1. Chemical Equilibrium CHAPTER 15 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop

2. CHAPTER 15 Chemical Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 2 Learning Objectives:  Reversible Reactions and Equilibrium  Writing Equilibrium Expressions and the Equilibrium Constant (K)  Reaction Quotient (Q)  K c vs K p  ICE Tables  Quadratic Formula vs Simplifying Assumptions  LeChatelier’s Principle  van’t Hoff Equation

3. CHAPTER 15 Chemical Equilibrium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 3 Lecture Road Map: ① Dynamic Equilibrium ② Equilibrium Laws ③ Equilibrium Constant ④ Le Chatelier’s Principle ⑤ Calculating Equilibrium

4. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 4 Calculating Equilibrium CHAPTER 15 Chemical Equilibrium

5. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5 Calculations Overview For gaseous reactions, use either K P or K C For solution reactions, must use K C Either way, two basic categories of calculations 1.Calculate K from known equilibrium concentrations or partial pressures 2.Calculate one or more equilibrium concentrations or partial pressures using known K P or K C

6. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6 Calculations K c with Known Equilibrium Concentrations When all concentrations at equilibrium are known –Use mass action expression to relate concentrations to K C Two common types of calculations A.Given equilibrium concentrations, calculate K B.Given initial concentrations and one final concentration Calculate equilibrium concentration of all other species Then calculate K

7. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7 Calculations Ex. 3 N 2 O 4 (g) 2NO 2 (g) If you place 0.0350 mol N 2 O 4 in 1 L flask at equilibrium, what is K C? [N 2 O 4 ] eq = 0.0292 M [NO 2 ] eq = 0.0116 M K C = 4.61  10 –3 K c with Known Equilibrium Concentrations

8. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8 For the reaction: 2A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A = 2.0 M, B = 1.0 M and C = 3.0 M. What is the expected value of K c at this temperature? A. 14 B. 0.15 C. 1.5 D. 6.75 Group Problem

9. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 9 Calculations Ex. 4 2SO 2 (g) + O 2 (g) 2SO 3 (g) At 1000 K, 1.000 mol SO 2 and 1.000 mol O 2 are placed in a 1.000 L flask. At equilibrium 0.925 mol SO 3 has formed. Calculate K C for this reaction. First calculate concentrations of each –Initial –Equilibrium K c with Known Equilibrium Concentrations

10. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10 Calculations Example Continued Set up concentration table –Based on the following: Changes in concentration must be in same ratio as coefficients of balanced equation Set up table under balanced chemical equation –Initial concentrations Controlled by person running experiment –Changes in concentrations Controlled by stoichiometry of reaction –Equilibrium concentrations Equilibrium Concentration = Initial Concentration – Change in Concentration

11. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11 Calculations Example Continued 2SO 2 (g) +O2(g)O2(g) 2SO 3 (g) Initial Conc. (M)1.000 0.000 Changes in Conc. (M) Equilibrium Conc. (M) [SO 2 ] consumed = amount of SO 3 formed = [SO 3 ] at equilibrium = 0.925 M [O 2 ] consumed = ½ amount SO 3 formed = 0.925/2 = 0.462 M [SO 2 ] at equilibrium = 1.000 – 0.975 = 0.075 [O 2 ] at equilibrium = 1.00 – 0.462 = 0.538 M –0.925–0.462+0.925 0.075 0.538 0.925

12. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 12 Calculations Overview Finally calculate K C at 1000 K K c = 2.8 × 10 2 = 280

13. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 Calculations ICE Table Summary ICE tables used for most equilibrium calculations: 1.Equilibrium concentrations are only values used in mass action expression  Values in last row of table 2.Initial value in table must be in units of mol/L (M)  [X] initial = those present when reaction prepared  No reaction occurs until everything is mixed

14. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 14 Calculations ICE Table Summary ICE tables used for most equilibrium calculations: 1.Equilibrium concentrations are only values used in mass action expression  Values in last row of table 2.Initial value in table must be in units of mol/L (M)  [X] initial = those present when reaction prepared  No reaction occurs until everything is mixed 3.Changes in concentrations always occur in same ratio as coefficients in balanced equation 4.In “change” row be sure all [reactants] change in same directions and all [products] change in opposite direction.  If [reactant] initial = 0, its change must be an increase (+) because [reactant] final cannot be negative  If [reactants] decreases, all entries for reactants in change row should have minus sign and all entries for products should be positive

15. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 Calculations Calculate [X ] equilibrium from K c and [X ] initial When all concentrations but one are known –Use mass action expression to relate K c and known concentrations to obtain missing concentrations Ex. 5 CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) At 1500 °C, K c = 5.67. An equilibrium mixture of gases had the following concentrations: [CH 4 ] = 0.400 M and [H 2 ] = 0.800 M and [CO] = 0.300 M. What is [H 2 O] at equilibrium ?

16. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Calculations Ex. 5 CH 4 (g) + H 2 O(g) CO(g) + 3H 2 (g) K c = 5.67 [CH 4 ] = 0.400 M; [H 2 ] = 0.800 M; [CO] =0.300 M What is [H 2 O] at equilibrium? First, set up equilibrium Next, plug in equilibrium concentrations and K c [H 2 O] = 0.0678 M Calculate [X ] equilibrium from K c and [X ] initial

17. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 17 Calculations Calculating [X ] Equilibrium from K c When Initial Concentrations Are Given Write equilibrium law/mass action expression Set up Concentration table –Allow reaction to proceed as expected, using “x” to represent change in concentration Substitute equilibrium terms from table into mass action expression and solve

18. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 18 Calculations Calculate [X] equilibrium from [X] initial and K C Ex. 6 H 2 (g) + I 2 (g) 2HI(g) at 425 ˚C K C = 55.64 If one mole each of H 2 and I 2 are placed in a 0.500 L flask at 425 °C, what are the equilibrium concentrations of H 2, I 2 and HI? Step 1. Write Equilibrium Law

19. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 19 Calculations Step 2: Construct an ICE table Initial [H 2 ] = [I 2 ] = 1.00 mol/0.500 L =2.00 M Amt of H 2 consumed = Amt of I 2 consumed = x Amount of HI formed = 2x Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial2.00 0.000 Change Equilibrium – x +2x – x +2x2.00 – x Calculate [X] equilibrium from [X] initial and K C

20. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 20 Calculations Step 3. Solve for x Both sides are squared so we can take square root of both sides to simplify Calculate [X] equilibrium from [X] initial and K C

21. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21 Calculations Step 4. Equilibrium Concentrations [H 2 ] equil = [I 2 ] equil = 2.00 – 1.58 = 0.42 M [HI] equil = 2x = 2(1.58) = 3.16 Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial 2.00 0.00 Change Equilibrium – 1.58 +3.16 – 1.58 +3.160.42 Calculate [X] equilibrium from [X] initial and K C

22. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 22 Calculations Calculate [X] equilibrium from [X] initial and K C Ex. 7 H 2 (g) + I 2 (g) 2HI(g) at 425 ˚C K C = 55.64 If one mole each of H 2, I 2 and HI are placed in a 0.500 L flask at 425 ˚C, what are the equilibrium concentrations of H 2, I 2 and HI? Now have product as well as reactants initially Step 1. Write Equilibrium Law

23. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 Calculations Conc (M)H 2 (g) +I2(g)I2(g) 2HI (g) Initial2.00 Change Equil’m – x +2x – x 2.00 + 2x 2.00 – x Step 2. Concentration Table Calculate [X] equilibrium from [X] initial and K C

24. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 24 Calculations Calculate [X] equilibrium from [X] initial and K C Step 3. Solve for x  [H 2 ] equil = [I 2 ] equil = 2.00 – x = 2.00 – 1.37 = 0.63 M  [HI] equil = 2.00 + 2x = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M

25. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 25 N 2 (g) + O 2 (g) 2NO(g) K c = 0.0123 at 3900 ˚C If 0.25 moles of N 2 and O 2 are placed in a 250 mL container, what are the equilibrium concentrations of all species? A. 0.0526 M, 0.947 M, 0.105 M B. 0.947 M, 0.947 M, 0.105 M C. 0.947 M, 0.105 M, 0.0526 M D. 0.105 M, 0.105 M, 0.947 M Group Problem

26. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 26 Conc (M) N 2 (g) + O 2 (g) 2NO (g) Initial 1.00 1.00 0.00 Change – x – x + 2x Equil 1.00 – x 1.00 – x + 2x Group Problem

27. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 27 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Ex. 8 CH 3 CO 2 H(aq) + C 2 H 5 OH(aq) CH 3 CO 2 C 2 H 5 (aq) + H 2 O(l) acetic acidethanol ethyl acetate K C = 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of 0.810 M, is heated at 100 °C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium?

28. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 28 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 1. Write equilibrium law Need to find equilibrium values that satisfy this Step 2: Set up concentration table using “x” for unknown –Initial concentrations –Change in concentrations –Equilibrium concentrations

29. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 29 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 2 Concentration Table Amt of CH 3 CO 2 H consumed = Amt of C 2 H 5 OH consumed = – x Amt of CH 3 CO 2 C 2 H 5 formed = + x [CH 3 CO 2 H] eq and [C 2 H 5 OH ] = 0.810 – x [CH 3 CO 2 C 2 H 5 ] = x (M)CH 3 CO 2 H (aq) +C 2 H 5 OH (aq) CH 3 CO 2 C 2 H 5 (aq) + H 2 O ( l ) I 0.810 0.000 C E –x–x+x+x– x +x+x0.810 – x

30. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 30 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 3. Solve for x Rearranging gives Then put in form of quadratic equation ax 2 + bx + c = 0 Solve for the quadratic equation using

31. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 31 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 3. Solve for x This gives two roots: x = 10.6 and x = 0.064 Only x = 0.064 is possible –x = 10.6 is >> 0.810 initial concentrations –0.810 – 10.6 = negative concentration, which is impossible

32. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 32 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 4. Equilibrium Concentrations [CH 3 CO 2 C 2 H 5 ] equil = x = 0.064 M [CH 3 CO 2 H] equil = [C 2 H 5 OH] equil = 0.810 M – x = 0.810 M – 0.064 M = 0.746 M CH 3 CO 2 H (aq) +C 2 H 5 OH (aq) CH 3 CO 2 C 2 H 5 (aq) + H 2 O I 0.810 0.000 C E –0.064+0.064– 0.064 +0.0640.746

33. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 33 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Cubic When K C is very small Ex. 9 2H 2 O(g) 2H 2 (g) + O 2 (g) At 1000 °C, K C = 7.3  10 –18 If the initial H 2 O concentration is 0.100 M, what will the H 2 concentration be at equilibrium? Step 1. Write Equilibrium Law

34. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 34 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Cubic Step 2. Concentration Table Cubic equation – tough to solve Make approximation –K C very small, so x will be very small –Assume we can neglect x –Must prove valid later Conc (M )2H 2 O (g) 2H 2 (g) +O2(g)O2(g) Initial0.1000.00 Change Equil’m – 2x +x+x +2x +x+x 0.100 – 2x

35. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 35 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Cubic Step 3. Solve for x Assume (0.100 – 2x)  0.100 Now our equilibrium expression simplifies to Conc (M)2H 2 O (g) 2H 2 (g) + O 2 (g) Initial0.1000.00 Change Equil’m – 2x +x+x +2x +x+x 0.100 = 7.3 × 10 –20

36. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 36 Calculations Calculate [X] equilibrium from [X] initial and K C Example: Cubic Step 3. Solve for x Now take cube root x is very small 0.100 – 2(2.6  10 –7 ) = 0.09999948 Which rounds to 0.100 (3 decimal places) [H 2 ] = 2x = 2(2.6  10 –7 ) = 5.2  10 –7 M

37. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 37 Calculations Simplifications: When Can You Ignore x In Binomial (C i – x)? If equilibrium law gives very complicated mathematical problems and if K is small –Then the change (x term) will also be small and we can assume it can be ignored when added or subtracted from the initial concentration, C i. How do we check that the assumption is correct? –If the calculated x is so small it does not change the initial concentration (e.g. 0.10 M initial – 0.003 M x-calc = 0.10) –Or if the answer achieved by using the assumption differs from the true value by less than five percent. This often occurs when C i > 100 x K c

38. Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 38 For the reaction 2A(g) B(g) given that K p = 3.5 × 10 –16 at 25 ˚C, and we place 0.2 atm A into the container, what will be the pressure of B at equilibrium? 2A  B I 0.2 0 atm C –2x +x E 0.2 – 2x x ≈0.2 x = 1.4 × 10 –17 [B]= 1.4 × 10 –17 atm Proof: 0.2 - 1.4 × 10 –17 = 0.2 Group Problem