Monday February 5, 2024 Day 20 General Equilibrium Concepts |
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Textbook Reading 13.1 Chemical Equilibria |
Course Lectures 7.1 pdf Video* Intro. to Chem. Equilibrium 7.2 pdf Video* Law of Mass Action |
| Equilibrium: Crash Course Chemistry |
Visual Demonstration of Chemical Equilibrium |
| Objectives 1. Describe what is meant by dynamic equilibrium and provide real-life examples. .....define phase equilibrium .....define "closed system" .....identify the forward process .....identify the reverse process 2. Write chemical equations as equilibrium reactions .....Use the double arrow in chemical equations to denote the existence of chemical equilibrium 3. Know the significance of Ratef = Rater 4. Draw/Interpret Conc. vs Time graphs |
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| Homework Problems 11.1 Refering to the diagram at right, answer the following questions (a - e) ... a. Why is the initial concentration of NO2 zero and why does it increase? b. Why do the concentrations of NO and O2 initially decrease? c. Both reactants begin with 0.0010 M concentrations. Why does the concentration of NO decrease faster than the concentration of O2? |
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d. When reactant and product concentrations no longer change, the system is at equilibrium. At approximately what time does this system reach equilibrium? e. In this reaction, are all reactants converted into products? How is this result different from what you would have expected as a Principles of Chemistry 1 student? f. Once equilibrium has been reached, how do the concentrations of O2, NO and NO2 change? 11.2 Equilibrium requires two competing processes. As a teacher, the equilibrium I maintain is the balance between the arrival of ungraded student papers and the departure of graded student work. What is an example, in your life, of two competing processes and their equilibrium? 11.3 The reaction of nitrogen gas with hydrogen gas produces gaseous ammonia. What is the forward reaction? What is the reverse reaction? Write the reaction as an equilibrium readtion. Click and drag the region below for correct answers 11.1 a. NO2 is a product and initially there is none present. The concentration of NO2 increases as the chemical reaction takes place making product from the two reactants. b. The reactants, NO and O2 are the only species initally present. As they react and form product, their concentrations decrease. c. NO levels decrease twice as fast as O2 levels because the balanced chemical reaction tells us that 2 NO molecules are required for every O2 molecule. d. After approximately 5000 seconds the reactant & product levels are observed to stabilize. e. In 1st semester chemistry, we dealt with "completion reactions" where it's assumed that reactants are entirely converted into products. The surprise here is that for equilibrium situations, the reverse reaction converts product back into reactants.... and this process competes with the forward reaction. f. Once equlibrium has been reached after 5000 seconds, the concentrations of products and reactants don't change (i.e. they're constant) 11.2 I'll be interested to see what you come up with for this question. 11.3 Forward reaction: N2(g) + 3 H2(g) → 2 NH3(g) Reverse reaction: N2(g) + 3 H2(g) ← 2 NH3(g) Equilibrium reaction: N2(g) + 3 H2(g) ↔ 2 NH3(g) .....OR... N2(g) + 3 H2(g) ⇌ 2 NH3(g) |
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Tuesday February 6, 2024 Day 21 Equilibrium: Reaction Quotients and Equilibrium Shifts |
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Textbook Readings 13.2 Equilibrium Constants |
Course Lectures 7.2 pdf Video* Law of Mass Action 8.2 pdf Video* The Reaction Quotient "Q" 7.3 pdf Video* Determining an Equil. Constants |
| The Equilibrium Constant |
K (Equilibrium Constant) vs Q (Reaction
Quotient) |
| Objectives 1. Given a balanced chemical reaction write out the Law of Mass Action 2. Given the Law of Mass Action (LMA) expression, insert equilibrium concentrations for products and reactants and calculate a value for the equilibrium constant Keq. 3. Identify pure solids (s) and liquids (l) and NOT include them when writing out the Law of Mass action for the equilibrium. 3. Given the LMA expression, insert non-equilibrium concentrations for products and reactants and calculate the reaction quotient "Q". 4. Use Q and Keq to determine which direction the reaction shifts. 5. Describe what is meant as a "shift right" or "shift left" in regards to equilibria. |
