Chemistry 30 – Unit 4 Part 1 Equilibrium Focusing on Acid-Base Systems To accompany Inquiry into Chemistry PowerPoint Presentation prepared by Robert Schultz

Chapter 16, Section 16.0 Review: Arrhenius theory of acids and bases Acids ionize in water to produce H + (aq) Bases ionize in water to produce OH ‾(aq) (Modified) Arrhenius Theory of Acids and Bases Acids react with water to produce H 3 O + (aq) (hydronium) Bases react with water to produce OH ‾(aq)

Chapter 16, Section 16.0 examples page 626 bottom Strong acids (A) ionize (or react with water) completely; weak acids (B) partially CH 3 COOH(aq) + H 2 O(l)  H 3 O + (aq) + CH 3 COO - (aq)HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl‾(aq) Figure 8.2, page 627

Chapter 16, Section 16.0 pH and acid-base titrations will be reviewed later in the unit

Chapter 16, Section 16.1 In Chemistry 20 you studied dynamic equilibrium in saturated solutions In dynamic equilibrium rate of forward and reverse processes is equal As a result, properties remain constant Read bridge illustration (under photo) page 634

Chapter 16, Section 16.1 Types of dynamic equilibrium: phase equilibrium solubility equilibrium chemical reaction equilibrium homogeneous equilibria – all equilibrium substances in same state heterogeneous equilibria – not all equilibrium substances in same state heterogeneous heterogeneous or homogeneous

Chapter 16, Section 16.1 At equilibrium, macroscopic properties remain constant Equilibrium can only be truly maintained in a closed system Read page 637

Chapter 16, Section 16.1 We’ll focus on chemical reaction equilibrium Discuss questions 3, 4, 7, 9, page 638

Chapter 16, Section 16.2 Law of Chemical Equilibrium: In a chemical reaction equilibrium represented by the equation: The following expression will be constant Called the equilibrium constant, K c a, b, c, d = balancing coefficients a A + b B  c C + d D

Chapter 16, Section 16.2 Examples page 641 question 2 You are not required to put states in, but don’t do a mixture of states and non-states – you will lose marks for communication Recommendation – no states

Chapter 16, Section 16.2 question 5 Special considerations about K c : Never include solids in a K c expression – their concentration (density) is constant Always include H 2 O(g) but include H 2 O(l) only if present in a mixture of liquids as this is only way its concentration can change

Chapter 16, Section 16.2 The expression of K c for is 3 NO 2 (g) + H 2 O(l)  2 HNO 3 (aq) + NO(g) Note that water is not included in the expression; you can’t change its concentration

Chapter 16, Section 16.2 Calculating the value of K c Example: Calculate the value of K c for N 2 (g) + 3 Cl 2 (g)  2 NCl 3 (g) given that [N 2 ] = 1.4 x mol/L [Cl 2 ] = 4.4 x mol/L [NCl 3 ] = 1.9 x mol/L note about units in calc and in answer

Chapter 16, Section 16.2 You must be able to enter these in your calculator Here’s how: You will need to use the ^ button for powers other than 2 or 3 For 2, use x 2 button; for 3 use math menu to do it directly

Chapter 16, Section 16.2 Do WS 61

Chapter 16, Section 16.2 Qualitative interpretation of K c Since [products] on top of the K c ratio, a large value indicates an equilibrium favouring products Generalizations: K > 1; products favoured K ≈1; approximately equal [reactants] and [products] K < 1; reactants favoured Remember “[ ]” = “concentration of” Not always true; can you think of why?

Chapter 16, Section 16.2 Le Châtelier’s Principle: a dynamic equilibrium tends to respond so as to relieve the effect of any change in the conditions that affect the equilibrium My translation: Do something to stress an equilibrium and

Chapter 16, Section 16.2 Le Châtelier’s Principle: A)Concentration Changes If concentration of a reactant is increased, rate of forward reaction increases, slowing as product concentrations increase thus reaching a new equilibrium Similar explanations can be used for other concentration changes leading to the generalizations:

Chapter 16, Section 16.2 Raise concentration of a reactant or lower concentration of a product, equilibrium shifts right Lower concentration of a reactant or raise concentration of a product, equilibrium shifts left Concentration changes shift equilibria but have NO effect on value of K c

Chapter 16, Section 16.2 The equilibrium, can be shifted to the right by: 2 SO 3 (g) + 97 kJ  2 SO 2 (g) + O 2 (g)

Chapter 16, Section 16.2 B)Temperature Changes Depends on whether reaction is endothermic or exothermic Endothermic: Reactants + energy  Products raise temperature shifts right lower temperature shifts left Changing temperature DOES have an effect on the value of K c treat energy like a reactant substance

Chapter 16, Section 16.2 Exothermic Reactants  Products + energy raise temperature shifts left lower temperature shifts right Remember that changing temperature DOES have an effect on the value of K c treat energy like a product substance

Chapter 16, Section 16.2 same example: Using temperature, shift right by: Using temperature, shift left by: 2 SO 3 (g) + 97 kJ  2 SO 2 (g) + O 2 (g)

Chapter 16, Section 16.2 C)Volume and Pressure Changes Equilibria involving gases may be affected by changes in volume leading to changes in pressure Pressure of a gas in a closed system is directly proportional to number of moles of the gas P V = n R T (Chem 20) If pressure of a gas is increased by

Chapter 16, Section 16.2 decreasing volume, equilibrium will shift towards side with fewer moles of gases (as this will decrease the pressure increase) If volume of system is decreased, thus causing a pressure increase in the example system, equilibrium will shift left Pressure/volume changes do NOT affect value of K c 2 SO 3 (g) + 97 kJ  2 SO 2 (g) + O 2 (g)

Chapter 16, Section 16.2 Equilibrium has shifted __________?? Does this agree with prediction?

