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Lecture 16.1 Chapter 16: Chemical Equilibria. All chemical reactions are reversible, at least in principle. Equilibrium is a dynamic situation. It appears.

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Presentation on theme: "Lecture 16.1 Chapter 16: Chemical Equilibria. All chemical reactions are reversible, at least in principle. Equilibrium is a dynamic situation. It appears."— Presentation transcript:

1 Lecture 16.1 Chapter 16: Chemical Equilibria

2 All chemical reactions are reversible, at least in principle. Equilibrium is a dynamic situation. It appears that no change is occurring. However, reactants are combining together to from products and the products are combining to reform the reactions. The two reactions, the forward one and the reverse one, occur at the same rate so there is no NET change in the concentrations. H 2 O (g) + CO (g) CO 2(g) + H 2(g)

3 Dynamite exploding-- equilibrium? The equilibrium position of a reaction– left right, or somewhere in between is determined by many factors: the initial concentrations, the relative energies of the products and reactants, and the relative degree of organization of the reactants and products. Energy and organization come into play because nature tries to achieve minimum energy and maximum disorder. (Chapter 20)

4 The Equilibrium Constant The Law of Mass Action Two Norwegian chemist, Cato Maximilian Guildberg ( ) and Peter Waage ( ) proposed in 1864 a general description of the equilibrium condition (called the Law of Mass Action). They proposed for the rxn of the type a A + b B c C + d D The equilibrium concentrations of rxn and products can be represented by the equilibrium constant expression Equilibrium Constant = K = [C] c [D] d PRODUCTS [A] a [B] b REACTANTS [ ] represent concentration at equilibrium A, B, C, and D represent chemical species and a,b, c, and d are their coefficients in the balanced equation.

5 Write the equilibrium expression for the rxn: 2NH 3(g) N 2(g) + 3H 2(g) K eq at a given T can be calculated if equilibrium concentrations of reaction components are known. NOTE: K eq constants are without units. The reason is beyond this course, but involves corrections for nonideal behavior of substances taking part in the reaction. When corrections are made, the units cancel out and the corrected K has no units. So, we will NOT use units for K. K eq = [N 2 ][H 2 ] 3 PRODUCTS [NH 3 ] 2 REACTANTS

6 Writing Equilibrium Constant Expressions Rule for Rxn Involving Solids R.O.T. K eq is in same phase In a heterogeneous rxn., equilibrium does NOT depend on amounts of pure solids or liquids present. The concentration of pure liquids and solids does NOT change. (C = constant) K eq =C CaO [CO 2(g) ] /C CaCO3 = [CO 2(g) ]

7 Rules for Rxns Involving Water R.O.T. K eq is in same phase However, if the rxn were carried out under conditions where the water is a gas rather than a liquid, that is then it is included in K eq b/c the concentration OR pressure of water vapor can change. If pure solids or pure liquids are involved in a chem. rxn, their concentrations are not included in the equilibrium expression. This simplification occurs ONLY with pure solids or liquids, not with solutions or gases, since the last two cases the concentrations vary. 2H 2 O (l) 2 H 2(g) + O 2(g) K eq = [H 2(g) ] 2 [O 2(g) ] 2H 2 O (g) 2 H 2(g) + O 2(g) K eq =[H 2(g) ] 2 [O 2(g) ]/[H 2 O (g) ]

8 Equilibrium Expressions Involving Pressures The relationship b/t P and concentration PV= nRT or P = (n/V)RT or P = CRT Write the equilibrium expression in terms of partial pressures and the equilibrium expression in terms of concentration for the rxn: 2NH 3(g) N 2(g) + 3H 2(g) K p = P N2 P H2 3 P NH3 2 K eq = [N 2 ][H 2 ] 3 [NH 3 ] 2

9 Write the expressions for K and K p for the following processes: PCl 5(s) PCl 3(l) + Cl 2(g) CuSO 4 5H 2 O (s) CuSO 4(s) + 5 H 2 O (g) K eq = [ Cl 2 ] K p = P Cl2 K eq = [H 2 O] 5 K p = P H2O 5

10 Calculating K p values The reaction for the formation of nitrosyl chloride 2NO (g) + Cl 2(g) 2 NOCl (g) was studied at 25 o C. The pressures at equilibrium were found to be P NOCl = 1.2 atm P NO = P Cl2 = 0.30 atm. Calculate the value of K p for the reaction. K p = P NOCl 2 /(P NO ) 2 (P Cl2 ) = (1.2) 2 /(0.05) 2 (0.3) = 1.9 x 10 3

11 The relationship between K and K p K p = K(RT) n For the derivation read A Closer Look p. 753 in text. Remember P = CRT or C = P/RT and must be substituted in equilibrium expression. The relationship is K p = K(RT) n where n is the difference in the sums of the coefficients for the gaseous products and reactants. NOTE: K p = K c when n = 0