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| Homework Problems 12.1. Write the mathematical expression for the reaction quotient, Qc, for each of the following reactions: a. N2(g) + 3 H2(g) ↔ 2 NH3(g) b. 4 NH3(g) + 5 O2(g) ↔ 4 NO(g) + 6 H2O(g) c. N2O4(g) ↔ 2 NO2(g) d. NH4Cl(s) ↔ NH3(g) + HCl(g) e. 2 H2(g) + O2(g) ↔ 2 H2O(l) 12.2 For each of the following, use the LMA expression you derived above and the equilibrium concentrations provided to determine a value for the equilibrium constant Keq. a. N2(g) + 3 H2(g) ↔ 2 NH3(g) Keq = _____ Equilibrium Concentrations: [N2]eq = 2.50 M [H2]eq = 1.95 M [NH3]eq = 0.45 M b. 4 NH3(g) + 5 O2(g) ↔ 4 NO(g) + 6 H2O(g) Keq = _____ Equilibrium Concentrations: [NH3]eq = 0.53 M [NO]eq = 0.22 M [O2]eq = 1.02 M [H2O]eq = 0.13 M c. N2O4(g) ↔ 2 NO2(g) Equilibrium Concentrations: [N2O4]eq = 1.78 M [NO2]eq = 2.55 x 10-4 M d. NH4Cl(s) ↔ NH3(g) + HCl(g) Equilibrium Concentrations: [NH3]eq = 4.89 x 10-4 M [HCl]eq = 9.44 x 10-1 M e. 2 H2(g) + O2(g) ↔ 2 H2O(l) Equilibrium Concentrations: [H2]eq = 6.88 x 10-18 M [O2]eq = 1.45 x 10-17 M 12.3 The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction the system shifts to reach equilibrium and the effect on reactant and product levels. a. 2 NH3(g) ↔ N2(g) + 3 H2(g) Kc = 17 initial molar concentrations: [NH3]i = 0.50 M, [N2]i = 0.15 M, [H2]i = 0.12 M b. 2 NH3(g) ↔ N2(g) + 3 H2(g) KP = 6.8×104 initial pressures (Pi): NH3 = 2.00 atm, N2 = 10.00 atm, H2 = 10.00 atm c. 2 SO3(g) ↔ 2 SO2(g) + O2(g) Kc = 0.230 initial molar concentrations: [SO3]i = 2.00 M, [SO2]i = 2.00 M, [O2]i = 2.00 M d. 2 SO3(g) ↔ 2 SO2(g) + O2(g) KP = 6.5atm initial pressures (Pi): SO2 = 1.00 atm, O2 = 1.130 atm, SO3 = 0 atm e. 2 NO(g) + Cl2(g) ↔ 2 NOCl(g) KP = 2.5×103 initial pressures (Pi): NO = 1.00 atm, Cl2 = 1.00 atm, NOCl = 0 atm f. N2(g) +O2(g) ↔ 2 NO(g) Kc = 50. initial molar concentrations: [N2]i = 0.100 M, [O2]i = 0.200 M, [NO]i = 1.00 M 12.4 What is meant when it's said a reaction shift right? ...or left? Click and drag the region below for correct answers 12.1 Answers click HERE 12.2 a. Keq = 0.0109 b. Keq = 1.29 x 10-7 c. Keq = 3.65 x 10-8 d. Keq = 4.62 x 10-4 e. Keq = 1.46 x 1051 12.3 a. Q = 1.036 x 10-3 Q < Kc Reaction shift right Products increase and reactants decrease b. Q = 2.50 x 103 Q < KP Reaction shift right Products increase and reactants decrease c. Q = 2 Q > Kc Reaction shifts left Products decrease and reactants increase d. Q = ∞ (infinity) Q > KP Reaction shifts left Products decrease and reactants increase (from zero) e. Q = 0 Q < KP Reaction shift right (initially, there are no products) Products increase and reactants decrease f. Q = 50 Q = Kc Products and reactants in dynamic equilibrium No change expected. 12.4 When a reaction is said to shift right, it means that reactants are converted into products. That is, product amounts increase and reactant amounts decrease. When a reaction shifts left, products are converted into reactants via the reverse reaction. Product levels drop and reactant levels increase. |
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Wednesday February 7, 2024 Day 22 Equilibrium Constant Manipulations |
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Textbook Readings 13.2 Equilibrium Constants |
Course Lectures 8.3 pdf Video* Equilibrium Constants: Recipes 8.4 pdf Video* Kc and Kp conversionss |
| Relating Kc and Kp |
Manipulating Kc and Kp |
| Objectives 1. Combine equilibrium reactions and constants to obtain new "net" reactions and its corresponding equilibrium constant. 2. Know that when reversing equilibrium reactions, the equilibrium constant is inverted. (inverted means to 1/x) 3. Know that when multiplying an equilibrium reaction by "n", the new equilibrium constant is determined as Kn 4. Know that when combining equilibrium constants from steps used to create the "net" reaction, the individual equilibrium constants are multiplied together. 5. Determine Δn from the balanced equlibrium equation. 6. Calculate Kp from Kc values (vice versa) using Δn and the following equation: Kp = Kc(RT)Δn |
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Homework Problems 13.1 Use the following equilibrium reaction and equilibrium constant: A + 2B ↔ 2C K = 10 ...to determine the equilibrium constant for this reaction: 2A + 4B ↔ 4C K = _______ 13.2 Use the following equilibrium reaction and equilibrium constant: A + 2B ↔ 2C K = 10 ...to determine the equilibrium constant for this reaction: 4C ↔ 2A + 4B K = _______ 13.3 Given the following information: HF(aq) ↔ H+(aq) + F-(aq) Kc = 6.8 x 10-4 H2C2O4(aq) ↔ 2 H+(aq) + C2O42-(aq) Kc = 3.8 x 10-6 Determine the equilibrium constant for the following reaction: 2HF(aq) + C2O42-(aq) ↔ 2 F-(aq) + H2C2O4(aq) Kc = ________ 13.4 Calculate the equilibrium constant for this reaction: 2A + B ↔ C + 3D given the following reaction steps: Rxn1: A + B ↔ D K1 = 0.40 Rxn2: A + E ↔ C + D + F K2 = 0.10 Rxn3: C + E ↔ B K3 = 2.0 Rxn4: F + C ↔ D + B K4 = 5.0 13.5 What is Δn for each of the following balanced equilibria? a. CO(g) + 3 H2(g) ↔ CH4(g) + H2O(g) b. CO(g) + H2O(g) ↔ CO2(g) + H2(g) c. CO(g) + 2H2(g) ↔ CH3OH(g) d. CO(g) + 1/2 O2(g) ↔ CO2(g) e. 2H2(g) + O2(g) ↔ 2H2O(l) 13.6 For the reaction 2NO2(g) ↔ N2O4(g) KP = 7.3 at 25 oC. What is Kc for this reaction? Answers: Click and drag in the space below 13.1 100 13.2 0.010 13.3 K = 0.12 13.4 K = 0.10 13.5 a. -2 b. 0 c. -2 d. -0.5 e. -3 Liquids don't count 13.6 179 |
