Initial amounts: mol Amounts at equilibrium: mol (5) Initial amounts: x mol Amounts at equilibrium: x mol

Transcription

1 4. CHEMICAL EQUILIBRIUM n Equilibrium Constants A Y + 2Z Initial amounts: mol Amounts at equilibrium: mol Concentrations at equilibrium: mol dm 3 K c (1.5/5) (3.0/5)2 (1/5) 2 mol dm 3 (1.5) (3)2 (5) 2.7 mol dm 3 mol dm A + B Y + Z Initial amounts: x mol Amounts at equilibrium: x mol ; ( x 2) 1 x 2 40 x 42 Thus, initially there must be 42 mol of A A + 2B Z Initial amounts: x 4 0 mol Amounts at equilibrium: x mol Concentrations at equilibrium: x mol dm 3

2 CHEMICAL EQUILIBRIUM n 73 1/5 [(x 1)/5] (2/5) (x 1) 25 x 1 x 26 Thus, initially there must be 26 mol of A Two moles of SO 3 (g) produce 3 mol of product; thus Σν +1 mol. Then, from Eq. 4.26, K P K c (RT) +1 ( mol dm 3 ) ( J K 1 mol K) J dm J m Pa 2.48 bar 4.5 I 2 2I Initial amounts: mol Equilibrium amounts: ( ) mol 4 2 ( / 05.) K c / mol mol dm mol dm 3 K P mol dm 3 ( J K 1 mol K) Pa bar 4.6. Addition of N 2 at constant volume and temperature necessarily requires the equilibrium to shift to the right. [NH 3 ] 2 n2 NH3 V2 K c [N 2 ][H 2 ] 3 n N2 nh 3 2 If n N2 is increased at constant V, the equilibrium must shift so as to produce more ammonia. If the pressure (as well as the temperature) is held constant, however, addition of N 2 requires that V is increased. If the proportional increase in V 2 is greater than the increase in n N2 the equilibrium will shift to the left when N 2 is added. The volume V is proportional to n N2 + n H2 + n NH3, and V 2 is proportional to (n N2 + n H2 + n NH3 ) 2 If n N2 is very much larger than n H2 + n NH3, V 2 will increase approximately with n 2 N and therefore 2 increases more strongly than n N2. If n N2 is not much larger than n H2 + n NH3, an increase in n N2 will have a relatively smaller effect on V 2. The increase in ammonia dissociation when N 2 is added is therefore expected when N 2 is in excess, but not otherwise.

3 74 n CHAPTER 4 On the other hand, n 3 H appears in the equilibrium expression; this varies more strongly than V 2, 2 and added H 2 therefore cannot lead to the dissociation of ammonia Suppose that in 1 dm 3 there are x mol of N 2 O 4 y mol of NO 2 Pressure bar (x + y)rt/v ( x+ y) mol bar dm K mol K 3 1 dm Therefore, x + y (1) Density g dm 3 (92.02x y) g (2x + y) g 1 dm 3 1 dm 3 Therefore, 2x + y (2) From Eqs. (1) and (2), we get x ; y N 2 O 4 (g) š 2NO 2 (g) P(1 α) 2 Pα Since partial pressures are proportional to the number of moles of each species present, x P(1 α) and y 2Pα, which means that y x from which we get α K c ( ) Pα P(1 α), y (2x + y) mol dm mol dm 3 K P K c RT (from Eq. 4.26, with Σν 1) mol dm dm 3 bar mol bar K x K P P 1 (from Eq. 4.32) K P (1 bar) Addition of He produces no effect, since concentrations, partial pressures, and mole fractions remain unchanged g NOBr g NOBr g mol 1 N g mol 1 O g mol 1 Br If α is the degree of dissociation, mol NOBr

