Combining reactions and equilibrium constants

Transcription

1 Combining reactions and equilibrium constants Rules for deriving equilibrium constants from other equilibrium constants. copyright by James B. Condon (2013) all rights reserved. Some of the graphics are supplied by Corel Corporation with their permission. Use of these materials by others is encouraged provided these notices are not altered. 1

2 By definition of the equilibrium expression for K º for the reaction: N 2 + 3H 2 º 2NH 3 1) is: (P ) 2 NH 3 K º 1 = (P ) 3 H 2 (P N 2 ) and the K º for the reaction: 2NH 3 º N 2 + 3H 2 2) is: (P K º H ) 3 (P N ) 2 = 2 2 (P ) 2 NH 3 So notice that: (K º 1)(K º 2) = 1 In other words, the K for the reverse reaction is the inverse for the K for the forward reaction, and, of course, vis-averse. 1 1 All Ks are stand conditions so for convenience the º will be left off. 2

3 It follows from what is on the last slide that if one were to subtract a reaction equation from a second reaction, then the K is obtained by dividing the K of the second reaction by that of the first. For example, using the reactions*: (P ) 2 PF 2P(s) + 3F 2 (g) º 2PF 3 (g) with K 1 = 3 1) (P ) 3 (P ) 2 PF 2P(s) + 5F 2 (g) º 2PF 5 (g) with K 2 = 5 2) (P ) 5 Subtracting reaction 1) from 2) yields: (P PF ) 2 2PF 3 (g) + 2F 2 (g) º 2PF 5 (g) with K 3 = 5 3) (P F ) 2 2 (P ) 2 PF 3 K Which implies: K 3 = 2 (Check it out by doing the division.) K 1 F 2 F 2 Notice also that for: PF 3 (g) + F 2 (g) º PF 5 (g) K = p K 3 3

4 The following table shows the relationship between the manipulation of the equilibrium reaction equations and what one needs to do for the equilibrium constants. Equilibrium Reaction Equation Addition reaction A + B Subtraction reaction A B Multiply reaction A by an integer, n Divide reaction A by an integer, m Equilibrium Constants Multiply Constants, K final = K A K B Divide Constants, K final = K A K B Raise the K A to a power n. K final = (K A ) n Take the mth root of K A. m K final = pk A A few more examples are given on the following slides. 4

5 What is the K for the reaction: Fe(OH) 2 + 6NH 3 º Fe(NH 3 ) OH with K 1 =? 1) This reaction may be constructed from the reactions: Fe(OH) 2 º Fe OH with a K sp 2) and: Fe NH 3 º Fe(NH 3 ) 6 2+ with a K f 3) ADDing the two equations yields equation 1). Thus the overall K of interest is the MULTPLICATION of the K sp and the K f : K 1 = K sp K f 5

6 What is the K for the reaction: Sn(OH) 4 + 2OH º Sn(OH) 6 with K 1 =? 1) This reaction may be constructed from the reactions: Sn(OH) 4 º Sn OH with a K sp 2) and: Sn OH º Sn(OH) 6 with a K f 3) ADDing the two equations yields equation 1). Thus the overall K of interest is the MULTPLICATION of the K sp and the K f : K 1 = K sp K f 6

7 The most insoluble form of nickel sulfide is the γ (gamma) form. The K sp for the NiS(γ) is 2.0 x What concentration of CN ion would be required to dissolve this sulfide to the extent of obtaining 0.10 M Ni(CN) 4 ions in solution? The K d for Ni(CN) 4 is 1.0 x Here the K d is given and not the K f. The overall equation is: NiS(γ) + 4CN º Ni(CN) 4 + S Which can be formed from: NiS(γ) º Ni 2+ + S 1) with a K sp and: Ni(CN) 4 º Ni CN 2) with a K d. SUBTRACTing 2) from 1) yields the overall reaction, therefore, to obtain the overall K one divides the K sp by K d. K K overall = sp Thus: K overall = K d 7

8 The most insoluble form of nickel sulfide is the γ (gamma) form. The K sp for the NiS(γ) is 2.0 x What concentration of CN ion would be required to dissolve this sulfide to the extent of obtaining 0.10 M Ni(CN) 4 ions in solution? The K d for Ni(CN) 4 is 1.0 x With the preliminaries for finished, that is, K overall = for the reaction: NiS(γ) + 4CN º Ni(CN) 4 + S, the problem may be solved: K = Substituting: [Ni(CN) 4 ][S ] [CN ] 4 (0.10)(0.10) = So that: [CN ] = [CN ] 4 8

9 Next is an example that is important for the concept for amphoteric compounds. Many of the slightly soluble (weak) bases (hydroxides) exhibit this behavior. 9

10 At what ph would Sn(OH) 2 dissolve to the Sn(OH) 4 ion to the extent of 0.10 M? The K sp for Sn(OH) 2 is and the K f for Sn(OH) 4 ion is The overall reaction is: Sn(OH) 2 + 2OH º Sn(OH) 4 with K total which can be constructed from the addition of the two reactions: Sn OH and º Sn(OH) 2 with K sp = Sn OH º Sn(OH) 4 with K f = Therefore the K total is given as: K total = K sp K f =

11 At what ph would Sn(OH) 2 dissolve to the Sn(OH) 4 ion to the extent of 0.10 M? The K sp for Sn(OH) 2 is and the K f for Sn(OH) 4 ion is The overall reaction is: Sn(OH) 2 + 2OH º Sn(OH) 4 with K total = Writing the equilibrium expression: K total = [Sn(OH) 4 ] [OH ] 2...and substituting: =...so [OH ]= 9.5 [OH ] Thus: poh = 2.02 and ph =

12 Combining reactions and equilibrium constants Rules for deriving equilibrium constants from other equilibrium constants. THE END 12