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Thursday February 8, 2024 Day 23 Equilibrium Constants and their Magnitudes. |
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Textbook Readings 13.2 Equilibrium Constants |
Course Lectures 8.1 pdf Video* What Equil. Constants tell us |
| Objectives 1. Describe what the magnitudes of equilibrium constants tell us about relative product and reactant amounts present at equilibrium. 2. Describe how reactions shift in each of the following four situations: i. Large Keq and small amount of product initially present |
The Magnitude of Equilibrium Constants |
ii. Large Keq and large amount of product initially present iii. Small Keq and small amount of product initially present iv. Small Keq and large amount of product initially present |
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Homework Problems 14.1. Kc = 1 x 1010 for an equilibrium reaction. What is favored? 14.2. Kc = 1.0 x 10-10 for an equilibrium reaction. What is favored? 14.3 Consider the following reaction, initial pressures and equilibrium constant: 2 NO(g) + O2(g) ↔ NO2(g) Kp = 1.00 x 104 initial 2.0 atm 1.0 atm 0.0 atm a. Can this reaction create products? b. Is the equilibrium constant large or small? Does this favor products or reactants? c. When equilibrium is reached, how will the amount of products compare to the reactants? d. In reaching equilibrium, will there have been large changes in reactant and product levels? e. Will there be much reactant left at equilibrium? 14.4 Consider the following reaction, initial pressures and equilibrium constant: 2 H2(g) + O2(g) ↔ 2H2O(g) Kp = 1.389 x 1080 1.00 atm 1.00 atm 0.00 atm a. Can this reaction make products? b. Is the equilibrium constant large or small? Does this favor products or reactants? c. When equilibrium is reached, how will the amount of products compare to the reactants? d. In reaching equilibrium, will there have been large changes in reactant and product levels? e. Will there be much reactant left at equilibrium? 14.5 Consider the following reaction, initial pressures and equilibrium constant: 2CO2(g) ↔ 2CO(g) + O2(g) Kp = 6.77 x 10-91 1.00 atm 1.00 atm 0.00 atm a. Can this reaction make products? b. Is the equilibrium constant large or small? Does this favor products or reactants? c. When equilibrium is reached, how will the amount of products compare to the reactants? d. In reaching equilibrium, will there have been large changes in reactant and product levels? e. Will there be much reactant left at equilibrium? Answers: Click and drag in the space below 14.1 This is a large equilibrium constant and products are favored. 14.2 This is a small equilibrium constant and reactants are favored. 14.3 a. Yes. With a lot of reactant and absolutely no product, the reaction can certainly make product. b. The Kp value is large and this suggests that the equilibrium favors product formation. c. At equilibrium, the amount of product will be much larger than any remaining reactant. Note, that in this equilibrium situation, a small amount of reactant must remain. d. Large changes have taken place. Initially, there was no product. However, conditions are right for a lot of product formation. We initially have a lot of reactant available and the equilibrium constant favors product formation. e. The initial pressures are in the correct stoichiometric ratio for a complete reaction 2.0 atm NO : 1.0 atm O2. Therefore, both reactants will be used up almost completely. However, small amounts of both reactants will be left to maintain the final equilibrium. 14.4 a. Yes. With a lot of reactant and absolutely no product, the reaction can certainly make product. b. The Kp value is large and this suggests that the equilibrium favors product formation. c. At equilibrium, there will be a large amount of product since we started entirely with reactant and the Kp value is large d. There will have been large changes to both reactants (decreased) and products (increased) e. Since the reactant stoichiometry is 2:1 we have twice as much O2 as is required. Therefore, when equilibrium is reached, H2 levels will be very small but about 0.50 atm of the O2 remains as an unreacted reactant. 14.5 a. Yes. With a lot of reactant and absolutely no product, the reaction can certainly make product. b. The Kp value is small and this suggests that the equilibrium favors reactant formation. c. The equilibrium constant supports reactant formation. Since reactants are all we have initially, we don't expect much change. Only a small amount of product will form; just enough to create the chemical equilibrium. d. As already mentioned, only a very small amount of product will form. Consequently any changes from the inital pressures will be very small. e. Reactant levels will be very near to their initial levels. |
