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Chem. 1B – 9/10 Lecture. Announcements Mastering Chemistry Online Homework for Ch. 14 due Tuesday (we still need to cover use of Q to predict reactions.

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Presentation on theme: "Chem. 1B – 9/10 Lecture. Announcements Mastering Chemistry Online Homework for Ch. 14 due Tuesday (we still need to cover use of Q to predict reactions."— Presentation transcript:

1 Chem. 1B – 9/10 Lecture

2 Announcements Mastering Chemistry Online Homework for Ch. 14 due Tuesday (we still need to cover use of Q to predict reactions and Le Châtelier’s Principle) Today’s Lecture –Equilibrium Problems: STARTING AT INITIAL CONDITIONS – only a couple of examples (one after Reaction Quotient) –Reaction Quotient and Reaction Direction –Le Châtelier’s Principle (direction of reaction following change) –Start to Chapter 15 (if time)

3 Chem 1B - Equilibrium Equilibrium Problems – Starting from initial conditions Example Problem 2: - cont. from last time A 10.0 L flask is filled with 0.0020 mol NO 2 (g) and it is expected to decompose as (ignoring the N 2 O 4 formation reaction previously mentioned): 2NO 2 (g) ↔ 2NO (g) + O 2 (g) With K C = 4.5 x 10 -16 Calculate the equilibrium concentration of each gas Got to determining x = 1.65 x 10 -8 (using assumption 2.0 x 10 -4 >> 2x)

4 Chem 1B - Equilibrium Equilibrium Problems – Overview Does problem ask to calculate K or an unknown concentration at equilibrium? KUnknown conc. Are concentrations of all species given at equilibrium? Yes No ICE table needed ICE table needed along with given equil. conc. No Are concentrations of all but 1 species given at equilibrium? Yes No ICE table needed ICE table needed No

5 Chem 1B - Equilibrium Equilibrium Problems – Example to Select Method Example Problem (somewhat tricky): CaCO 3 (s) is placed in a sealed vessel and heated to 800 K and establishes the following equilibrium: CaCO 3 (s) ↔ CaO(s) + CO 2 (g) If K C at this temperature is 6.4 x 10 -4, determine the equilibrium concentration of CO 2. Go back to flow diagram – what is being asked for?, is an ICE table needed? Is it possible to solve this problem – if yes, what is the answer?

6 Chem 1B – Equilibrium The Reaction Quotient and Reaction Direction For a given “system” (e.g. closed flask containing chemicals), the system can either be AT EQUILIBRIUM or under some other conditions (e.g. initial conditions) The equilibrium equation and constant only applies to equilibrium conditions A second quantity, the REACTION QUOTIENT = Q, can be calculated under any conditions (also Q C and Q P ) For generic reaction: aA + bB ↔ cC + dD Q for above reaction, note: for this reaction, [C] = conc. C (but not necessarily at equilibrium conditions)

7 Chem 1B – Equilibrium The Reaction Quotient and Reaction Direction When Q > K, we are too heavy on products, so reaction would proceed toward reactants (loss of C and D and gain of A and B) When Q < K (e.g. initial conditions if A and B are mixed and [C] = [D] = 0 or Q = 0), reaction proceeds toward products Q = K indicates we are at equilibrium

8 Chem 1B - Equilibrium The Reaction Quotient and Reaction Direction Example: An air resource board employee is studying the effects of car exhaust pipe length on pollution concentrations Air leaving the engine has both NO and NO 2 (NO 2 is a worse pollutant produced from NO + O 2 ) In the exhaust pipe, the reaction can continue toward equilibrium: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) with K P = 4.2 x 10 8 at the exhaust pipe temperature The gas partial pressures are measured just leaving the engine (start of exhaust pipe) and found to be: P NO = 1.0 x 10 -4 atm, P O2 = 0.030 atm, and P NO2 = 2.2 x 10 -7 atm. In which direction will this reaction proceed?

9 Chem 1B - Equilibrium Equilibrium Problems – Large K Value Questions The example covered last time for calculation of Q is an example of this type In the exhaust pipe (if long enough), the reaction could reach equilibrium: 2NO (g) + O 2 (g) ↔ 2NO 2 (g) with K P = 4.2 x 10 8 From the initial gas partial pressures: P NO = 1.0 x 10 -4 atm, P O2 = 0.030 atm, and P NO2 = 2.2 x 10 -7 atm, let’s calculate equilibrium partial pressures show why use of standard ICE table fails and how to add full right and backwards to get it to work (NOTE: THIS IS NOT COVERED IN THE TEXT – NOT SURE IF THIS WILL BE TESTED ON)

10 Chem 1B - Equilibrium Le Châtelier’s Principle Le Châtelier’s Principle is used to determine how a reaction at equilibrium will shift due to a change in conditions Continuing the past example (reaction in a car’s exhaust pipe: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) If we can change conditions (e.g. fuel/oxygen ratio or temperature) we may be able to limit formation of NO 2 in the exhaust pipe Le Châtelier’s Principle: When a chemical system at equilibrium is disturbed, the system shifts in a direction that minimizes the disturbance