Chapter 16, Section 16.2 Adding an inert gas will change total pressure but not individual pressures of other gases in system Equilibrium won’t shift The only way to shift equilibrium position using pressure is to change the system volume WS 62

Chapter 16, Section 16.2 Equilibrium graphs - handout shift direction? stress? shift direction? Why no CCl 4 (l) on graphs? right  [HF] right  [HCl] if no shift

Chapter 16, Section 16.2 shift direction? stress? shift direction? stress? spikies and roundies right kJ  temperature right  volume   pressure kJ if no shift

Chapter 16, Section 16.2 A single stress takes place at each of times A, B, C, and D on the graph Identify each stress ABCDABCD  volume   pressure  shift left  temperature  shift left  [C 2 H 6 ]  shift left addition of inert gas  no shift

Chapter 16, Section 16.2 Review Diploma exam sample questions re equilibrium graphs Catalysts have NO effect on equilibrium position – recall activation energy – unit 1 Activation energy is reduced for both forward and reverse reactions Both forward and reverse go faster

Chapter 16, Section 16.2 Demos: N 2 O 4 (g) + energy  2 NO 2 (g) colourlessbrown Predictions: Raise temperature Lower temperature Observe

Chapter 16, Section 16.2 Predictions: (starting with intermediate purple colour) Raise temperature Lower temperature Add water Add HCl(aq) Add AgNO 3 (s) Observe CoCl 4 2- (al) + 6 H 2 O(al)  Co(H 2 O) 6 2+ (al) + 4 Cl - (al) + energy blue pink

Chapter 16, Section 16.3

Chapter 16, Section 16.3

Chapter 16, Section 16.3 Try The Ammonia Factory!The Ammonia Factory!

Chapter 16, Section 16.2 Equilibrium calculations: ICE Tables (also called ICE Boxes) Examples: (Handout) The equilibrium: 2 NO 2 (g)  N 2 O 4 (g) was established when mol of NO 2 (g) was placed in a 2.00 L flask. At equilibrium, the concentration of N 2 O 4 (g) was mol/L. What is the equilibrium constant for the reaction?(25.2) Initial Change Equilibrium

Chapter 16, Section 16.2 Initial [NO 2 ] = 2 NO 2 (g)  N 2 O 4 (g) I mol/L0 C E mol/L ‾2 x mol/L mol/L mol/L stoichiometry line! It’s okay to leave out units in K c calculations

Chapter 16, Section 16.2 At a temperature of 300°C and a pressure of 40.5 MPa, 90.0 mol of H 2 (g) and 80.0 mol of N 2 (g) are injected into a reaction vessel. When equilibrium is established 37.0 mol of NH 3 (g) are present. What is the number of moles of H 2 (g) present in the equilibrium mixture?(34.5 mol) N 2 (g) + 3 H 2 (g)  2 NH 3 (g) I 80.0 mol 90.0 mol 0 C E 37.0 mol mol 34.5 mol

Chapter 16, Section 16.2 Consider the following chemical equilibrium: Initially, mol NO(g), mol H 2 (g), and mol H 2 O(g) are placed in a 1.0 L container. At equilibrium, [H 2 O(g)] = mol/L. What is the value of K c ? (This one’s a little tougher!) (6.5 x 10 2 ) 2 NO(g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O(g)

Chapter 16, Section NO(g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O(g) I 0.10 mol/L mol/L mol/L C E mol/L ½ x mol/L0.012 mol/L0.019 mol/L Ice Box Calcs II WS

Chapter 16, Section 16.2 Using the approximation method Sometimes using an ICE Box leads to a quadratic equation You will never have to solve a quadratic equation on my exam or on the Diploma Exam so the only time this will happen is if K c is very small compared to the concentrations and an approximation can be made Consider the example:

Chapter 16, Section 16.2 Practice Problem 25, page 660 COCl 2 (aq)  CO(g) + Cl 2 (g) I 1.5 mol/L 0 0 C E ? ? +x 1.5-x mol/L -x x x looks quadratic it is!

Chapter 16, Section 16.2 Since K c is smaller than we can assume that x will be very small and that Remember that you will not have to solve quadratics – you will always be able to make this approximation

Chapter 16, Section 16.2 Try Practice Problem 26, page H 2 S(g)  2 H 2 (g) + S 2 (g) I 1.3 mol/L 0 0 C-x +x E 1.3–x mol/L x

Chapter 16, Section 16.3 Applications of Equilibrium Systems Read pages Do Applications of Equilibrium Systems Worksheet

Chapter 16, Section 16.3