Download presentation

Presentation is loading. Please wait.

Published bySean Schultz Modified over 4 years ago

1
Equilibrium

2
Reactions are reversible Z A + B C + D ( forward) Z C + D A + B (reverse) Z Initially there is only A and B so only the forward reaction is possible Z As C and D build up, the reverse reaction speeds up while the forward reaction slows down. Z Eventually the rates are equal

3
Reaction Rate Time Forward Reaction Reverse reaction Equilibrium

4
What is equal at Equilibrium? Z Rates are equal Z Concentrations are not. Z Rates are determined by concentrations and activation energy. Z The concentrations do not change at equilibrium. Z or if the reaction is verrrry slooooow.

5
Law of Mass Action Z For any reaction Z jA + kB lC + mD Z K = [C] l [D] m PRODUCTS power [A] j [B] k REACTANTS power Z K is called the equilibrium constant. Z is how we indicate a reversible reaction

6
Playing with K Z If we write the reaction in reverse. Z lC + mD jA + kB Z Then the new equilibrium constant is Z K = [A] j [B] k = 1/K [C] l [D] m

7
Playing with K Z If we multiply the equation by a constant Z njA + nkB nlC + nmD Z Then the equilibrium constant is Z K = [C] nl [D] nm ( [C] l [D] m ) n = K n [A] nj [B] nk = ([A] j [B] k ) n

8
The units for K Z Are determined by the various powers and units of concentrations. Z They depend on the reaction.

9
K is CONSTANT Z At any temperature. Z Temperature affects rate. Z The equilibrium concentrations dont have to be the same, only K. Z Equilibrium position is a set of concentrations at equilibrium. Z There are an unlimited number.

10
Equilibrium Constant One for each Temperature

11
Calculate K Z N 2 + 3H 2 2NH 3 Z Initial At Equilibrium Z [N 2 ] 0 =1.000 M [N 2 ] = 0.921M Z [H 2 ] 0 =1.000 M [H 2 ] = 0.763M Z [NH 3 ] 0 =0 M [NH 3 ] = 0.157M

12
Calculate K Z N 2 + 3H 2 2NH 3 Z Initial At Equilibrium Z [N 2 ] 0 = 0 M [N 2 ] = 0.399 M Z [H 2 ] 0 = 0 M [H 2 ] = 1.197 M Z [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M Z K is the same no matter what the amount of starting materials

13
Equilibrium and Pressure Z Some reactions are gaseous Z PV = nRT Z P = (n/V)RT Z P = CRT Z C is a concentration in moles/Liter Z C = P/RT

14
Equilibrium and Pressure Z 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z Kp = (P SO3 ) 2 (P SO2 ) 2 (P O2 ) Z K = [SO 3 ] 2 [SO 2 ] 2 [O 2 ]

15
Equilibrium and Pressure Z K = (P SO3 /RT) 2 (P SO2 /RT) 2 (P O2 /RT) Z K = (P SO3 ) 2 (1/RT) 2 (P SO2 ) 2 (P O2 ) (1/RT) 3 Z K = Kp (1/RT) 2 = Kp RT (1/RT) 3

16
General Equation Z jA + kB lC + mD Z K p = (P C ) l (P D ) m = (C C xRT ) l (C D xRT ) m (P A ) j (P B ) k (C A xRT ) j (C B xRT ) k Z K p = (C C ) l (C D ) m x(RT) l+m (C A ) j (C B ) k x(RT) j+k K p = K (RT) (l+m)-(j+k) = K (RT) n n= (l+m)-(j+k) = Change in moles of gas

17
Homogeneous Equilibria Z So far every example dealt with reactants and products where all were in the same phase. Z We can use K in terms of either concentration or pressure. Z Units depend on reaction.

18
Heterogeneous Equilibria Z If the reaction involves pure solids or pure liquids the concentration of the solid or the liquid doesnt change. Z As long as they are not used up we can leave them out of the equilibrium expression. Z For example

19
For Example Z H 2 (g) + I 2 (s) 2HI(g) Z K = [HI] 2 [ H 2 ][ I 2 ] Z But the concentration of I 2 does not change. Z K [ I 2 ] = [HI] 2 = K [ H 2 ]

20
Write the equilibrium constant for the heterogeneous reaction

21
The Reaction Quotient Z Tells you the direction the reaction will go to reach equilibrium Z Calculated the same as the equilibrium constant, but for a system not at equilibrium Z Q = [Products] coefficient [Reactants] coefficient Z Compare value to equilibrium constant

22
What Q tells us Z If Q

23
Example Z for the reaction Z 2NOCl(g) 2NO(g) + Cl 2 (g) Z K = 1.55 x 10 -5 M at 35ºC Z In an experiment 0.10 mol NOCl, 0.0010 mol NO(g) and 0.00010 mol Cl 2 are mixed in 2.0 L flask. Z Which direction will the reaction proceed to reach equilibrium?