4 CHEMICAL EQUILIBRIUM n 75 n NOBr 0.01 (1 α) mol; n NO 0.01 α mol n Br α mol Total amount, n α 0.861; n n NOBr mol; n Br mol α mol PV 3 RT bar 1dm bar dm mol mol n NO mol K c ( mol dm 3 ) 2 ( mol dm 3 ) ( mol dm 3 ) mol dm 3 K P K c RT (from Eq. 4.26) mol dm dm 3 atm mol bar K x K P (0.355 bar) 1 (from Eq. 4.32) Suppose that, if there were no dissociation, the partial pressure of COCl 2 was P; then the actual partial pressures are COCl 2 (g) CO(g) + Cl 2 (g) P(1 α) Pα Pα P + Pα 2 bar With α P 2 bar 1 + ( ) 2 bar K P ( ) 2 2[1 ( )] 2 ( ) 2 bar bar K c K P (RT) 1 (from Eq. 4.26) bar ( dm 3 bar mol 1 ) mol dm 3 K x K P P 1 (from Eq. 4.32) bar (2 bar)

5 76 n CHAPTER H 2 + I 2 2HI Initially: mol At equilibrium: 1 x 2 3 x 2 x mol After addition of 2 mol H 2 : 3 x 3 x 2x mol K x 3/2 x 2 1 x 2 3 x 2 4x 2 (3 x) 2 K 4(3/2)2 (3/2) g iodine 0.05 mol I 2 When all of the solid iodine has just gone, the iodine pressure is 0.10 atm. The consumption of 0.05 mol I 2 leads to the formation of 0.10 mol HI, which exerts a pressure of P HI 0.1 mol dm3 bar K 1 mol K 10 dm bar Then, if P H2 is the partial pressure of H 2 after equilibrium is established, (0.260 bar) 20 P 0.10 bar H2 P H bar bar 10 dm n H bar dm 3 mol mol Thus, mol of H 2 is produced in the equilibrium mixture, and 0.05 mol of H 2 is required to remove the 0.05 mol of I 2. Therefore, mol of H 2 must be added Assuming that we start with one mole of N 2 O 4 (g), N 2 O 4 (g) š 2NO 2 (g) 1 α 2α We obtain (1+α) moles of gas at equilibrium. Therefore, equilibrium partial pressures are 1 α 2α PN2O P; P 4 NO P, 2 1+ α 1+ α and

6 CHEMICAL EQUILIBRIUM n α 1 P 2 c P x P K P; K K ( RT) ; K K / P. 1 α Therefore, at bar, K P 1.08 bar, K c mol dm 3, K x 1.81, and at 6.18 bar, K P 1.45 bar, K c mol dm 3, K x Rewriting the reaction in terms of one mole of HCl, we get HCl(g) O 2 (g) 1 2 Cl 2 (g) H 2 O(g) (1 y) P O2 y/2 y/2 From examining the equation above, it is possible to establish the following relationships: x x Cl2 HCl y ; 2(1 y) H2O Cl2 1. It is important to remember that the same ratios hold for partial pressures also. Now, 1/ Assuming ideal behavior, the partial pressure of oxygen is 0.51 bar. Therefore, Since no reactants are initially present, we write x x PCl P H P 2 2 Cl2 y 1 K P. 1/ 4 1/ 4 1/ 4 P P P P 2(1 y) P K P HCl 1/ 2 O O2 HCl (1 0.76) 0.51 O2 0 1/ bar 0.25 O2 H 2 (g) + I 2 (g) HI(g) x x n 2x. where n is the initial amount of HI: n 10.0 g/( g mol 1 ) mol. 2 ( n 2 x) n 2x Since KP 65.0, or 65, we can solve for x. The solution is x x x 10 3 mol. Therefore, the equilibrium mole fractions are x H2 x I2 x/n ; x HI (n 2x)/n

7 78 n CHAPTER 4 n Equilibrium Constants and Gibbs Energy Changes a. G /J mol ln ( ) G J mol kj mol 1 b. (C 6 H 5 COOH) 2 2C 6 H 5 COOH Initially: mol dm 3 At equilibrium: x 0.1 2x mol dm 3 (0.1 2x) 2 x x x ± ± x or mol Only the second answer is possible, and this leads to Dimer: mol dm 3 ; Monomer: 0.01 mol dm K P at 3000 K (0.4)2 0.2 (0.6) atm G /J mol ln G J mol kj mol a. ln K c ; K c b. ln K c ; K c 408 c. K c G kj mol a. G G Products G Reactants 2( 16.63) 0 (3 0) kj mol 1 K P exp( G /RT) bar 2 b. G kj mol 1 K P bar 2 c. G kj mol 1 K P bar 1

