Monday February 12, 2024 Day 25 Equilibrium: ICE and Determining Equilibrium Concentrations |
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Textbook Readings 13.4 Equilibrium Calculations |
Course Lectures 9.1 pdf Video* Det. Equil. Conc. (perfect square) |
| How to Generate an ICE Table |
Equilibrium Concentration (Perfect Square) |
| Objectives 1. Verify balanced chemical equation 2. Record the ICE table beneath the balanced chemical equation (Initial, Change, Equilibrium) 3. Determine the Law of Mass action expression from the ICE table 4. Substitute equilibrium concentration expressions into the Law of Mass Action 5. Solve for "x" 6. Determine equilibrium concentrations using "x" determined above 7. Double check equilibrium concentrations by substituting into the LMA expression to verify "K" |
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Homework Problems 16.1 1.00 mole of H2 and 2.00 moles of I2 are placed in a sealed, 1.00 L reaction container and the following reaction takes place. ... Calculate initial concentrations, complete the ICE table below and write out the law of mass action (LMA). H2(g) + I2(g) ↔ 2HI(g) Keq = 3.5 Initial _______ _______ _______ Change _______ ________ _______ Equilibrium _______ ________ _______ 16.2 0.500 mole of H2, 2.00 moles of CO2 and 0.75 mole of CO are placed in a sealed, 2.00 L reaction container and the following reaction takes place. ... Calculate initial concentrations, complete the ICE table below and write out the LMA. H2O(g) + CO(g) ↔ CO2(g) + H2(g) Keq = 0.015 Initial _______ _______ _______ _______ Change _______ ________ _______ _______ Equilibrium _______ ________ _______ _______ 16.3 2.00 moles of H2 and 2.00 moles of I2 are placed in a sealed, 1.00 L reaction container and the following reaction takes place. ... H2(g) + I2(g) ↔ 2HI(g) Keq = 3.5 Calculate initial concentrations, create/complete an ICE table and use the Law of Mass action to determine values for "x" and the equilibrium concentrations. 16.4 1.500 mole of H2 and 1.50 moles of CO2 are placed in a sealed, 2.00 L reaction container and the following reaction takes place. ... H2O(g) + CO(g) ↔ CO2(g) + H2(g) Keq = 0.015 Calculate initial concentrations, create/complete an ICE table and use the Law of Mass action to determine values for x and the equilibrium concentrations. 16.5 Consider the following ICE problem and determine a value for "x" and the equilibrium constant. HC2H3O2(aq) + H2O(l) ↔ H3O+(aq) + C2H3O2-(aq) Initial 1.00 M 0.00 M 0.00 M Change _______ 1.6 x 10-5 M _______ Equilibrium _______ _______ _______ 16.6 Consider the following ICE problem and determine a value for the equilibrium constant. A (g) + 2 B(g) ↔ 3 C(g) Initial 1.00 M 0.0500 M 0.00 M Change _______ _______ 3.5x 10-2 M Equilibrium _______ _______ _______ Click and drag the region below for correct answers 16.1 H2(g) + I2(g) ↔ 2HI(g) Initial 1.0 M 2.0 M 0.0 M Change -x -x +2x Equilibrium (1.0 - x) (2.0 - x) 2x LMA: (2x)2 = 3.5 (1.0 -x) (2.0 - x) 16.2 H2O(g) + CO(g) ↔ CO2(g) + H2(g) Initial 0.00 M 0.3735 M 1.00 M 0.250 M Change +x +x -x -x Equilibrium x 0.375 + x (1.00 - x) (0.250 - x) LMA: (1.00 - x) (0.250 - x) = 0.015 x (0.375 + x) 16.3 [H2]eq = [I2]eq = 1.033 M [HI]eq = 1.933 M 16.4 [H2O]eq = [CO]eq = 0.668 M [CO2]eq = [H2]eq = 0.0818 M 16.5 x = 1.6 x 10-5 Keq = 2.56 x 10-10 16.6 x = 1.1666 x 10-2 Keq = 6.100 x 10-2 |
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Tuesday February 13, 2024 Day 26 ICE and Determining Equilibrium Concentrations |
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Textbook Readings 13.4 Equilibrium Calculations |
Course Lectures 9.2 pdf Video* Det. Equil. Conc. (quadratic) |
| Equilibrium Equations (Quadratic Solutions) |
Practice Problem: ICE Box Calculations |
