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Le Chatelier’s Principle

Le Chatelier’s Principle. Illustrating the effects of increasing the amount of Carbon Dioxide in the air . a A + b B c C + d D. Equilibrium Law and the Equilibrium Constant. We can describe a chemical system in equilibrium in 2 ways: Qualitatively  Le Chatelier’s Principle

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Le Chatelier’s Principle

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  1. Le Chatelier’s Principle • Illustrating the effects of increasing the amount of Carbon Dioxide in the air

  2. aA + bBcC + dD Equilibrium Law and the Equilibrium Constant • We can describe a chemical system in equilibrium in 2 ways: • Qualitatively Le Chatelier’s Principle • Quantitatively = MATHEMATICALLY • The “Law of Mass Action” and the equilibrium constant Keq • For the reaction • The ratio of products over reactants at equilibrium is constant…

  3. Equilibrium Law and the Equilibrium Constant aA + bBcC + dD [C]c[D]d CONSTANT K = [A]a[B]b • Definition of Equilibrium Law (The Law of Mass Action): • If the system is at equilibrium at a given temperature, then the ratio of the concentration of the product to the concentration of the reactant is a constant. • In other words, for the reaction: WHY?

  4. aA + bBcC + dD MATH • What do we know about a system at equilibrium? 1. RATEforward = RATEreverse • In other words, the speed of making products is equal to the speed of making reactants • What does the rate/speed of reaction depend on? 2.  RATEforward [A]a[B]b RATEforward= kf[A]a[B]b

  5. aA + bBcC + dD [C]c[D]d = Keq= kf [A]a[B]b kr MORE MATH 3.  RATEreverse [C]c[D]d RATEreverse= kr[C]c[D]d 4.  RATEforward = RATEreverse kf[A]a[B]b = kr[C]c[D]d

  6. [C]c[D]d [C]c[D]d [NH3]2 = Keq= kf [A]a[B]b [A]a[B]b [N2][H2]3 kr Equilibrium Law and the Equilibrium Constant • …means that if we know the equilibrium constant, we can figure out equilibrium concentrations and vice versa • Example 1: N2 (g) + 3H2(g) 2NH3(g) Keq= Keq=

  7. [C]c[D]d [CH3OH] [0.00261] [A]a[B]b [H2]2[CO] [0.250]2[0.105] Equilibrium Law and the Equilibrium Constant-Calculating Keq CO(g) + 2H2(g)  CH3OH(g) • Calculate Keq for the above reaction if the equilibrium concentration of the reactants and products respectively are [CH3OH]= 0.00261M, [CO]= 0.105 M and [H2] = 0.250 M Keq= Keq= Keq= Keq = 0.398

  8. [C]c[D]d [CH3OH] [x] [A]a[B]b [H2]2[CO] [0.100]2[0.210] Equilibrium Law and the Equilibrium Constant-Calculating Equilibrium Concentrations CO(g) + 2H2(g)  CH3OH(g) • What is the equilibrium concentration of methanol if Keq = 0.500 and the equilibrium concentration of CO is 0.210M and H2 is 0.100M? x = 0.00105M = 1.05 * 10-3 M Keq= Keq= 0.500 =

  9. Calculating Equilibrium Concentrations From Initial concentrations N2 (g) + 3H2(g) 2NH3(g) • A reaction vessel containing 1.00mol/L of N2(g) and 1.00mol/L H2(g) was heated to 5000C. The equilibrium concentration of NH3(g) was found to be 0.25mol/L • Determine the equilibrium concentrations of N2(g) and H2(g) • Determine the equilibrium constant, Keq for the formation of NH3(g) • Determine the equilibrium constant for the decomposition of NH3(g) • What is the relationship between the equilibrium constant for the forward and reverse reactions?

