For the reaction: A B R f = R r. Chemical Equilibrium Chapter The Concept of Equilibrium. The Concept of Equilibrium

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1 Chemical Equilibrium Chapter This is the last unit of the year, and it contains quite a lot of material. Do not wait until the end of the unit to begin studying. Use what you have learned about reading, taing notes and exam preparation to help you. You will tae a test over unit 5, not a quiz! The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate in a closed system. The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate. The Concept of Equilibrium CO + H 2 O CO 2 + H 2 Equilibrium achieved Product conc. increases and then becomes constant at equilibrium Reactant conc. declines and then becomes constant at equilibrium Used to show equilibrium Properties of a System at Equilibrium 1. Appear from outside to be inert or non changing 2. Can be initiated from both directions 3. Only exist in a closed system 4. Based on ratios of reactants to products Kinetics Definition For the reaction: A B R f = R r

2 Rxn A B Rxn B A Rate = f [A] Rate = r [B] At equilibrium, forward and reverse reaction occur at the same rate, or: Rearranging: Kinetics Definition f [A] = r [B] f r B A Where: Kinetics Definition f r K K eq is the equilibrium constant. eq B A K K c is used to denote that the equilibrium is expressed using molar concentrations So, from inetics, we see that the equilibrium constant, describing the rate when the forward and reverse reactions are occurring at the same rate, is a function of reactant and product concentration c The Equilibrium Constant For Example: Forward reaction: N 2 O 4 (g) 2NO 2 (g) Rate law: Rate = f [N 2 O 4 ] At equilibrium: Rate f = Rate r f [N 2 O 4 ] = r [NO 2 ] 2 Rewriting this, it becomes: f r Reverse reaction: 2 NO 2 (g) N 2 O 4 (g) Rate law: Rate = r [NO 2 ] 2 = [NO 2 ] 2 [N 2 O 4 ] The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = f r = [NO 2 ] 2 [N 2 O 4 ] Remember from inetics f and r are temperature dependant. Thus the equilibrium of a reaction, or K eq, is also temperature dependant. Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are. Stoichiometric Connection to (K eq ) Law of Mass Action (Guldberg-Waage, 1864) Postulates that the coefficients of a balanced chemical equation can be used to describe the ratio of reactants to products for a system at equilibrium.

3 From the law of mass action, the equilibrium for the reaction; aa + bb cc + dd Can be expressed as: K This is called an equilibrium expression c c d C D A a B b (AP equation sheet) The Haber Process (Fritz Haber 1912) Developed a method to produce NH 3 from N 2 and H 2. The ammonia was needed for German explosive production for WWI. As a result of the process being developed, much needed ammonia was could now be produced for agriculture. The Haber Process The Equilibrium Constant Haber discovered that under extreme pressure and temperature conditions, the equilibrium between ammonia and its free elements could be shifted to favor the production of ammonia. Since pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C c ) (P Dd ) (P Aa ) (P Bb ) Relationship Between K c and K p From the ideal-gas law we now that: PV = nrt Rearranging it, we get:; For gas A: P = n V RT = MRT P A [A]RT Relationship Between K c and K p Plugging this into the expression for K p,, for each substance, the relationship between K c and K p becomes Where K p = K c (RT) n n = (moles of gaseous product) (moles of gaseous reactant)

4 1. Write the equilibrium expression for the reaction used in the Haber process in terms of both concentration and pressure. 2. Calculate the equilibrium constant for the haber process at 500 o C, if the following concentrations were determined at equilibrium: [NH 3 ] = 3.1 x 10-2 mol/l [N 2 ] = 8.5 x 10-1 M [H 2 ] = 3.1 x 10-3 M 3. Determine the equilibrium constant in terms of partial pressures. 4. Write the equilibrium expression for the following reactions A) N 2 O 4 (g) 2NO 2 (g) B) HSO 4 - (aq) + H 2 O SO 4 2- (aq) + H 3 O + (aq) C) AgNO 3(s) Ag + (aq) + NO 3 - (aq) Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibriumconstant expression due to the fact that their concentrations are constant. The concentration of a solid or pure liquid, not a solution, is equal to its density(g/cm 3 ) divided by its molecular weight (g/mol). However, the concentration of pure liquids can not be ignored in homogeneous equilibriums. The result is mol/cm 3, which is constant at any temperature. Therefore, for the reaction: AgNO 3(s) Ag + (aq) + NO 3 - (aq) 3 K eq [Ag ][NO constant ] Because the concentration of a solid or liquid is constant, it is omitted from the equilibrium expression. K eq [Ag ][Cl Even though pure substances are omitted from the expression, they must still be present in the reaction for equilibrium to be established. ]

5 5. Identify as homogeneous or heterogeneous and write the equilibrium expressions for: The magnitude of the equilibrium constant provides important information about the reaction at equilibrium. K c >>> 1 For the Haber Process, there was a much larger number in the numerator of the equilibrium expression (or much higher concentration of product) so, the equilibrium lies to the right. Favors products K c <<< 1 When there is a much larger number in the denominator (or much greater concentration of reactants), the reaction lies to the left. 6. Using the data from problem number 2, describe the Haber process in terms of the magnitude of K c. What if: Favors Reactants K c = 1 As we have seen, K can be used to describe the concentration of reactants and products at equilibrium, or the magnitude of the reaction: i.e. Lies to the right or left. Substituting the reactant and product concentrations into the equilibrium expression for a reaction NOT in equilibrium, you get a reaction quotient, Q. But, what about a system NOT at equilibrium? Where: Q = K c only when the reaction is at equilibrium

6 Using this, and by nowing the equilibrium constant for a reaction at a given temperature, the reaction quotient can be used to predict the direction that a reaction not at equilibrium will move. Q > K eq Reaction proceeds toward reactants Q < K eq Reaction proceeds toward products If Q > K, there is too much product and the equilibrium shifts to the left. If Q < K, there is too much reactant, and the equilibrium shifts to the right. If Q = K, the system is at equilibrium. Expression K c >>> 1 K c <<< 1 K c = 1 Q > K eq Q < K eq Q = K eq Summary Meaning rxn favors products rxn favors reactants roughly equal [ ] s rxn proceeds left rxn proceeds right equilibrium 7. For the reaction 2NO 2(g) N 2 O 4(g), K c = 8.8 at 25 o C. If reaction analysis shows that 2.0 x 10-3 mol of NO 2 and 1.5 x 10-3 mol of N 2 O 4 are present in a 10.0 L flas, is the reaction at equilibrium? Defend your answer.