| Objectives 1. Correctly calculate initial concentrations of reactants and products. 2. Construct ICE tables for equilibrium reactions complete with initial concentrations, "x" values and equilibrium concentration expressions. 3. Set up LMA for equilibrium ICE problems 4. Determine "x" values using the Quadratic Equation and use "x" to determine equilibrium concentrations. 5. Verify calculated equilibrium concentrations by substituting them into the LMA expression and comparing the result to Keq. |
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Homework Problems 17.1 Phosphorus trichloride and chlorine gas react to form phosphorus pentachloride via the following reaction: PCl3(g) + Cl2(g) ↔ PCl5(g) Kc = 16.0 Initially 5.00 moles of PCl5(g) are placed in a 5.00 L reaction container. a. Complete the ICE table for the equilibrium and use the LMA and the quadratic equation to determine "x" and the equilibrium concentrations of all species. b. Substitute your equilibrium concentrations back into the LMA and verify that it's value equals the equilibrium constant given above. (Note that this step is a requirement on all exam equilibrium problems) 17.2 Hydrogen and fluorine gas react to form hydrogen fluoride via the following reaction (source) H2(g) + F2(g) ↔ 2 HF(g) Kc = 1.15 x 102 Initially, 0.250 moles of H2 and 0.500 moles of F2 are placed in a 250. mL reaction container. a. Complete the ICE table for the equilibrium and use the LMA and the quadratic equation to determine "x" and the equilibrium concentrations of all species. b. Substitute your equilibrium concentrations back into the LMA and verify that it's value equals the equilibrium constant given above. (Note that this step is a requirement on all exam equilibrium problems) Answers: Click and drag in the space below 17.1 x = 0.221 [PCl3(g) ]eq = 0.221 M [Cl2(g)]eq = 0.221 M [PCl5(g) ]e = 0.779 M 17.2 x = 0.968 [H2]eq = 3.2 x 10-2 M [F2]eq = 1.03 M [HF]eq = 1.94 M |
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Wednesday February 14, 2024 Day 27 Simplifying Equilibrium Assumptions |
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Textbook Readings 13.4 Equilibrium Calculations |
Course Lectures 9.3 pdf Video* Equilibria: Simplifying Assumptions 9.4 pdf Video* Simplifying Assumptions Concluded |
| Making LMA math EASIER! |
When to make simplifying assumptions |
| Objectives 1. Use initial concentrations and Keq values to determine whether the x ~ 0 assumption is valid. |
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Homework Problems 18.1 For each of the following use the equilibrium constant to determine: i. products or reactants are favored ii. direction of the equilibrium shift iii. strength of the shift consider initial reactant amounts iv. if the x ~ 0 assumption is valid. Example H2(g) + N2(g) ↔ 2 NH3(g) Kc = 3.7 x 108 Initial 6.0 M 2.0 M 1.0 M Answer: i. Products favored ii. Equilibrium shifts right iii. Strong shift iv. x~0 approximation is not valid. a. 3 H2(g) + N2(g) ↔ 2 NH3(g) Kc = 3.7 x 108 Initial 0.00 M 0.00 M 0.75 M b. O3(g) + NO(g) ↔ O2(g) + NO2(g) Kc = 6.00 x 1034 Initial 3.75 M 0.50 M 1.00 M 1.00 M c. 3 H2(g) + N2(g) ↔ 2 NH3(g) Kc = 3.7 x 108 Initial 0.00 M 0.90 M 0.50 M d. 3 H2(g) + N2(g) ↔ 2 NH3(g) Kc = 3.7 x 108 Initial 0.00 M 0.00 M 22.5 M e. O3(g) + NO(g) ↔ O2(g) + NO2(g) Kc = 6.00 x 1034 Initial 3.75 M 0.00 M 0.00 M 5.00 M f. N2(g) + O2(g) ↔ 2NO(g) Kc = 4.10 x 10-31 Initial 2.00 M 2.00 M 2.00 M g. O3(g) + NO(g) ↔ O2(g) + NO2(g) Kc = 6.00 x 1034 Initial 0.50 M 1.00 M 12.00 M 15.00 M h. 3 H2(g) + N2(g) ↔ 2 NH3(g) Kc = 3.7 x 108 Initial 2.5 M 0.00 M 1.5 M i. 2 H2S(g) ↔ 2H2(g) + S2(g) Kc = 1.67 x 10-7 Initial 0.00 M 1.00 M 1.00 M j. 3 H2(g) + N2(g) ↔ 2 NH3(g) Kc = 3.7 x 108 Initial 0.00 M 0.15 M 0.00 M Answers: Click and drag in the space below 18.1 a. i. Products favored ii. Equilibrium shifts left because there is initially no H2 or N2 iii. Slight shift since products present initially iv. x~0 approximation VALID b. i. Products favored ii. Equilibrium shifts right since both reactants initially present iii. Strong