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13.1 Equilibrium Conditions When a system is at equilibrium it may appear that everything has stopped; however, this is NOT the case. Think of chemical.

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Presentation on theme: "13.1 Equilibrium Conditions When a system is at equilibrium it may appear that everything has stopped; however, this is NOT the case. Think of chemical."— Presentation transcript:

1 13.1 Equilibrium Conditions When a system is at equilibrium it may appear that everything has stopped; however, this is NOT the case. Think of chemical equilibrium like the cars on the Golden Gate Bridge and that the traffic flow in both directions is the same. No net change

2 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.

3 What is equilibrium? Equal rates Concentrations are not Rates are determined by concentrations and activation energy Concentrations do not change at equilibrium or if the reaction is verrrrrry slooooooow.

4

5 13.2 The Equilibrium Constant For any reaction Where K is the equilibrium constant A,B,C & D are the concentrations of the “chemical equilibrium j,k,l & m are the coefficients in the balanced equation

6 Try this: Write the equilibrium expression for 4NH 3 + 7O 2  4NO 2 + 6H 2 O

7 And if I knew… …the equilibrium concentrations for each of the reaction components, I could calculate K. Try this reaction: N 2 + 3H 2 ↔ 2NH 3 given: [NH 3 ]=3.1 x mol/L [N 2 ]=8.5 x mol/L [H 2 ]=3.1 x mol/L

8 What would K be for the reverse reaction? 2NH 3 ↔ N 2 + 3H 2 K’ = 2.6x10 -5 How about if only 1mol of NH 3 was produced? N 2 + H 2 ↔ 1NH 3 K’’=1.9 x 10 2

9 Big conclusions!! By comparing K for the forward reaction to K for the reverse reaction, notice that And if I multiply (or divide) my reaction by a factor then where n is the factor

10 K is constant…??? The equilibrium constant K always has the same value at a given temperature. A change in the temperature will change the rate and thus a new K. A set of concentrations at equilibrium is called an equilibrium position. There are an unlimited (infinite) number of equilibrium positions.

11 13.3 Equilibrium Expressions Involving Pressures So far we’ve been talking about equilibrium in terms of concentration… But gases can also be described by pressures as well. PV = nRT Or Where C is molar concentration of the gas

12 For: N 2 + 3H 2 ↔ 2NH 3

13 The relationship The relationship between K and K p comes from the fact that for an ideal gas C = P/RT. But for the general reaction the relationship between K and K p is Where ∆n is the sum of the coefficients of the gaseous products MINUS the sum of the coefficients of the gaseous reactants. ∆n = change in moles of gas

14 13.4 Heterogeneous Equilibria Remember that homogeneous equilibria is when all the products and reactants are gases 2SO 2 (g) + O 2 (g) ↔ 2SO 3 (g)

15 Let’s look at a problem with homogeneous equilibria The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl 2 (g) at 74 0 C are [CO] = M, [Cl 2 ] = M, and [COCl 2 ] = 0.14 M. Calculate the equilibrium constants K c and K p. CO (g) + Cl 2 (g) ↔ COCl 2 (g) Kc =Kc = [COCl 2 ] [CO][Cl 2 ] = x = 220 K p = K c (RT)  n K p = 220 x ( x 347) -1 = 7.7

16 Heterogeneous equilibria Heterogeneous equilibria is when you have reaction components that may not be in the gaseous state. CaCO 3 (s) ↔CaO(s) + CO 2 (g)

17 17 P CO 2 = K p CaCO 3 (s) CaO (s) + CO 2 (g) P CO 2 does not depend on the amount of CaCO 3 or CaO

18 More importantly remember… …that the concentrations of pure solids and liquid can not change therefore they are not included in the expression for the equilibrium constant CaCO 3 (s) CaO (s) + CO 2 (g) [CaCO 3 ] = constant [CaO] = constant K p = P CO 2 [CaO][CO 2 ] [CaCO 3 ] K c = [CO 2 ] K c =

19 Write the equilibrium constant for the heterogeneous reaction

20 Try this one. Consider the following equilibrium at 295 K: The partial pressure of each gas is atm. Calculate K p and K c for the reaction? Hint: this time start with K p and then find K c NH 4 HS (s) ↔ NH 3 (g) + H 2 S (g) K p = P NH 3 H2SH2S P = x = K c = x ( x 295) -2 = 1.20 x 10 -4

21 13.5 Applications of the Equilibrium Constant Knowing the equilibrium constant for a reaction allows us to predict several important features of the reaction: – Tendency of the reaction to occur – If concentrations given represent equilibrium – Equilibrium position from the initial concentrations

22 Important feature 1:Tendency of a reaction to occur The tendency of a reaction to occur can be indicated by the magnitude of K K >> 1 K << 1 Lie to the rightFavor products Lie to the leftFavor reactants Reaction goes to completion Reaction does not occur to any significant extent

23 Remember The size of K and the time it takes for the reaction to reach equilibrium are NOT directly related. The time it takes to reach equilibrium depends on the reaction rate and therefore ultimately the activation energy

24 Reaction Quotient - Q Tells us the direction the reaction needs to go in order to reach equilibrium It is calculated the same as the equilibrium concentration; however, it uses initial concentrations rather than equilibrium concentrations.

