Unit A: Equilibrium General Outcomes: Explain that there is a balance of opposing reactions in chemical equilibrium systems. Determine quantitative relationships in simple equilibrium systems.
2 Read p. 670 Unit Introduction Chemistry involves the study of change with chemicals. Lots happening at a molecular level. chemical rxn equilibrium is always a dynamic balance between opposing changes rxns occur at different but equivalent rates within a closed system observe net effect directly i.e. measurable property does not fluctuate Unit A explores the nature of dynamic equilibrium in chemical systems by focussing on acid-base solutions, allowing you to describe, explain, and predict new chemical systems
Prior Knowledge Activity p. 672 #1-7 3
4 EQUILIBRIUM SYSTEMS Chapter 15
5 Static Equilibrium no movement b/c opposing forces act simultaneously Example: book sitting on a level desk gravity pulls down and desk pushes up Chemical Equilibrium always in a dynamic system since balance exists between opposing agents of change (opposite processes occurring at the same rate) even though entities are too small to see Example: juggler some objects move upwards and some downwards No net change rate of upward and downward movement equal at any given time
6 Today s Objectives Define equilibrium and state the criteria that apply to a chemical system in equilibrium i.e. closed system, constancy of properties, as well as equal rates of forward and reverse reactions. Section 15.1 (pp. 676-689)
7 Closed Systems At Equilibrium matter cannot enter or leave the system Example: test tube or beaker (no gas produced in rxn) easier to study chemical properties when separate rxn from its surroundings. The use of a controlled system is an integral part of scientific study Example: soft drink in closed bottle (closed system in equilibrium) See Figure 1 p. 676 nothing appears to change until opened where reduced pressure releases CO 2(g) Can re-carbonate a flat drink by adding pressurized CO 2(g) (reverse rxn)
8 Closed Systems At Equilibrium Fundamental Collision-Reaction Theory initially used four assumptions for Stoich calculations fast, spont, quant, stoich Not always true Example: corrosion not instantaneous Chapter 15 examines assumption of quantitative reaction
9 Prelab INVESTIGATION 15.1 p. 700 Discuss procedure determine quantity to mix same concentrations used select different volumes to react include choice in prediction Record Observations using ink submit original observations sheet with report (initialed by instructor) informal lab report
10 HOMEWORK INV 15.1 Prelab p. 700 prepare procedure and observation table
11 Post-lab INVESTIGATION 15.1 p. 700 Discuss anomaly that occurred unexpected result that contradicts previous rules or experience Direct evidence of both reactants present after rxn appears to have stopped i.e. rxn not quantitative Read Did You Know? p. 677 (Anomalies Signals for Change) Discuss Practice Q p. 677 #2
12 HOMEWORK INV 15.1 Informal Report p. 700 DUE: W, Sept 23 by 4pm
13 Today s Objectives Define equilibrium and state the criteria that apply to a chemical system in equilibrium i.e. closed system, constancy of properties, as well as equal rates of forward and reverse reactions Identify, write, and interpret chemical equations for systems at equilibrium. Define K c to predict the extent of the reaction and write equilibrium-law expressions for given chemical equations, using lowest whole-number coefficients. Section 15.1 (pp. 676-689)
14 >99% Chemical Equilibrium Not all reactions are quantitative (reactants products) Evidence: For many reactions reactants are present even after the reaction appears to have stopped Recall the necessary conditions for a chemical reaction: Particles must collide with the correct orientation and have sufficient energy If product particles can collide effectively, a reaction is said to be reversible Reaction Rate depends on temperature, surface area, and concentration
15 Chemical Equilibrium Consider the following reversible reaction: The final state of this chemical system can be explained as a competition between: The collisions of reactants to form products (forward) The collisions of products to re-form reactants (reverse) Assuming that this system is closed (so the reactants and products cannot escape) and will eventually reach a Dynamic Equilibrium opposing changes occur simultaneously at the same rate
16 Chemical Equilibrium INV 15.1 anomaly explained by this theory with the idea that reverse rxn occurs, where products react to re-form original reactants Equilibrium established from the balance between forward (written left to right) and reverse rxns (written right to left) competing for collisions use double headed, equilibrium arrow the final state of this chemical system can be explained as a competition between collisions of reactants to form products and collisions of products to re-form reactants This competition requires a closed system here bound by the volume of the liquid phase in the beaker