8 CHEMICAL EQUILIBRIUM n 79 d. G ( ) kj mol 1 K P K P Σv K c K P (RT) Σv K x K P P Σv a bar (mol dm 3 ) b bar (mol dm 3 ) c bar (mol dm 3 ) d a. G RT ln K ln J mol kj mol 1 H G + T S J mol kj mol 1 b. CO + H 2 O CO 2 + H 2 Initially: mol At equilibrium: 2 x 2 x x x mol K P K c x 2 (2 x) (Note that the total volume cancels out and so need not be considered.) x 2 x ; x The amounts are therefore mol (CO); mol (H 2 O) For reaction (1) mol (CO 2 ); mol (H 2 ) K 1 exp( / ) For reaction (2) dm 3 mol 1 K 2 exp(31 000/ ) mol dm 3 For the coupled reaction (3)

9 80 n CHAPTER 4 K 3 K 1 K K K K(overall) K 1 K G RT ln K ln J mol kj mol Let the solubility be c moles per dm 3. Then Cr(OH) 3 (s) Cr OH c 3c Then the value of the solubility product is K sp c(3c) 3 27c 4. Therefore, c (K sp /27) 1/ mol dm 3. n Temperature Dependence of Equilibrium Constants a. Zero b. G 0; H > 0; S > 0 c. K c K P (RT) Σν ( ) mol dm 3 G ln J mol 1 d. K P > 1 bar e. G < a. Yes b. Yes c. S > kj mol a. G kj mol 1 H kj mol 1 S Standard state is 1 bar J K 1 mol 1

10 CHEMICAL EQUILIBRIUM n 81 b. ln (K P /bar 1 ) G RT K P bar 1 c. K c K P (RT) dm 3 mol 1 d. G RT ln K c ln ( ) e. S J mol kj mol 1 H G T J K 1 mol 1 f. G (100 C) H T S ( ) J mol 1 ln (K P /bar 1 ) G RT K P (100 C) bar a. 2H 2 (g) + O 2 (g) 2H 2 O(g) f G 2 f G H2 O 2 f G H 2 f G O2 2( ) 2(0) kj mol 1 f H 2 f H H2 O 2 f H H 2 f H O2 2( ) 2(0) kj mol 1 G H T S f S b. G RT ln K P ( ) kj mol J K 1 mol 1 ln (K P /bar J mol ) J K 1 mol K K P bar 1 c. G(2000 C)/kJ mol 1 f H f S ( ) kj mol 1 ln (K P /bar ) K P bar 1

11 82 n CHAPTER r G 2 f G (O 3, g) 3 f G (O 2, g) 2(163.2) 3(0) kj mol 1. K exp( r G /RT) exp[ /( )] a. S H G T 35.1 J K 1 mol b. ln K c K c mol dm 3 c. G (25 C) ( ) J mol ln K c K c mol dm a. H kj mol 1 S J K 1 mol 1 G ( ) 6839 J mol kj mol b. ln K P ; K P If partial pressure of neopentane x bar, partial pressure of n-pentane (1 x) bar x 1 x x x; x x Thus P(neopentane) bar and P(n-pentane) bar a. Slope of plot of ln K c against 1/T is ln From Eq. 4.73, H J mol 1

12 CHEMICAL EQUILIBRIUM n 83 b kj mol kj mol a. Slope of plot of ln K c against 1/T is ln K H J mol kj mol 1 At 25 C, G RT ln S ( ) J mol 1 H G T J K 1 mol 1 b. The difference between the reciprocals of the absolute temperatures corresponding to 25 C and 37 C is a. At 400 C, The slope of the plot ln K w against 1/T was K, and in going from 25 C to 37 C ln K w is thus increased by At 25 C, ln K w is and at 37 C it is therefore K w at 37 C is therefore mol 2 dm 6. log 10 (K P /bar) K P bar G /kj mol ln G 4549 J mol kj mol 1 From Figure 4.2 (b), H J mol 1 S J mol kj mol J K 1 mol 1

13 84 n CHAPTER J K 1 mol 1 b. K c ( ) mol dm 3 G /kj mol ln G J mol kj mol 1 c. I 2 + cyclopentene 2HI + cyclopentadiene 0.1 x 0.1 x 2x x 4x 3 (0.1 x) For a very approximate solution, neglect x in comparison with 0.1: 4x (0.1) x x For a better solution, calculate 4x 3 /(0.1 x) 2 at various x values: x Final concentrations are [I 2 ] M; x x 3 (0.1 x) [cyclopentene] M; [HI] M; [cyclopentadiene] M H kj mol 1 G kj mol 1 S ln K P K P bar J K 1 mol H kj mol 1 G kj mol 1 S J K 1 mol 1