25 Compare Q to K c Q c < K c system proceeds from left to right to reach equilibrium Not enough products Forward reaction occurs Q c = K c the system is at equilibrium No net change Q c > K c system proceeds from right to left to reach equilibrium Too many products Reverse reaction occurs

26 Try this… N 2 (g)+ 3H 2 (g)↔ 2NH 3 (g) At the start of a reaction, there are mol N 2, 3.21x10 -2 mol H 2, and 6.42x10 -4 mol NH 3 in a 3.50 L reaction vessel at 375 o C. If the equilibrium constant (K c ) for the reaction is 1.2 at this temperature, decide whether the system is at equilibrium. IF it is not, predict which way the net reaction will proceed.

27 How did you do? First: find the molarity of each gas. Then find Q Not at equilibrium Shift left to right

28 Important feature 2: given the equilibrium constant and initial concentrations, find equilibrium concentrations/pressures of products & reactants Let’s do a few problems Gaseous N 2 O 4 was placed in a flask and allowed to reach a temperature where K p = At equilibrium, the pressure of N 2 O 4 was found to be 2.71atm. Calculate the equilibrium pressure of NO 2. N 2 O 4 (g)↔NO 2 (g)

29 Here’s how First: Given K p and P N2O4 = 2.71atm

30 Here’s another type of problem At a certain temperature a 1.00L flask initially contained 0.298M PCl 3 and 8.7x10 -3 M PCl 5. After the system had reached equilibrium, 2.00x10 -3 M Cl 2 was found in the flask. Gaseous PCl 5 decomposes according to the following reaction: PCl 5 (g) ↔ PCl 3 (g) + Cl 2 (g) Calculate the equilibrium concentrations for all species and K c

31 The steps to calculating equilibrium concentrations – ICE diagrams 1.Express the equilibrium concentrations of all species in terms of the initial concentrations and a single unknown x, which represents the change in concentration. 2.Write the equilibrium constant expression in terms of the equilibrium concentrations. Knowing the value of the equilibrium constant, solve for x. 3.Having solved for x, calculate the equilibrium concentrations of all species.

32 Let’s set one up ConcentrationPCl 5 ↔PCl 3 + Cl 2 Initial8.7x Change-2x x10 -3 Equilibrium6.7x x10 -3 Since the concentration of Cl 2 we know has changed by 2x10 -3, the balanced reaction tells us that since it is a reaction that the reactants decompose by 2x10 -3 and the products concentration increases by 2x10 -3

33 Once we know the equilibrium concentrations… …we can find the equilibrium constant.

34 Let’s try another using an ICE diagram There are some instances when you may not be given ANY equilibrium concentrations. Here’s how to do those. Let’s get started!!

35 H 2 (g) + I 2 (g) ↔ 2HI (g) A mixture of M H 2, M I 2 and M HI was placed in a 1.00L stainless-steel flask at 430 o C. The equilibrium constant for the reaction is 54.3 at that temperature. Calculate the concentrations of each species at equilibrium.

36 The first step Calculate Q Since 19.5<<<54.3, the reaction will proceed from left to right. The hydrogen and iodine will be depleted and there will be a gain in HI. H 2 (g) + I 2 (g) ↔ 2HI (g)

37 Set up an ICE diagram Now for the Algebra!! ConcentrationH2H2 + I 2 ↔ 2HI Initial Change-x +2x Equilibrium x x x

38 Next: the equilibrium constant is… After multiplying all of this out… Now do the quadratic formula…

39 And the x values are… x = M and x = M The first value of x can NOT be correct since it is larger than the original concentration of hydrogen and iodine. Therefore x = This is the change in concentration. Now back to the ICE diagram!!

40 You can check your work by plugging the values in to find K c. Remember K c = 54.3 ConcentrationH2H2 + I 2 ↔ 2HI Initial Change ( ) Equilibrium M M0.0255M

41 13.6: Solving equilibrium problems A more complete set of steps to solve equilibrium problems. 1.Write the balanced equation for the reaction. 2.Write the equilibrium expression using the law of mass action 3.List the initial concentrations 4.Calculate Q, and determine the direction of the shift to equilibrium 5.Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6.Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown 7.Check your calculated equilibrium concentrations by making sure the give the correct value of K.

42 Let’s do a few more ICE problems Consider the reaction represented by the equation Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial # M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction? You’ll need this for the next few problems!!

