0.399M [N 2 ] = M [H 2 ] = M [H 2 ] = M [NH 3 ] = M [NH 3 ] = 0.304M Equilibrium constant does not change 25 [N 2 ][H 2 ] 3 [NH 3 ] 2 K = [N 2 ][H 2 ] 3 [NH 3 ] 2 K = [.399][1.197] 3 [.203] 2 K = [1.044][.304] 3 [1.348] 2 K = =.0602

Le Chatelier’s Principle Consider a change in volume for the reaction of nitrogen with hydrogen to form ammonia gas. N 2 (g) + 3H 2 (g) 2NH 3 (g) When the volume of a gas is decreased, the pressure increases. By Le Chatelier’s Principle, the system will shift in the direction that reduces pressure. The reaction will shift to the right to reduce the number of molecules (4 on the left, 2 on the right). 26

Le Chatelier’s Principle Consider a change in temperature for the reaction of nitrogen with hydrogen to form ammonia gas. N 2 (g) + 3H 2 (g) 2NH 3 (g) + 92 kJ This is an exothermic reaction with heat being released. If heat is added, the reaction will shift to the left to consume the heat. In an endothermic reaction, the shift will be to the right. 27

Equilibrium Constant The size of the equilibrium constant K tells us the tendency of the reaction to occur. If K is much larger than 1 at equilibrium, the reaction system will consist of mostly products. A (g) B (g) where If K =10,000 then at equlibrium, [B] is 10,000 times greater than [A]. If K is small, the reaction system will consist of mostly reactants. 28 [A] [B] K =

Solubility Equilibria When a typical ionic solid dissolves in water, it completely dissociates into cations and anions. Calcium fluoride dissolves in water as follows: CaF 2 (s) → Ca 2+ (aq) + 2 F – (aq) When the solid is first added, no ions are present. As dissolving occurs, the ion concentrations increase until the start to reform the solid and equilibrium is reached. No more solid dissolves and the solution is saturated. The equilibrium expression for this process is K sp = [Ca 2+ ] [F – ] 2 where [Ca 2+ ] and [F – ] 2 are in mol/L K sp is called the solubility product. 29

Solubility Equilibria Copper bromide, CuBr, has a measured solubility of 2.0 x mol/L at 25 o C. That is, when excess CuBr (s) is placed in 1.0 L of water, we can determine that 2.0 x mol of the solid dissolves to produce a saturated solution. Calculate the solid’s K sp value. CuBr (s) → Cu + (aq) + Br – (aq) K sp = [Cu + ] [Br – ] 2.0 x mol/L CuBr → 2.0 x mol/L Cu + (aq) x mol/L Br – (aq) So [Cu + ] = 2.0 x mol/L and [Br – ] = 2.0 x mol/L K sp = [2.0 x ] [2.0 x ] = 4.0 x