24
Solving Equilibrium Problems Z Given the starting concentrations and one equilibrium concentration. Z Use stoichiometry to figure out other concentrations and K. Z Learn to create a table of initial and final conditions.

25
Z Consider the following reaction at 600ºC Z 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z In a certain experiment 2.00 mol of SO 2, 1.50 mol of O 2 and 3.00 mol of SO 3 were placed in a 1.00 L flask. At equilibrium 3.50 mol of SO 3 were found to be present. Calculate Z The equilibrium concentrations of O 2 and SO 2, K and K P

26
Z Consider the same reaction at 600ºC 2SO 2 (g) + O 2 (g) 2SO 3 (g) Z In a different experiment.500 mol SO 2 and.350 mol SO 3 were placed in a 1.000 L container. When the system reaches equilibrium 0.045 mol of O 2 are present. Z Calculate the final concentrations of SO 2 and SO 3 and K

28
Solving Equilibrium Problems Type 1

29
What if youre not given equilibrium concentration? Z The size of K will determine what approach to take. Z First lets look at the case of a LARGE value of K ( >100). Z Allows us to make simplifying assumptions.

30
Example Z H 2 (g) + I 2 (g) 2HI(g) Z K = 7.1 x 10 2 at 25ºC Z Calculate the equilibrium concentrations if a 5.00 L container initially contains 15.8 g of H 2 294 g I 2. Z [ H 2 ] 0 = (15.8g/2.02)/5.00 L = 1.56 M Z [I 2 ] 0 = (294g/253.8)/5.00L = 0.232 M Z [HI] 0 = 0

31
Z Q= 0

32
Z Choose X so it is small. Z For I 2 the change in X must be X-.232 M Z Final must = initial + change H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change final X

33
Z Using to stoichiometry we can find Z Change in H 2 = X-0.232 M Z Change in HI = -twice change in H 2 Z Change in HI = 0.464-2X H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M final X

34
Z Now we can determine the final concentrations by adding. H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M X-0.232 M 0.464-2X final X

35
Z Now plug these values into the equilibrium expression Z K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M X-0.232 M 0.464-2X final 1.328+X X 0.464-2X

36
Why we chose X Z K = (0.464-2X) 2 = 7.1 x 10 2 (1.328+X)(X) Z Since X is going to be small, we can ignore it in relation to 0.464 and 1.328 Z So we can rewrite the equation Z 7.1 x 10 2 = (0.464) 2 (1.328)(X) Z Makes the algebra easy

37
Z When we solve for X we get 2.3 x 10 -4 Z So we can find the other concentrations Z I 2 = 2.3 x 10 -4 M Z H 2 = 1.328 M Z HI = 0.464 M H 2 (g) I 2 (g) 2 HI(g) initial 1.56 M 0.232 M 0 M change X-0.232 M X-0.232 M 0.464-2X final 1.328+X X 0.464-2X

38
Checking the assumption Z The rule of thumb is that if the value of X is less than 5% of all the smallest concentrations, our assumption was valid. Z If not we would have had to use the quadratic equation Z More on this later. Z Our assumption was valid.

39
Practice Z For the reaction Cl 2 + O 2 2ClO(g) K = 156 Z In an experiment 0.100 mol ClO, 1.00 mol O 2 and 0.0100 mol Cl 2 are mixed in a 4.00 L flask. Z If the reaction is not at equilibrium, which way will it shift? Z Calculate the equilibrium concentrations.

40
At an elevated temperature, the reaction: has a value of K eq = 944. If 0.234 mol IBr is placed in a 1.00 L. flask and allowed to reach equilibrium, what is the equilibrium concentration in M. of I 2 ?

41
Type 2 Problems with small K K<.01

42
Process is the same Z Set up table of initial, change, and final concentrations. Z Choose X to be small. Z For this case it will be a product. Z For a small K the product concentration is small.