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Friday February 9, 2024 Day 24 Le Châtelier’s Principle |
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Textbook Readings 13.3 Shifting Equilibria: Le Châtelier’s Principle |
Course Lectures |
| Objectives 1. Utilize Le Chatelier's Principle to predict reaction shifts that occur when reactant or product concentrations change 2. Le Chatelier's Principle to predict reaction shifts that occur when pressure or volume changes 3. Le Chatelier's Principle to predict reaction shifts that occur when temperature changes. |
Le Chatelier's Principle |
| Homework problems. 15.1 Consider the following equilibrium reaction: H2(g) + O2(g) ↔ H2O2((l) ΔHrxn = - 187 kJ/mol a. How does the equilibrium shift if H2 is added? How are the O2 levels affected by the shift. b. Where does the word "heat" belong in this reaction? c. How does the equilibrium shift if the temperature is increased? d. How do the number of reactant particles compare to the number of product particles? e. How does the equilibrium shift if the pressure is increased? f. How does the equilibrium shift if the volume is decreased? 15.2 Consider the following equilibrium reaction: 2 NH3(g) ↔ N2(g) + 3 H2(g) ΔHrxn = + 92 kJ/mol a. How does the equilibrium shift if NH3 is removed from the reaction container? b. How does the reaction shift if the volume of the reaction container is decreased? c. How does the reaction shift if the temperature of the reaction is cooled? 15.3 Consider the following equilibrium reaction: 2 SO3(g) ↔ 2 SO2 (g) + O2 (g) ΔHrxn = 197.78 kJ List all steps you can take to maximize product yield. 15.4 Consider the following equilibrium reaction: Reactants (red) ↔ Products (blue) The reaction mixture is heated and observed to change from red to blue. a. In which direction (right or left) has the reaction shifted to achieve a new equilibrium? b. How has the shift affected the reactant and product levels? c. How has heating the reaction mixture changed the equilibrium constant? d. Where does the word "heat" belong (reactant or product)? e. Is the reaction exothermic or endothermic? f. If an increased in pressure shifts the color back to red, what do you know? Answers: Click and drag in the space below. 15.1 a. Adding hydrogen gas (a reactant) causes the reaction to shift right as the reaction attempts to convert the excess H2 into product. O2 levels will decrease as the reaction shifts right even though they weren't responsible for the shift initially. b. Since ΔHrxn < 0 (neg) the reaction is exothermic and the word "heat" belongs on the product side of the reaction c. Increasing temperature is the same as adding heat. Since heat is a product in this reaction, additional heat shifts the reaction left. d. There are two reactant particles and one product particle. e. Increasing the pressure (equivalently decreasing the volume) has the effect of increasing the density of particles. The reaction shifts in a direction to reduce the particle crowding. A rightward shift comverts two particles into one, effectively lowering the particle crowding. f. A decrease in volume is equivalent to an increase in pressure and the reaction shifts right. 15.2 a. Removing reactant causes the reaction to shift left to replace the missing reactant. b. 2:4 ...the reaction shifts left to reduce particle denstiy. c. This is an endothermic reaction and "heat" is a reactant. Cooler temps will produce a leftward shift. 15.3 i. Increase temperature ii. Lower pressure or increase volume iii. Add SO3 , a reactant, to the reaction chamber. iv. Remove SO2 or O2 product from the reaction chamber. 15.4 a. Shifts right b. Reactant level decreases, Product level increases. c. Since a rightward shift increased product levels and decreased reactant levels to reach a new equilibrium, the equilibrium constant (Product/Reactant ratio) has increased. d. "Heat" is a reactant e. The reaction is endothermic f. The color change tells us that the equilibrium shifts left when pressure is increased. This is only possible if there are fewer reactant particles than product particles in the balanced reaction. |
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End Week 5