shift since reaction favors products iv. x~0 approximation is NOT VALID. c. i. Products favored ii. Equilibrium shifts left since there is no H2 initially iii. Slight shift since reaction favors products iv. x~0 approximation VALID d. i. Products favored ii. Equilibrium shifts left since there is no H2 or N2 initially iii. Slight shift since reaction favors products iv. x~0 approximation VALID e. Neither the forward nor the reverse reaction can occur since we're missing both a reactant AND a product. No shift occurs. f. i. Reactants favored ii. Equilibrium shifts left since there's a lot of product present initially. iii. Strong shift since reaction favors reactants iv. x~0 approximation is NOT VALID. g. i. Products favored ii. Equilibrium shifts right since both reactants initially present iii. Strong shift since reaction favors products iv. x~0 approximation is NOT VALID. h. i. Products favored ii. Equilibrium shifts left since there is initially no N2 iii. Slight shift since products are favored iv. x~0 approximation is VALID. i. i. Reactants favored ii. Equilibrium shifts left since there is no reactant initially present. iii. Strong shift iv. x~0 approximation is NOT VALID. j. Neither the forward nor the reverse reaction can occur since we're missing both a reactant AND a product. No shift occurs. |
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Thursday February 15, 2024 Day 28 Simplifying Assumptions Part 2 |
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Textbook Readings 13.4 Equilibrium Calculations |
Course Lectures 9.3 pdf Video* Equilibria: Simplifying Assumptions 9.4 pdf Video* Simplifying Assumptions Concluded |
| Making LMA math EASIER! |
When to make simplifying assumptions |
| Objectives 1. Use initial concentrations and Keq values to determine whether the x ~ 0 assumption is valid. 2. Use the LMA expression and the x ~ 0 to solve for "x" and then determine the equilibrium concentrations 3. Verify the x~0 assumption is valid by performing a 5% test. 4. Verify calculated equilibrium concentrations by substituting them into the LMA expression and comparing the result to Keq. |
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Homework Problems 19.1 Examine the Law of Mass Action expression below and solve for "x" using then = 0 approximation. Check the x~ assumption with the 5% rule for both "1.75" and "2.25". (2x)2 5.88 x 10-8 = ---------------------------- (1.75 - x)(2.25 - 2x)2 19.2 Consider the following equilibrium reaction: A(aq) + 3 B(aq) ↔ C(aq) Kc = 2.9 x 10 -10 ...where initially [A] = 5.00 M [B] = 4.00 M and [C] = 0.00 M i. Construct an ICE table for the problem and record the LMA expression ii. The x ~ 0 assumption is valid for this problem. Why? iii. Solve for x using the x ~ 0 assumption iv. Determine the equilibrium concentrations for all species. v. Verify your equilibrium concentrations reproduce the Kc value vi. Perform all 5% tests. 19.3 Consider the following equilibrium 2 H2S(g) + 3 O2(g) ↔ 2 H2O(g) + 2 SO2(g) Kc = 7.69 x 108 ...where initially [H2S] = 0.00 M [O2 ] = 0.00 M [H2O] = 2.00 M and [SO2] = 3.00 M i. Construct an ICE table for the problem and record the LMA expression ii. The x ~ 0 assumption is valid for this problem. Why? iii. Solve for x using the x ~ 0 assumption iv. Determine the equilibrium concentrations for all species. v. Verify your equilibrium concentrations reproduce the Kc value vi. Perform all 5% tests. Answers: Click and drag in the space below... 