14 CHEMICAL EQUILIBRIUM n a. K c 95/5 19 G /J mol ln 19 G 7299 J mol kj mol 1 b. G/J mol ln G ( ) J mol J mol kj mol 1 The reaction will therefore go from left to right a. H kj mol 1 G kj mol 1 S J K 1 mol 1 b. ln K P K P c. From the data in Table 2.1, d J K 1 mol 1 e 10 3 ( ) J K 2 mol 1 f 10 4 ( ) J K 1 mol 1 Then, from Eq. 2.52, H T /J mol [(T/K) ] [(T/K) ] T/K (T/K) (T/K) /(T/K) H d. ln K dt 2 RT / K 3 ( T / K) T ( T / K) d ( T / K) (T/K) ln (T/K) (T/K) (T/K) 2 + I

15 86 n CHAPTER T/K 1.51 ln (T/K) (T/K) + (T/K) 2 + I I is obtained from the fact that at , ln K I ln K T/K 1.51 ln (T/K) (T/K) + (T/K) 2 e. ln K P at 1000 K K P (1000 K) ln Partial pressures at 1395 K are: CO: atm; CO 2 : ( ) atm; O 2 : atm K P ( ) (0.9999) atm bar At 1443 K, K P ( ) ( ) atm bar At 1498 K, K P ( ) ( ) atm bar Then T/K K P /atm 10 4 /(T/K) ln(10 12 K P /atm) From a plot of ln (K P /atm) against 1/(T/K), slope K and H R slope 609 kj mol 1. G (1395 K) ln ( ) kj mol 1 S H G T (standard state 1 atm) 209 J K 1 mol 1

16 CHEMICAL EQUILIBRIUM n Suppose that there are present x mol of I 2 and y mol of I: x + y P At 800 C (x + y) mol RT V atm (x + y) dm3 atm K 1 mol dm K x + y y ; y x Degree of dissociation, x α At 1000 C, x + y y ; y x ; α At 1200 C, x + y y ; y x ; α a. α , 0.202, at the three temperatures b. At 800 C, ( K c 4 mol) mol dm mol dm 3 At 1000 C, ( K c 4 ) mol dm 3 At 1200 C, ( K c 3 ) mol dm 3

17 88 n CHAPTER 4 c. K P K c (RT) Σν (Eq. 4.26) K c RT since Σν 1 At 800 C, K P mol dm dm 3 m J K 1 mol K kpa 9.12 Torr bar At 1000 C, K P kpa At 1200 C, K P kpa d. T/K 10 4 K c /mol dm /(T/K) 10 4 K c ln mol dm Slope of a plot of ln(k c /mol dm 3 ) against 1/(T/K) is K 1. U 139 kj mol 1 H U + RT ( ) J mol kj mol 1 e. At 1000 C, K P kpa bar. G ( J mol 1 ) ln(0.169/bar) J mol kj mol 1 (standard state: 1 bar) S H G T J K 1 mol G RT ln K ln J mol kj mol 1 Slope of plot ln ( / ) (1/300) (1/340) K H J mol kj mol 1 S ( H G )/T 183 J K 1 mol G H T S J mol 1

18 CHEMICAL EQUILIBRIUM n 89 K c exp(34 140/ ) The equilibrium constant is equal to unity when G is equal to zero T T K G ln S J mol 1 H G T J K 1 mol 1 At 25 C, G H T S ( ) J mol G ln S J mol 1 H G T ln K c ln K c K c mol dm J K 1 (standard state: 1 mol dm 3 ) The equilibrium constant is unity when G is zero: 0 H T S T H S 740 K G RT ln K c ln ( ) S J mol 1 H G T J K 1 mol 1 The equilibrium constant is unity when G 0.