17 Chemical Equilibrium Consider the following hypothetical system: AB + CD AD + BC forward reaction, therefore AD + BC AB + CD reverse reaction Initially, only AB and CD are present. The forward reaction is occurring exclusively at its highest rate. As AB and CD react, their concentration decreases. This causes the reaction rate to decrease as well. As AD and BC form, the reverse reaction begins to occur slowly. As AD and BC s concentration increases, the reverse reaction speeds up. Eventually, both the forward and reverse reaction occur at the same rate Dynamic Equilibrium
18 Conditions of Dynamic Equilibrium 1. Can be achieved in all reversible reactions when the rates of the forward and reverse reaction become equal Represented by rather than by 2. All observable properties appear constant colour, ph, etc. 3. Can only be achieved in a closed system no exchange of matter and must have a constant temperature 4. Equilibrium can be approached from either direction. This means that the equilibrium concentrations will be the same regardless if you started with all reactants, all products, or a mixture of the two
19 Equilibrium reached in closed chemical system (no matter in or out) with constant macroscopic properties (no observable change occurring, i.e. always looks the same) despite continued microscopic changes
20 Types Of Equilibrium all types are explained by a theory of dynamic equilibrium Phase Equilibrium single chemical substance exists in more than one phase, or state of matter, in a closed system Example: water in a sealed container evaporates until reaches the maximum vapour pressure, assuming that is remains at a constant temperature (See Figure 2 p. 677)
21 Types Of Equilibrium Solubility Equilibrium single chemical solute interacting with solvent, where excess solute is in contact with the saturated solution (See Figure 3 p. 677) i.e. a saturated solution where rate of dissolving = rate of recrystallization Chemical Reaction Equilibrium involves several substances reactants and products in a closed system
Mini Investigation p. 678 Modelling Dynamic Equilibrium 22 Cylinder #1 Volume (ml) Cylinder #2 Volume (ml) 25.0 0.0 20.0 5.0 17.0 8.0 14.0 11.0 11.0 14.0 8.0 17.0 5.0 20.0 2.0 23.0 2.0 23.0 2.0 23.0 2.0 23.0 Assume that each time the large straw transfers 5 ml and the smaller straw transfers 2 ml.
23 Chemical Reaction Equilibrium Combines ideas from different theories (atomic, kinetic molecular, collision-reaction) as well as the concepts of reversibility and dynamic equilibrium Describes and explains, but limited in predicting quantitative properties of equilibrium systems Read Did You Know? p. 680 (Lavoisier and Closed Systems)
24 Hydrogen-Iodine Reaction Equilibrium System Figure 4 p. 679 Studied extensively b/c simple structure molecules and gas phase rxn NOTE: use term reagent instead of reactant b/c of rxn reversibility
25 Hydrogen-Iodine Reaction Equilibrium System Figure 4 p. 679 rxn rapid at first, then dark purple color of iodine vapor gradually fades and remains constant Table 1 compiles data from three experiments that manipulate concentrations of the initial products and reactants, as well as react at 448 C so that the system quickly reaches an observable equilibrium each time.
26 Hydrogen-Iodine Reaction Equilibrium System Figure 4 p. 679 quantitatively graph the reaction progress by plotting the concentration/quantity of the reagents versus time (Figure 5) The rate of reaction of the reactants decreases as the number of reactant molecules decrease. The rate that the product turns back to reactants increases as the number of product molecules increases. These two rates become equal at some point, after which the quantity of each will not change. use data to describe equilibrium as a percent rxn and by the equilibrium constant Percent Reaction describes equilibrium for only one specific system example Equilibrium Constant describes all systems of the same reaction, at a given temperature
27 Percent Yield Percent YIELD product quantity at equilibrium compared to theoretical maximum Synonymous with percent reaction useful for communicating the equilibrium position Maximum possible yield is calculated using stoich theoretical value assume quantitative forward rxn with no reverse rxn Use %Yield to reference quantities of chemicals present in equilibrium system write %Yield above equilibrium arrows ( ) in chemical equation