11 Chem 1B - Equilibrium Le Châtelier’s Principle Changes in Conditions – Types: –Changes in reactant or product concentrations (or partial pressures) –Effect of a change in volume (compression/expansion or dilution/concentration) –Change in temperature Top two changes affect Q; bottom change affects K, in all cases with the “stress” Q ≠ K Given the above changes, we should be able to determine if, under new conditions, the system will re- establish equilibrium by shifting to reactants or products Can take intuitive (all) or mathematical approaches (1 st two changes) to solving problems

12 Chem 1B - Equilibrium Le Châtelier’s Principle Intuitive Method –Addition to one side results in switch to other side –Example: Mathematical Method AgCl(s) ↔ Ag + + Cl - Addition of Ag + When Q>K, reaction goes toward reactants When Q<K, reaction goes toward products Example: Q = [Ag + ][Cl - ] As [Ag + ] increases, Q>K reaction shifts to reactants (more AgCl(s))

13 Chem 1B - Equilibrium Le Châtelier’s Principle Stress Number 1 Reactant/Products: Addition of reactant: shifts toward product Removal of reactant: shifts toward reactant Addition of product: shifts toward reactant Removal of product: shifts toward product

14 Chem 1B - Equilibrium Le Châtelier’s Principle – Product/Reactant Stress Example: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) reaction in tailpipe Change:Expectation –Increase O 2 (“lean” conditions) –Decrease O 2 (fuel rich conditions) –Increase NO (run engine hotter) –Remove NO 2 (NO 2 trap??) Right pointing arrow means more NO 2 produced (not desired)

15 Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress For Gases – a decrease in volume will cause a simultaneous increase in concentration and pressure The system copes with this stress by minimizing the space/pressure taken up by molecules and switching to the side with fewer moles of gas molecules Example: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) –has 3 moles gas in reactants and 2 in products, so decreasing volume shifts system to products

16 Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress Example: 2NO(g) + O 2 (g) ↔ 2NO 2 (g) –Mathematical explanation: –Initially at equilibrium K C = 10 5 and [NO] = 0.0010 M, [O 2 ] = 0.0010 M and [NO 2 ] = 0.010 M –Now we reduce the volume from 10.0 to 1.00L –New concentrations: [NO] = 0.010 M, [O 2 ] = 0.010 M and [NO 2 ] = 0.10 M (same number of moles in 1/10 th the volume so 10X more concentrated) –Q = (0.10 M) 2 /[(0.010 M) 2 (0.010 M)] = 10 2 < K C, so products favored

17 Chem 1B - Equilibrium Le Châtelier’s Principle – Volume Stress Note: in aqueous solutions, dilution works in the same way (increase in space due to dilution favors side with more moles) a 1 M HC 2 H 3 O 2 (acetic acid) solution is diluted by adding an equal volume of water. How does this reaction change? HC 2 H 3 O 2 (aq) ↔ H + (aq) + C 2 H 3 O 2 - (aq)

18 Chem 1B - Equilibrium Le Châtelier’s Principle – Temperature Stress Note: change in T changes K (while initial K becomes Q) If ΔH>0, as T increases, products favored - this also means K increases with T If ΔH<0, as T increases, reactants favored Easiest to remember by considering heat a reactant or product Example: OH - + H + ↔ H 2 O(l) + heat (reaction  H < 0) Increase in T

19 Chem 1B - Equilibrium Le Châtelier’s Principle Looking at the reaction below, that is initially at equilibrium, AgCl(s) ↔ Ag + (aq) + Cl - (aq) (ΔH°>0) determine the direction (toward products or reactants) each of the following changes will result in a)increasing the temperature b)addition of water (dilution of system) c)addition of AgCl(s) d)addition of NaCl

20 Chem 1B – Aqueous Chemistry Overview: –Both Chapters 15 and 16 apply equilibrium theory to aqueous systems (that is one reason we have been focusing on gases in Chapter 14) –Chapter 15 is focused on acids and bases (including qualitative understanding), while Chapter 16 continues with acid/base chemistry and proceeds to quantitative treatment of solubility and metal – ligand complexation

21 Chem 1B – Chapter 15 Topics Definitions of Acids/Bases Acid Strength Water Autoprotolysis pH Scale Weak Acids in Water Bases/Base Strength Conjugate Acids and Bases: pH of Salt Solutions Polyprotic Acids Molecular Scale View of Acids and Bases

22 Chem 1B – Aqueous Chemistry Overview: –Both Chapters 15 and 16 apply equilibrium theory to aqueous systems (that is one reason we have been focusing on gases in Chapter 14) –Chapter 15 is focused on acids and bases (including qualitative understanding), while Chapter 16 continues with acid/base chemistry and proceeds to quantitative treatment of solubility and metal – ligand complexation


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