43 Fe 3+ (aq) + SCN - (aq) ↔ FeSCN 2+ (aq) Initial Change Equilibrium

44 Consider the reaction represented by the equation Fe 3+ (aq) + SCN - (aq) ↔ FeSCN 2+ (aq) Trial #2: Initial:10.0 M Fe 3+ (aq) and 8.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq)

45 Consider the reaction represented by the equation Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #3: Initial:6.00 M Fe 3+ (aq) and 6.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq)

46 Consider the reaction represented by the equation Fe 3+ (aq) + SCN - (aq) ↔ FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial #19.00 M5.00 M1.00 M Trial #23.00 M2.00 M5.00 M Trial #32.00 M9.00 M6.00 M Find the equilibrium concentrations for all species.

47 Let’s see what happen here… H 2 (g) + I 2 (g) ↔ 2HI(g) K=7.1 x 10 2 Calculate the equilibrium concentrations if a 5.00L container initially contains 15.8g of H 2 and 294g of I 2 [HI] 0 = zero

48 Since Q = 0 and Q < K, more product will be formed. We know that: K is large so the rxn will almost go to completion I 2 is the limiting reagent. It will be the smallest at equilibrium…let it’s equilibrium concentration = x Concentrati on H2H2 + I 2 ↔ 2HI Initial Change Equilibrium x

49 Fill in ICE based on stoichiometry Now use K = 7.1 x 10 2 and algebra to find the value of x. BUT…we know x will be very small…sooo.. ConcentrationH2H2 + I 2 ↔ 2HI Initial Change x – x (x – 0.232) or -2x Equilibrium x x-2x

50 Because x is so small, it is negligible when added to another number and will essentially have little effect on the outcome. Giving us the following for our equation… Making the algebra MUCH, MUCH easier. Now, find the other concentrations.

51 The end result! Whew!! ConcentrationH2H2 + I 2 ↔ 2HI Initial Change x – x (x – 0.232) or -2x Equilibrium x

52 Checking the assumption: (x is negligible) The rule of thumb is that if the value of x is less than 5% of all the smallest concentrations, our assumption is valid. If not, we would have used the quadratic equation to solve the problem.

53 What if K is really small? Begin the problem exactly the same way. For small K values, the product concentration is small Choosing a product to be x.

54 Here we go… Gaseous NOCl decomposes to form the gases NO and Cl 2. At 35 o C the equilibrium constant is 1.6x If 1.20 mol NOCl, 0.45 mol of NO, and 0.87 mol Cl 2 are mixed into a 1.0L flask, what are the equilibrium concentrations? 2NOCl(g) ↔ 2NO(g) + Cl 2 (g) First find Q

55 Choose a product to be x at equilibrium NO will be the limiting reagent. It will be the smallest at equilibrium…let it’s equilibrium concentration = x Concentration2NOCl↔ 2NO+ Cl 2 Initial Change Equilibrium x

56 Fill in ICE based on stoichiometry And add to get the equilibrium “values” Use these values and K c = 1.6 x to find the value of x. Concentration2NOCl↔ 2NO+ Cl 2 Initial Change - (x – 0.45) x – (x – 0.45) Equilibrium xx0.5x

57 Let’s again assume that x is negligible and ignore it. Now use this to find final concentrations

58 Concentration2NOCl↔ 2NO+ Cl 2 Initial Change - (x – 0.45) x – (x – 0.45) Equilibrium x

59 13.7 Le Chatlier’s Principle If there is a change imposed (a stress) on a system at equilibrium, the position of equilibrium will shift in a direction that tends to reduce that change (or stress)

60 Concentration Change the amounts of products or reactants…

61 Le Châtelier’s Principle Concentration 61 Change Shifts the Equilibrium Increase concentration of product(s) Makes Q > K left Decrease concentration of product(s) Makes Q < K right Decrease concentration of reactant(s) Makes Q > K Increase concentration of reactant(s) Makes Q < K right left aA + bB cC + dD Add Remove Remember the effect on Q will tell you the direction of the shift

62 The effect of pressure There are three ways to change pressure 1.Add or remove a gaseous product or reactant 2.Add an inert gas (one not involved in the reaction 3.Change the volume of the container We already looked at #1, and #2 won’t have any effect on equilibrium, so let’s look at effects of pressure by changing volume.

63 Changes in volume and pressure A (g) + B (g) C (g) ChangeShifts the Equilibrium Increase pressureSide with fewest moles of gas Decrease pressureSide with most moles of gas Decrease volume Increase volumeSide with most moles of gas Side with fewest moles of gas

64 Changes in temperature The big key to all of the previous changes are that they effect the equilibrium position but NOT the equilibrium constant. However, K, the equilibrium constant is changed with temperature. And it affects both the forward and the reverse reactions. The direction of the shift depends on whether or not it is exo- or endothermic

65 Changes in temperature ChangeExothermic Rxn Heat is product Increase temperature K decreases Shift to the left Decrease temperature K increases Shift to the right Endothermic Rxn Heat is reactant K increases Shift to the right K decreases Shift to the left equilibrium video

66 Catalyst & Le Châtlier Adding a catalyst – Does not change K – Does NOT shift the position of an equilibrium system – System will reach equilibrium sooner! Catalyst lowers Ea for both forward and reverse reactions Catalyst does NOT change equilibrium constant or shift equilibrium.


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