28 Percent Yield Example: If 2.50 mol of hydrogen gas reacts with 3.0 mol of iodine gas in a 1.00L vessel, what is the percent yield if 3.90 mol of hydrogen iodide is present at equilibrium % Yield = products experimental products theoretical max 100% 2.5 mol H 2 2 mol HI 1 mol H 2 = 5 mol HI % Yield = 3.9 mol 5 mol 100% = 78%
29 Scientist consider all rxns occurring in both directions fwd Reactants Products rev Think of all chemical equations as two sets of reagents that create two opposing reactions with different %Yields at equilibrium Reactants Products both reagents
30 Categories of Equilibria non-spont rxn strongly favors reactants (%Yield << 1%) Quantitative rxn favors products (%Yield >> 99%) observed to be complete generally written as a single arrow b/c reverse reaction negligible technically not wrong with have equilibrium arrow Redefine understanding of quantitative: rxn >99% complete, meaning that the reverse rxn happens so little that it can be ignored for all normal purposes since reverse rxn is non-spont consider for our purposes that if a rxn is quantitative, then (arbitrarily) assume that less than 0.1% of the original reactants remain un-reacted at equilibrium
p. 680
32 Categories of Equilibria if no L.R. then can NOT assume complete (quantitative) rxn stoich calculations differ slightly arrange values in an ICE table (initial, change, equilibrium)
33 Sample Problem 15.1 p. 681 Demonstrate all substances gas, therefore can apply mole ratio for stoich calculations with concentrations instead of chemical amounts (moles) b/c same volume for every gaseous substance in the closed container i.e. closed system volume common factor same applies to rxn with all (aq) entities dissolved in the same volume of solvent
34 Example: Analyze Trial 2 data from Table 1 p. 679 balance rxn set up ICE table beneath solve for the change
35 HOMEWORK Practice Qs - p. 682 #3-7 Lab Exercise 15.A p. 683 (Synthesis of Equilibrium Law)
36 Equilibrium Constant, K c Formally known as K eq Mathematical relationship between amount concentrations (mol/l) in a rxn constant value derived for a chemical system over a range of amount concentrations
Equilibrium Law p. 684 generalized expression derived from analysis of numerous equilibrium systems upper case = entities lower case = coefficients of balanced rxn eq n aa + bb cc + dd K c = [C]c [D] d [A] a [B] b use consistent units so they can be ignored in the K c expression 37
38 Equilibrium Law Consider the convention of Products over Reactants Let x = coeff. K c = [Products]x [Reactants] x only applies if amount concentrations are observed to remain constant (at equilibrium), in a closed system, and at a given temperature very valuable constant in chemistry all K c values (or calculations using it) must specify the rxn temperature at equilibrium
39 Communication Example 1 p. 684 Write the equilibrium law expression for the reaction of nitrogen monoxide gas with oxygen gas to form nitrogen dioxide gas.
40 Equilibrium Law K c describes the extent of the forward rxn reverse rxn written as reciprocal 1 K c Example: K c for decomposition is reciprocal of K c for formation rxn
41 Communication Example 2 p. 684 The value of Kc for the formation of HI (g) from H 2(g) and I 2(g) is 40, at a given temperature. What is the value of Kc for the decomposition of HI (g) at the same temperature? OR K c, reverse = H 2(g) + I 2(g) 2 HI (g) K c = 1 K c,forward = 1 40 = 0.025 2 HI (g) H 2(g) + I 2(g) [HI]2 [H 2 ][I 2 ] = 40 K c = [H 2][I 2 ] [HI] 2 = 1 40 = 0.025 Formation (forward rxn) Decomposition (reverse rxn)
42 Equilibrium Law higher numerical value of K c implies a greater tendency of the system to favor the forward rxn K c, products i.e. products favored at equilibrium K c > 1 equilibrium favors products K c < 1 equilibrium favors reactants K c gives info on equilibrium position of rxn, NOT on rxn rate consider in Thermochemistry Unit (Energetics) moderate concentration change of reactants or products results in small concentration change is all other entities, therefore K c remains same K c is affected by a large change in concentrations
43 Equilibrium Law Adjust K c to reflect pure substances, i.e. (s) or condensed (l) have fixed amount concentration concentration of condensed (s) and (l) states are NOT included in K c calculations only include variable concentrations of dissolved (aq) solutions and gaseous mixtures consider net ionic equation
44 Communication Example 3 p. 685 Write the equilibrium law expression for the decomposition of solid ammonium chloride to gaseous ammonia and gaseous hydrogen chloride. Omit solid from the Equilibrium Law expression
45 Communication Example 4 p. 685 Write the equilibrium law expression for the reaction of zinc in copper(ii) chloride solution. Zn (s) + CuCl 2(aq) ZnCl 2(aq) + Cu (s) Omit solids and spectator ions from the Equilibrium Law expression