  10. Initial-Change-Equilibrium Calculations (I.C.E. Calculations) Question 3a-Determine the concentration for all species at equilibrium N2 (g) + 3H2(g) 2NH3(g) “The equilibrium concentration of NH3(g) was found to be 0.250mol/L” ∴ 0.250 = 2x x = 0.125 mol/L At Equilibrium: [N2(g)] = 1.00 – 0.125 = 0.875 mol/L [H2(g)] = 1.00 – 3(0.125) = 0.625 mol/L [NH3(g)] = 0 + 2(0.125) = 0.250 mol/L

  11. [0.250]2 [NH3]2 [N2][H2]3 [0.875][0.625]3 KeqCalculations Question 3b-Determine Keq for the formation of NH3(g) N2 (g) + 3H2(g) 2NH3(g) At Equilibrium: [N2(g)] = 0.875 mol/L [H2(g)] = 0.625 mol/L [NH3(g)] = 0.250 mol/L Keq = 0.293 Keq= Keq=

  12. [0.875][0.625]3 [0.250]2 Keq Calculations Question 3c-Determine Keq for the decomposition of NH3(g) N2 (g) + 3H2(g) 2NH3(g) At Equilibrium: [N2(g)] = 0.875 mol/L [H2(g)] = 0.625 mol/L [NH3(g)] = 0.250 mol/L [N2][H2]3 [NH3]2 Keq =3.42 Keq= Keq= Is there a faster way we could have done this question?

  13. [0.875][0.625]3 [0.250]2 [NH3]2 [0.875][0.625]3 [N2][H2]3 [0.250]2 Keq Calculations Question 3d-What is the relationship between Kf and Kr? N2 (g) + 3H2(g) 2NH3(g) [N2][H2]3 Forward Reverse [NH3]2 Keq= Keq= Keq= Keq= Keq = 0.293 Keq = 3.42

  14. [NH3]2 1 Keq (forward) [N2][H2]3 Keq Calculations • Because…. [N2][H2]3 [NH3]2 1 = Keq (reverse)=

  15. [C] [C]2 [C] [A][B]2 [A]2[B]4 [A][B]2 Characteristics of Keq-More Math • Look at the 2 reactions below…what happens to Keq when we double the coefficients of the equation? Rxn 1 A + 2B C Keq= 8 Rxn 2 2A + 4B 2C Keq = ? 2 = 82 Keq = 8 = Keq = ? = = = 64

  16. [C] [C]1/2 [C] [A]1/2[B] [A][B]2 [A][B]2 Characteristics of Keq-More Math = (Keq)3 • If doubling the coefficients changed Keq by a POWER OF 2 then… • How would Keq change if we tripled the coefficients? • How would Keq change if we HALVED the coefficients? Rxn 1 A + 2B C Keq= 8 Rxn 2 1/2A + B 1/2C Keq = ? 1/2 Keq = ? = = Keq = 8 = = 81/2 = √8

  17. [SO3] [SO3] [SO3] [O2]1/2[NO] [O2]1/2[NO] [O2]1/2[SO2] [O2]1/2[SO2] [NO2] [NO2] [SO2] Characteristics of Keq-More Math Look at the following two reactions: • SO2(g) + 1/2O2(g) SO3(g)      Keq1 = 2.2 • NO2(g) NO(g) + ½ O2(g)      Keq2 = 4.0 What is the equilibrium constant for the reaction • SO2(g) + NO2(g) SO3(g) + NO(g) Keq3 = ? [NO] * = Keq2 = 4.0 = Keq1 = 2.2 = Keq3 = ? = [NO2]

  18. [SO3] [SO3] [O2]1/2[NO] [O2]1/2[SO2] [NO2] [SO2] Characteristics of Keq-More Math Keq1 Keq2 [NO] = [NO2] = Keq1 *Keq2 = 2.2 *4.0 = 8.8 Concentration versus Pressure We can write Keqin terms of concentrations (mol/L) or in terms of pressure, Kp if all species are in the gas phase Kp= Keq(RT)∆n Dn = moles of gaseous products – moles of gaseous reactants * Keq3 = ? =