43
For example Z For the reaction 2NOCl 2NO +Cl 2 Z K= 1.6 x 10 -5 Z If 1.20 mol NOCl, 0.45 mol of NO and 0.87 mol Cl 2 are mixed in a 1 L container Z What are the equilibrium concentrations Z Q = [NO] 2 [Cl 2 ] = (0.45) 2 (0.87) = 0.15 M [NOCl] 2 (1.20) 2

44
Z Choose X to be small Z NO will be LR Z Choose NO to be X 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change Final

45
Z Figure out change in NO Z Change = final - initial Z change = X-0.45 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change Final X

46
Z Now figure out the other changes Z Use stoichiometry Z Change in Cl 2 is 1/2 change in NO Z Change in NOCl is - change in NO 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change X-.45 Final X

47
l Now we can determine final concentrations l Add 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final X

48
l Now we can write equilibrium constant l K = (X) 2 (0.5X+0.645) (1.65-X) 2 l Now we can test our assumption X is small ignore it in + and - 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final 1.65-X X 0.5 X +0.645

49
l K = (X) 2 (0.645) = 1.6 x 10 -5 (1.65) 2 l X= 8.2 x 10 -3 l Figure out final concentrations 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final 1.65-X X 0.5 X +0.645

50
l [NOCl] = 1.64 l [Cl 2 ] = 0.649 l Check assumptions l.0082/.649 = 1.2 % OKAY!!! 2NOCl 2NO + Cl 2 Initial1.20 0.45 0.87 Change 0.45-X X-.45 0.5X -.225 Final 1.65-X X 0.5 X +0.645

51
Practice Problem Z For the reaction 2ClO(g) Cl 2 (g) + O 2 (g) Z K = 6.4 x 10 -3 Z In an experiment 0.100 mol ClO(g), 1.00 mol O 2 and 1.00 x 10 -2 mol Cl 2 are mixed in a 4.00 L container. Z What are the equilibrium concentrations?

52
Type 3 Mid-range Ks.01

53
No Simplification Z Choose X to be small. Z Cant simplify so we will have to solve the quadratic (we hope) Z H 2 (g) + I 2 (g) 2HI(g) K=38.6 Z What is the equilibrium concentrations if 1.800 mol H 2, 1.600 mol I 2 and 2.600 mol HI are mixed in a 2.000 L container?

54
Problems Involving Pressure Z Solved exactly the same, with same rules for choosing X depending on K P Z For the reaction N 2 O 4 (g) 2NO 2 (g) K P =.131 atm. What are the equilibrium pressures if a flask initially contains 1.000 atm N 2 O 4 ?

55
Le Châteliers Principle Z If a stress is applied to a system at equilibrium, the position of the equilibrium will shift to reduce the stress. Z 3 Types of stress a Concentration a Pressure a Temperature

56
Change amounts of reactants and/or products Z Adding product makes Q>K Z Removing reactant makes Q>K Z Adding reactant makes Q

57
Change Pressure Z By changing volume Z System will move in the direction that has the least moles of gas. Z Because partial pressures (and concentrations) change, a new equilibrium must be reached. Z System tries to minimize the moles of gas if volume is reduced Z And visa versa

58
Change in Pressure Z By adding an inert gas Z Partial pressures of reactants and product are not changed Z No effect on equilibrium position

59
Change in Temperature Z Affects the rates of both the forward and reverse reactions. Z Doesnt just change the equilibrium position, changes the equilibrium constant. Z The direction of the shift depends on whether it is exo- or endothermic

60
Exothermic DH<0 Z Releases heat Z Think of heat as a product Z Raising temperature push toward reactants. Z Shifts to left.

61
Endothermic DH>0 Z Produces heat Z Think of heat as a reactant Z Raising temperature push toward products. Z Shifts to right.

Similar presentations

OK

TOPIC A: EQUILIBRIUM Equilibrium, Le Chatelier’s Principle, Acid- Base Equilibrium, Ksp, pH.

TOPIC A: EQUILIBRIUM Equilibrium, Le Chatelier’s Principle, Acid- Base Equilibrium, Ksp, pH.

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on election in india 2012 Ppt on endangered species tiger Ppt on microsoft excel tutorial Ppt on isobars and isotopes of oxygen Ppt on testing of circuit breaker Ppt on review of literature sample Ppt on different occupations for kids Ppt on california academy of sciences Authority of the bible ppt on how to treat Ppt on job satisfaction