46 Equilibrium Law Equilibrium depends on [reacting substances] Represent in expression as actually exists Represent ions as individual entities Always write expression from balanced net ionic rxn equation with simplest whole number coefficients, unless otherwise stated i.e. K c Excludes: (s) & (l) pure substances spectator ions units
48 HOMEWORK Lab Exercise 15.B p. 686 (Determining an Equilibrium Constant) DUE: R, Sept 24
49 Today s Objectives Calculate equilibrium constants and concentrations when concentrations at equilibrium are known initial concentrations and one equilibrium concentration are known equilibrium constant and one equilibrium concentration are known Section 15.1 (pp. 676-689)
50 Calculations in Equilibrium Systems Example: In the following system: N 2(g) + 3H 2(g) 2NH 3(g) K c = 1.2 at 375 o C 0.249 mol N 2(g), 3.21x10-2 mol H 2(g), and 6.42x10-4 mol NH 3(g) are combined in a 1.00 L vessel. Is the system at equilibrium? If not, predict the direction in which the reaction must proceed. K c = [NH 3] 2 [N 2 ][H 2 ] 3 = (6.42 x 10 4 ) 2 = 0.0500 1.2 0.249 (3.21 X 10 2) 3 Value < K c more reactants than products reaction must proceed to the right
51 Calculations in Equilibrium Systems Example: Find the equilibrium concentration of the ions that are formed when solid silver chloride is dissolved in water. (K c = 5.4x10-4 ) AgCl (s) Ag + (aq) + Cl (aq) K c = [Ag + ][Cl ] = 5.4x10 4 at equilibrium: [Ag + ] = [Cl ] = x x 2 = 5.4x10 4 x = 0.023 mol L
52 Predict Final Equilibrium Concentrations examine simple homogenous systems and algebraically predict reagent concentrations at equilibrium using a known value for K c and initial concentration values usually requires using an ICE table
53 ICE Tables Example: Consider the following equilibrium at 100 o C: N 2 O 4(g) 2 NO 2(g) 2.0 mol of N 2 O 4(g) was introduced into an empty 2.0 L bulb. After equilibrium was established, only 1.6 mol of N 2 O 4(g) remained. What is the value of K c? 2.0 mol = 1.0 mol/l (I) 2.0L 1.6 mol = 0.8 mol/l (E) 2.0L mol/l N 2 O 4(g) 2 NO 2(g) I: 1.0 0 C: x + 2x E: 1.0 x = 0.80 2x Solve for x: 1 x = 0.8 x = 0.2, 2x = 0.4 K c = 0.42 0.8 = 0.2
54 ICE Tables STEPS: Always write out the equilibrium reaction and equilibrium law expression if not given. Draw an ICE Table (Initial, Change in, and Equilibrium concentrations) I + C = E Substitute known values Solve for x Solve for equilibrium concentrations
55 ICE Tables Example: A 10 L bulb is filled with 4.0 mol of SO 2(g), 2.2 mol of O 2(g), and 5.6 mol of SO 3(g). At equilibrium, the bulb was found to contain 2.6 mol of SO 2(g). Calculate K c for this reaction: 2 SO 2(g) + O 2(g) 2 SO 3(g) 4.0 mol = 0.40 mol/l (I) 10.0L 5.6 mol = 0.56 mol/l (I) 10.0L 2.2 mol = 0.22 mol/l (I) 10.0L 2.6 mol = 0.26 mol/l (E) 10.0L
56 ICE Tables Example: A 10 L bulb is filled with 4.0 mol of SO 2(g), 2.2 mol of O 2(g), and 5.6 mol of SO 3(g). At equilibrium, the bulb was found to contain 2.6 mol of SO 2(g). Calculate K c for this reaction: 2 SO 2(g) + O 2(g) 2 SO 3(g) I 0.4 0.22 0.56 C 2x x + 2x E 0.26 0.22 x 0.56 + 2x 0.4 2x = 0.26 x = 0.07 K c = 0.7 2 (0.15)(0.26) 2 = 48
57 ICE Tables Example: Determine the equilibrium concentrations for all species present given that the initial concentration of each reactant is 0.200 mol/l in the following reaction: N 2(g) + O 2(g) 2 NO (g) K c = 0.00250 I 0.2 0.2 0 C x x + 2x E 0.2 x 0.2 x 2x 0.195 mol/l 0.195 mol/l 0.00976 mol/l K c = 0.0025 = (2x)2 square root both sides 0.005 = 2x 0.01 0.05x = 2x (0.2 x) 2 0.2 x 0.01 = 2.05x x = 0.00488
58 Sample Problem 15.2 p. 686 ICE table required to solve see handout
When [reactants] > 1000K c disregard the change in concentration ICE Tables Approximation Rule 59 Example: Calculate the concentration of gases produced when 0.100 mol/l COCl 2(g) decomposes into carbon monoxide and chlorine gas. (K c for this reaction is 2.2x10-10 ) 0.1 > 1000 x 2.2x10-10 assume 0.1 x 0.1 i.e. negligible change COCl 2(g) CO (g) Cl 2(g) I 0.1 0 0 C x + x + x E 0.1 x x x 0.100 mol/l 4.7x10-6 mol/l 4.7x10-6 mol/l K c = 2.2 10 10 = x2 0.1 x = 4.7 10 6 mol L
60 Predicting Equilibrium NOTE: If the initial reactant concentrations are not equal, solving for unknown change in concentration requires the application of the quadratic formula not seen in Nelson textbook can find quadric formula in data booklet x 1,2 = b ± b2 4ac 2a
61 Predicting Equilibrium predicting equilibrium by applying the quadratic formula Example: In a closed, 500mL reaction vessel at a temperature of 198 C phosphorus trichloride gas and chlorine gas combined to form phosphorus pentachloride gas. Initially 0.015mol of each reactant is placed in the vessel. The K c for this reaction is 2.00 at 198 C. What is the amount concentration of the product present at equilibrium? See Handout
62 HOMEWORK Section 15.1 Review Qs p. 688 #1-11