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

Transcription

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

2 CHAPTER 15 Chemical Equilibrium Learning Objectives: q Reversible Reactions and Equilibrium q Writing Equilibrium Expressions and the Equilibrium Constant (K) q Reaction Quotient (Q) q K c vs K p q ICE Tables q Quadratic Formula vs Simplifying Assumptions q LeChatelier s Principle q van t Hoff Equation

3 CHAPTER 15 Chemical Equilibrium Lecture Road Map: Dynamic Equilibrium Equilibrium Laws Equilibrium Constant Le Chatelier s Principle Calculating Equilibrium 3

4 CHAPTER 15 Chemical Equilibrium Calculating Equilibrium 4

5 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. Calculate one or more equilibrium concentrations or partial pressures using known K P or K C 5

6 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 6

7 K c with Known Equilibrium Concentrations Ex. 3 N O 4 (g) NO (g) If you place mol N O 4 in 1 L flask at equilibrium, what is K C? [N O 4 ] eq 0.09 M [NO ] eq M K c [NO ] [N O 4 ] K c [0.0116] [0.09] K C

8 Group Problem For the reaction: A(aq) + B(aq) 3C(aq) the equilibrium concentrations are: A.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 C. 1.5 D K c [C ]3 [A] [B] K c [3.0] 3 [.0] [1.0] 8

9 K c with Known Equilibrium Concentrations Ex. 4 SO (g) + O (g) SO 3 (g) At 1000 K, mol SO and mol O are placed in a L flask. At equilibrium 0.95 mol SO 3 has formed. Calculate K C for this reaction. First calculate concentrations of each Initial Equilibrium [SO ] [O ] 1.00 mol 1.00 L 1.00 M [SO 3 ] 0.95 mol 1.00 L 0.95 M 9

10 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 ConcentraKon IniKal ConcentraKon Change in ConcentraKon 10

11 Example Continued SO (g) + O (g) SO 3 (g) Initial Conc. (M) Changes in Conc. (M) Equilibrium Conc. (M) [SO ] consumed amount of SO 3 formed [SO 3 ] at equilibrium 0.95 M [O ] consumed ½ amount SO 3 formed 0.95/ 0.46 M [SO ] at equilibrium [O ] at equilibrium M 11

12 Overview Finally calculate K C at 1000 K K c [SO [SO ] 3 ] [O ] K c [0.95] [0.075] [0.538] K c

13 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. 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 13

14 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. 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 14

15 Calculate [X ] equilibrium from K c and [X ] inikal 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 O(g) CO(g) + 3H (g) At 1500 C, K c An equilibrium mixture of gases had the following concentrations: [CH 4 ] M and [H ] M and [CO] M. What is [H O] at equilibrium? 15

16 Calculate [X ] equilibrium from K c and [X ] inikal Ex. 5 CH 4 (g) + H O(g) CO(g) + 3H (g) K c 5.67 [CH 4 ] M; [H ] M; [CO] M What is [H O] at equilibrium? First, set up equilibrium K c [CO][H ]3 [CH 4 ][H O] [H O] [CO][H ]3 [CH 4 ]K c Next, plug in equilibrium concentrations and K c [0.300][0.800] [H O] [0.400](5.67) [H O] M

17 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 17

18 Calculate [X] equilibrium from [X] initial and K C Ex. 6 H (g) + I (g) K C HI(g) at 45 C If one mole each of H and I are placed in a L flask at 45 C, what are the equilibrium concentrations of H, I and HI? Step 1. Write Equilibrium Law K c [HI] [H ][I ]

19 Calculate [X] equilibrium from [X] initial and K C Step : Construct an ICE table Conc (M) H (g) + I (g) HI (g) Initial Change Equilibrium x x +x.00 x.00 x Initial [H ] [I ] 1.00 mol/0.500 L.00 M Amt of H consumed Amt of I consumed x Amount of HI formed x +x (x ) (.00! x )(.00! x ) (x ) (.00! x ) 19

20 Calculate [X] equilibrium from [X] initial and K C Step 3. Solve for x Both sides are squared so we can take square root of both sides to simplify K x (.00 x ) (x) (.00! x) 7.459(.00 x ) x x x x x 0

21 Calculate [X] equilibrium from [X] initial and K C Step 4. Equilibrium Concentrations Conc (M) H (g) + I (g) HI (g) Initial Change Equilibrium [H ] equil [I ] equil M [HI] equil x (1.58)

22 Calculate [X] equilibrium from [X] initial and K C Ex. 7 H (g) + I (g) K C HI(g) at 45 C If one mole each of H, I and HI are placed in a L flask at 45 C, what are the equilibrium concentrations of H, I and HI? Now have product as well as reactants initially Step 1. Write Equilibrium Law K c [HI] [H ][I ] 55.64