  19. Summary of Mathematics of Keq • The Keq for forward and reverse reactions are reciprocals (the inverse) of each other • If you multiply the stoichiometry of a reaction by n, Keq is changed by the power of n • If you add chemical equations together, you multiply their Keq’s together • Kp= Keq(RT)∆n (Kp is for gases only) • If K>>1, then the concentration of products is large compared to reactants (reaction heads towards completion) • If K<<1, then the concentration of reactants is large compared to products (very small amount of product formed)

  20. Characteristics of Keq- No More Math • The equilibrium constant is a dimensionless quantity (no units). • In quoting a value for the equilibrium constant, you must specify the balanced equation and the temperature. • The concentrations of pure solids, pure liquids and solvents (like H2O) do not appear in the equilibrium constant expressions... • Because the concentrations of pure solids, liquids, and solvents are constant • Concentrations are expressed as mol/L . For gases, concentrations can be expressed as either mol/L or atm

  21. Keq • Playing with Keq-Computer Animation • For homework, please follow the instructions on the site..for each run, calculate the concentrations of A and B using the Keq value given in the plot

  22. aA+ bBcC + dD [C]c[D]d [A]a[B]b Reaction Quotients (Q) • Keq is the constant for a reaction AT EQUILIBRIUM, but how do we mathematically determine if a reaction is at equilibrium or which way it must “shift” to get there? • We use the “reaction quotient” Q • For Q=

  23. aA+ bBcC + dD [C]c[D]d [A]a[B]b Reaction Quotients (Q) • When? • When we are unsure where a system is in reference to its equilibrium • When there are both reactants and products present but the system is not specified to be at equilibrium • How? • Step 1: Input the given initial concentrations of products and reactants into the equation for Q • Step 2: Compare the resulting value of Q to the equilibrium constant for that reaction at that temperature • Step 3: Determine which was the reaction must “shift” to reach equilibrium Q=

  24. aA+ bBcC + dD [C]c[D]d [A]a[B]b Reaction Quotients (Q) • 3 Possibilities for Q in reference to Keq • Q > Keq • Q <Keq • Q =Keq • Q >Keq • Means that there is a greater number of products present than when the system is at equilibrium • The equilibrium must “shift left” and turn product into reactant to reach equilibrium Q=

  25. aA+ bBcC + dD [C]c[D]d [A]a[B]b Reaction Quotients (Q) • Q <Keq • Means that there fewer products present than when the system is at equilibrium • The equilibrium must “shift right” and turn reactant into product to reach equilibrium • Q =Keq • The system is at equilibrium Q=

  26. [C]c[D]d [A]a[B]b Reaction Quotients (Q) • Q <Keq • Q =Keq • Q >Keq Q=

  27. CO(g) + H2O(g) CO2(g) + H2(g) [CO2][H2] [4.000][4.000] [CO][H2O] [4.000][4.000] Reaction Quotients (Q) • Example Using Q: • Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 9000C, Keq is 4.200. Calculate the concentrations of all species at equilibrium if 4.000 mol of each species are initially placed in a 1.000L closed container Q= Q= Q = 1

  28. CO(g) + H2O(g) CO2(g) + H2(g) [CO2][H2] [CO][H2O] Reaction Quotients (Q) Q = 1 Q < Keq Rxn Q=

  29. CO(g) + H2O(g) CO2(g) + H2(g) [4.000+x][4.000+x] [4.000+x]2 [CO2][H2] [4.000-x][4.000-x] [4.000-x]2 [CO][H2O] Reaction Quotients (Q) x = 1.377 4.200 = 4.200 = Keq=

  30. CO(g) + H2O(g) CO2(g) + H2(g) Reaction Quotients (Q) x = 1.377 [CO(g)] =4.000-x [CO(g)] = 2.623 mol/L [H2O(g)] = 4.000-x [H2O(g)] = 2.623 mol/L [CO2(g)] = 4.000+x [CO2(g)] = 5.377 mol/L [H2(g)] = 4.000+x [H2(g)] = 5.377 mol/L Last step is taking the determined equilibrium concentrations and subbing them back into the expression for Q  should result in Q = Keq

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