19.1 Letting x = 0 in the denominator values we find x = 3.61 x 10-4 (3.61 x 10-4 / 1.75) x 100 = 0.02% < 5% ...x ~ 0 assumption is valid 19.2 x LMA 2.9 x 10-10 = -------------------------------------- (5.00 - x) (4.00 - 3x)3 x = 9.28 x 10-8 [A]eq = 5.00 M [B]eq = 4.00 M [C]eq = 9.28 x 10-8 M 5% rule checks.... For "A" 0.0000018 % < 5% :) For "B" 0.0000069% < 5% :) Note that in this problem the very small equilibrium constant tells us that reactants are favored. Since initially there are nothing but reactants, it is possible to use the x = 0 approximation since required right shift will be very small keeping most of the reactants on the left. 19.3 (2.00 - 2x)2 (3.00 - 2x)2 LMA 7.69 x 108 = ------------------------------- (2x)2 (3x)3 x = 1.36 x 10-2 [H2S]eq = 2.73 x 10-2M [O2]eq = 4.09 x 10-2 M [H2O]eq = 1.97 M [SO2]eq = 2.97 M Note that in this problem the very large equilibrium constant tells us that products are favored. Since initially there is nothing but products present, it is possible to use the x = 0 approximation since a change from inital conditions will be very small. 5% rule checks.... For H2O 1.36 % < 5% :) For SO2 0.91 % < 5% :) |
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Friday February 16, 2024 Day 29 Material Shifts |
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Textbook Readings 13.4 Equilibrium Calculations |
Course Lectures 9.45 pdf Video* Predicting Reaction Shifts 9.47 pdf Video* Introduction to Material Shift 9.51 pdf Video* Material shifts problem 1 & 2 9.52 pdf Video* Material shifts problem 3 9.53 pdf Video* Material shifts problem 4s |
| Objectives 1. Correctly determine if the x ~ 0 assumption is valid: a. Valid .... Set up ICE problem and solve completely using x ~ 0 assumption b. NOT Valid .... Perform a "Material Shift" and then solve completely using x ~ 0 assumption |
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Homework Problems 20.1 Given the following equilibria and initial concentrations, determine whether a Material Shift is necessary a. A + 2B ↔ C Kc = 2.5 x 10-6 initial 1.00 M 1.00 M 0.00 M b. A + 2B ↔ C Kc = 2.5 x 10-6 initial 0.00 M 0 .00 M 1.00 M c. A + 2B ↔ C Kc = 2.5 x 10-6 initial 0.00 M 1 .00 M 1.00 M d. A ↔ 2D + 3C Kc = 9.8 x 1010 initial 0.00 M 1 .00 M 1.00 M e. A ↔ 2D + 3C Kc = 9.8 x 1010 initial 1.00 M 0 .00 M 0.00 M 20.2 The following equilibria have very small Kc values and require Material Shifts to the left. What are the new concentrations after the M.S. is done? a. A + 2B ↔ C Kc = 2.5 x 10-6 initial 0.00 M 0 .00 M 3.00 M b. A ↔ 2B + C Kc = 2.5 x 10-6 initial 0.00 M 2.00 M 1.00 M c. A ↔ 2B + C Kc = 2.5 x 10-6 initial 0.00 M 2.00 M 0.50 M 3. 0.400 moles HCl, 0.400 moles of HI, 0.400 moles of Cl2 are placed in a 300.0 mL container. There is no I2 initially present. 2 HCl(g) + I2(g) ↔ 2 HI(g) + Cl2(g) Kc = 3.50 × 10-32 a. Perform the correct material shift. b. Construct an ICE table and Law of Mass Action c. Solve for "x" using the x~0 assumption. d. Check your equilibrium concentrations for correctness. e. Perform all 5% tests. Answers: Click and drag in the space below 20.1 a. Kc is small favoring reactants. Initially all material is on the reactant side. No Material Shift required and the x ~ 0 assumption is likely valid. b. Kc is small favoring reactants. Initially, all material is on the product side. At equilibrium most of C will have been turned into A and B Do a material shift before attempting the ICE problem. c. Kc is small favoring reactants. Initially ,there is both B and C present. At equilibrium most of C will have been turned into A and additional B. Do a material shift before attempting the ICE problem. d. Kc is large favoring products. Initially, all material is on the product side. No Material Shift is required and the x ~ 0 assumption is likely valid. e. Kc is large favoring products. Initially, all material is on the reactant side. Do a material shift before attempting the ICE problem. 20.2 a. After Material Shift concentrations are: [A] = 3.00 M [B] = 6.00 M [C] = 0.00M b. After Material Shift concentrations are: [A] = 1.00 M [B] = 0.00 M [C] = 0.00M c. After Material Shift concentrations are: [A] = 0.50 M [B] = 1.00 M [C] = 0.00M 20.3 x = 2.494438 x 10-16 [HCl]eq = 2.67 M [I2]eq = 0.667 M [HI]eq = 4.99 x 10-16 M [Cl2]eq = 0.667 M |
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