23 Calculate [X] equilibrium from [X] initial and K C Step. Concentration Table Conc (M) H (g) + I (g) HI (g) Initial Change Equil m x x +x.00 x.00 x.00 + x (.00 + x ) (.00 x )(.00 x ) (.00 + x ) (.00 x ) K (.00 (.00 + x ) x ) 3

24 Calculate [X] equilibrium from [X] initial and K C Step 3. Solve for x.00 + x (.00 x ) 7.459(.00 ).00 + x x x.00 + x x x 1.37 [H ] equil [I ] equil.00 x M [HI] equil.00 + x.00 + (1.37) M 4

25 Group Problem N (g) + O (g) K c at 3900 C NO(g) If 0.5 moles of N and O are placed in a 50 ml container, what are the equilibrium concentrations of all species? A M, M, M B M, M, M C M, M, M D M, M, M 5

26 Group Problem Conc (M) N (g) + O (g) NO (g) Initial Change x x + x Equil 1.00 x 1.00 x + x [N ] [O ] mol 0.50 L 1.00 M (x ) x (1! x ) 1! x x M [NO] x M 6

27 Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Ex. 8 CH 3 CO H(aq) + C H 5 OH(aq) CH 3 CO C H 5 (aq) + H O(l) acetic acid ethanol ethyl acetate K C 0.11 An aqueous solution of ethanol and acetic acid, each with initial concentration of M, is heated at 100 C. What are the concentrations of acetic acid, ethanol and ethyl acetate at equilibrium? 7

28 Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 1. Write equilibrium law K c [C [CH COC H OH][CH 5 H5 ] CO 3 3 H] 0.11 Need to find equilibrium values that satisfy this Step : Set up concentration table using x for unknown Initial concentrations Change in concentrations Equilibrium concentrations 8

29 Step Concentration Table (M) CH 3 CO H(aq) + C H 5 OH(aq) Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation I C E Amt of CH 3 CO H consumed Amt of C H 5 OH consumed x Amt of CH 3 CO C H 5 formed + x [CH 3 CO H] eq and [C H 5 OH ] x [CH 3 CO C H 5 ] x x x +x x x 0.11 CH 3 CO C H 5 (aq) + H O(l) +x x (0.810! x )(0.810! x ) 9

30 Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 3. Solve for x Rearranging gives 0.11 ( x + x ) x Then put in form of quadratic equation ax + bx + c x 0.178x x 1.178x x 0 0 Solve for the quadratic equation using x b ± b 4ac a 30

31 Step 3. Solve for x Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation x ( 1.178) ± ( 1.178) (0.11) 4(0.11)(0.0717) x ± (1.388) (0.03) ± This gives two roots: x 10.6 and x Only x is possible x 10.6 is >> initial concentrations negative concentration, which is impossible 31

32 Calculate [X] equilibrium from [X] initial and K C Example: Quadratic Equation Step 4. Equilibrium Concentrations CH 3 CO H(aq) + C H 5 OH(aq) CH 3 CO C H 5 (aq) + H O I C E [CH 3 CO C H 5 ] equil x M [CH 3 CO H] equil [C H 5 OH] equil M x M M M 3

33 Calculate [X] equilibrium from [X] initial and K C Example: Cubic When K C is very small Ex. 9 H O(g) H (g) + O (g) At 1000 C, K C If the initial H O concentration is M, what will the H concentration be at equilibrium? Step 1. Write Equilibrium Law K c [H ] [O] [H O] 18 33

34 Step. Concentration Table Calculate [X] equilibrium from [X] initial and K C Example: Cubic Conc (M ) H O(g) (0.100 x ) 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 H (g) + O (g) Initial Change Equil m x +x +x x 18 (x ) x +x 4x ( x x ) 34

35 Calculate [X] equilibrium from [X] initial and K C Example: Cubic Step 3. Solve for x Assume (0.100 x) Conc (M) H O (g) H (g) + O (g) Initial Change Equil m x +x +x x Now our equilibrium expression simplifies to 3 18 (x ) x 4x (0.100) x 0.010( ) x 35

36 Calculate [X] equilibrium from [X] initial and K C Example: Cubic Step 3. Solve for x x 3 Now take cube root x x is very small ( ) Which rounds to (3 decimal places) [H ] x ( ) M 7 36

37 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 M initial 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 37

38 Group Problem For the reaction A(g) B(g) given that K p at 5 C, and we place 0. atm A into the container, what will be the pressure of B at equilibrium? Q K A D B P P B P A I 0. 0 atm C x +x E 0. x x x [B] atm 3.5!10-16 Proof: x (0.) 38