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Equilibrium Chapter 15 Chemical Equilibrium John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Chemistry, The Central.

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Presentation on theme: "Equilibrium Chapter 15 Chemical Equilibrium John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Chemistry, The Central."— Presentation transcript:

1 Equilibrium Chapter 15 Chemical Equilibrium John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2 Equilibrium The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

3 Equilibrium The Concept of Equilibrium As a system approaches equilibrium, both the forward and reverse reactions are occurring. At equilibrium, the forward and reverse reactions are proceeding at the same rate.

4 Equilibrium A System at Equilibrium Once equilibrium is achieved, the amount of each reactant and product remains constant.

5 Equilibrium Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow N 2 O 4 (g) 2 NO 2 (g)

6 Equilibrium The Equilibrium Constant

7 Equilibrium The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ]

8 Equilibrium The Equilibrium Constant Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2

9 Equilibrium The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

10 Equilibrium The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

11 Equilibrium The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

12 Equilibrium What Are the Equilibrium Expressions for These Equilibria?

13 Equilibrium The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b

14 Equilibrium Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV

15 Equilibrium Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes Where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant)

16 Equilibrium Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are.

17 Equilibrium Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

18 Equilibrium Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N 2 and H 2 or whether we start with NH 3. We will have the same proportions of all three substances at equilibrium.

19 Equilibrium What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium.

20 Equilibrium What Does the Value of K Mean? If K >> 1, the reaction is product-favored; product predominates at equilibrium. If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

21 Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction = K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = 4.72 at 100  C [N 2 O 4 ] [NO 2 ] 2 N2O4 (g)N2O4 (g) 2 NO 2 (g)

22 Equilibrium Manipulating Equilibrium Constants The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number. K c = = at 100  C [NO 2 ] 2 [N 2 O 4 ] N2O4 (g)N2O4 (g) 2 NO 2 (g) K c = = (0.212) 2 at 100  C [NO 2 ] 4 [N 2 O 4 ] 2 2 N 2 O 4 (g) 4 NO 2 (g)

23 Equilibrium Manipulating Equilibrium Constants The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps.

24 Equilibrium Heterogeneous Equilibrium

25 Equilibrium The Concentrations of Solids and Liquids Are Essentially Constant Both can be obtained by dividing the density of the substance by its molar mass—and both of these are constants at constant temperature.

26 Equilibrium The Concentrations of Solids and Liquids Are Essentially Constant Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)

27 Equilibrium As long as some CaCO 3 or CaO remain in the system, the amount of CO 2 above the solid will remain the same. CaCO 3 (s) CO 2 (g) + CaO (s)

28 Equilibrium Equilibrium Calculations

29 Equilibrium Equilibrium Calculations A closed system initially containing x 10 −3 M H 2 and x 10 −3 M I 2 At 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 −3 M. Calculate K c at 448  C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g)

30 Equilibrium What Do We Know? [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change At equilibrium 1.87 x 10 -3

31 Equilibrium [HI] Increases by 1.87 x M [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change+1.87 x At equilibrium 1.87 x 10 -3

32 Equilibrium Stoichiometry tells us [H 2 ] and [I 2 ] decrease by half as much [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change-9.35 x x At equilibrium 1.87 x 10 -3

33 Equilibrium We can now calculate the equilibrium concentrations of all three compounds… [H 2 ], M[I 2 ], M[HI], M Initially1.000 x x Change-9.35 x x At equilibrium 6.5 x x x 10 -3

34 Equilibrium …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x ) 2 (6.5 x )(1.065 x )

35 Equilibrium

36 The Reaction Quotient (Q) To calculate Q, one substitutes the initial concentrations on reactants and products into the equilibrium expression. Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium.

37 Equilibrium If Q = K, the system is at equilibrium.

38 Equilibrium If Q > K, there is too much product and the equilibrium shifts to the left.

39 Equilibrium If Q < K, there is too much reactant, and the equilibrium shifts to the right.

40 Equilibrium Le Châtelier’s Principle

41 Equilibrium Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

42 Equilibrium What Happens When More of a Reactant Is Added to a System?

43 Equilibrium The Haber Process The transformation of nitrogen and hydrogen into ammonia (NH 3 ) is of tremendous significance in agriculture, where ammonia-based fertilizers are of utmost importance.

44 Equilibrium The Haber Process If H 2 is added to the system, N 2 will be consumed and the two reagents will form more NH 3.

45 Equilibrium The Haber Process This apparatus helps push the equilibrium to the right by removing the ammonia (NH 3 ) from the system as a liquid.

46 Equilibrium The Effect of Changes in Pressure

47 Equilibrium The Effect of Changes in Temperature Co(H 2 O) 6 2+ (aq) + 4 Cl (aq) CoCl 4 (aq) + 6 H 2 O (l)

48 Equilibrium The Effect of Changes in Temperature

49 Equilibrium Catalysts increase the rate of both the forward and reverse reactions.

50 Equilibrium Equilibrium is achieved faster, but the equilibrium composition remains unaltered.


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