12 Example: The value of K p was 1.9 x 10 3 in the previous example. Calculate the value of K at 25 o C for the reaction 2NO (g) + Cl 2(g) 2NOCl (g) 1.9 x 10 3 = K(( Latm/mol K)( )) (2-3) 1.9 x 10 3 = K (24 –1 ) K = = 4.6 x 10 4 NOTE: If n = 0, then K p = K c K p = K(RT) n

13 Manipulating Equilibrium Expressions 2NO (g + Cl 2(g) 2NOCl (g) K = 4.6 x 10 4 K forward = [NOCl] 2 /[NO] 2 [Cl 2 ] = 4.6 x 10 4 K reverse = [NO] 2 [Cl 2 ]/[NOCl] 2 = 1/4.6 x 10 4 If stoichiometric coefficients of balanced equations are multiplied by some factor, n, the equilibrium constant for new equation is raised to the power of the new factor K new = (K old ) n K forward = [NOCl] 6 /[NO] 6 [Cl 2 ] 3 = (4.6 x 10 4 ) 3 Forward and reverse equilibrium expressions 3 (2 NO (g) + Cl 2(g) 2 NOCl (g) )

14 Write the equilibrium expression for the reverse direction for the following reaction Br 2(g) 2 Br (g) K = 2.2 x 10 –15 K eq = [Br 2 ]/[Br] 2 = 1/ 2.2 x 10 –15 Write the equilibrium expression for the reverse direction for the following reaction ½ (Br 2(g) 2 Br (g) )K = 2.2 x 10 –15 K new = (K old ) n note: n is multiplication factor K eq = [Br 2 ] 1/2 /[Br] = 1/(2.2 x 10 –15 ) 1/2

15 In general, when two or more equations are added to produce a net equation, K for the net equation is the product of the K for added equations. AgCl (s) is dissolved in water (to a small extent). Ammonia is then added to form the complex Ag(NH 3 ) 2 + AgCl (s) Ag + (aq) + Cl - (aq) K 1 = [Ag + ][Cl - ] = 1.8 x 10 –10 Ag + (aq) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) K 2 =[Ag(NH 3 ) 2 + (aq) ] = 1.6 x 10 7 [Ag + ][NH 3 ] 2 AgCl (s) + 2 NH 3(aq) Ag(NH 3 ) 2 + (aq) + Cl - (aq) K net =[Ag(NH 3 ) 2 + ][Cl - ] =K 1 K 2 =(1.6 x 10 7 )(1.8 x 10 –10 )=2.9x [NH 3 ] 2 NET

16 Exercise 16.3 p. 756 Manipulating Equilibrium Constant Expressions The following equilibrium constants are given at 500 K. H 2(g) + Br 2(g) 2 HBr (g) K = 7.9 x H 2(g) 2 H (g) K = 4.8 x 10 –41 Br 2(g) 2 Br (g) K = 2.2 x 10 –15 Calculate K for the reaction of H and Br atoms to give HBr H (g) + Br (g) HBr (g) K = ? H 2(g) + Br 2(g) 2 HBr (g) K 1 = 7.9 x H (g) H 2(g) K 2 = 1/4.8 x 10 –41 2 Br (g) Br 2(g) K 3 = 1/2.2 x 10 –15 2 H (g) + 2 Br (g) 2 HBr K net = [HBr] 2 /[H] 2 [Br] 2 Want K net = [HBr]/[H][Br] so… K net =(K 1 K 2 K 3 ) ½ =((7.9 x10 11 )(1/4.8 x10 –41 )(1/2.2 x10 –15 )) ½ = (7.5 x ) ½ = 2.7 x 10 33

17 Work on HW

18 H 2 O + CO CO 2 + H 2 (a) H 2 O and CO are mixed in equal numbers and begin to react (b) to from CO 2 and H 2. After time has passed, equilibrium is reached (c) and the number of reactant and product molecules then remain constant over time (d) a.b.c.d.

19 Demonstration: Limewater and CO 2 Saturated solution of limewater Ca(OH) 2(s) Ca(OH) 2(aq) CO 2 + Ca(OH) 2(aq) Add MORE CO 2(g) to the CaCO 3 (s) CO 2 + H 2 O + CaCO 3(s) Add CO 2(g) CaCO 3(s) + H 2 O Ca(HCO 3 ) 2(aq)

20 Equilibria in CO 2 /Ca 2+ /H 2 O System CaCl 2 + 2NaHCO 3 Add CO 2(s ) Ca 2+ + HCO 3 – CaCO 3(s) + H 2 O + CO 2 2NaCl + Ca(HCO 3 ) 2 CaCO 3(s) + 2 H 2 O + 2 CO 2 Net: Ca 2+ + HCO 3 - CaCO 3(s) + H 2 O +CO 2


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