19 90 n CHAPTER 4 G (44.9 T) T 895 K For the equilibrium Br 2 (l) Br 2 (g) the vapor pressures are the equilibrium constants. The corresponding G values (standard state: 1 bar) are G ( K) RT ln 1 0 G ( K) (J mol 1 ) ln J mol 1 We can therefore set up two simultaneous equations: 0 H ( K) S J mol 1 H ( K) S Subtraction gives S J mol 1 /48.9 K J K 1 mol 1 At 25 C, H J K 1 mol K J mol 1 G ( ) J mol J mol 1 The vapor pressure at 25 C is the corresponding equilibrium constant: P K P exp( 3212/ ) bar For the reaction, G kj mol 1 Then K P exp( G /RT) exp( / ) bar 1 (The unit arises from the standard state of 1 bar.) Σν for the reaction is From Eq. 4.26, K c From Eq. 4.32, K P (RT) Σν bar 1 ( J) Pa bar m 3 mol 1 (since J Pa 1 m 3 ) dm 3 mol 1

20 CHEMICAL EQUILIBRIUM n 91 K x K P P Σν, so that at 2 bar pressure, K x bar 1 2 bar This equilibrium constant (Eq. 4.31) is x(o 3 ) 2 /x(o 2 ) 3, and since it is so small, x(o 2 ) is almost exactly unity. Thus x(o 3 ) and x(o 3 ) At 500 K, by Eq. 2.46, H H (300 K) + C P ( ) 9600 ( ) J mol 1 For the entropy change at 500 K, the corresponding equation is S (T 2 ) S (T 1 ) + T 2 C P dt T 1 T S (T 1 ) + C P ln (T 2 /T 1 ) At 500 K, S is therefore S ln(500/300) J K 1 mol 1 Therefore G J mol 1 The equilibrium constant K P is therefore exp(20 297/ ) Since the reaction involves no change in the number of molecules, K c and K x also have this value (see Eqs and 4.32), and they are not affected by the pressure. Let the mole fraction of HI be x; then the mole fractions of H 2 and I 2 are both (1 x)/2, and the expression for K x is therefore K x 4x 2 /(1 x) Taking square roots x/(1 x) whence x 0.85 Pressure has no effect on the above values Values of G and of K[ exp( G /RT) u ] are Temperature θ / C T/K G /kj mol 1 K

21 92 n CHAPTER G H T S 0 when T and K 44.4 C At this temperature there will be equal concentrations of P and D. n Binding to Protein Molecules If the concentration of M is [M], that of sites occupied and unoccupied is n[m]. The association reactions may be formulated in terms of the sites, S, S + A K s SA K s [SA] [S][A] where [S] is the concentration of unoccupied sites and [SA] is the concentration of occupied sites. The total concentration of sites, n[m], is n[m] [S] + [SA] [SA] 1 K s [A] + 1 The average number of occupied sites per molecule is the total concentration of occupied sites divided by the total concentration of M: ν [SA] [M] n nk s [A] 1 K s [A] K s [A] The total concentration of the molecule M is [M] 0 [M] + [MA] + [MA 2 ] [MA n ] The total concentration of occupied sites is the total concentration of bound A molecules: [A] b [MA] + 2[MA 2 ] n[ma n ] Expressing every term in terms of [A]: Thus [M] 0 [M]{1 + K 1 [A] + K 1 K 2 [A] (K 1 K 2... K n )[A] n } [A] b [M]{K 1 [A] + 2K 1 K 2 [A] n(k 1 K 2... K n )[A] n }

22 CHEMICAL EQUILIBRIUM n 93 ν [A] b [M] 0 K 1 [A] + 2K 1 K 2 [A] n(k 1 K 2... K n )[A] n 1 + K 1 [A] + K 1 K 2 [A] (K 1 K 2... K n )[A] n With K 1 nk s, K 2 (n 1)K s /2, K 3 (n 2)K s /3... K n K s /n, the preceding equation becomes ν nk s [A] + n(n 1)K2 s [A] nk n s [A]n 1 + nk s [A] + n(n 1)K 2 s [A]2 / K n s [A]n nk s [A] {1 + (n 1)K s [A] K n s 1 [A] n 1 } 1 + nk s [A] + n(n 1)K 2 s [A]2 / K n s [A]n The coefficients are the binomial coefficients, ν nk s [A](1 + K s [A])n 1 (1 + K s [A]) n nk s [A] 1 + K s [A] which are the expressions obtained in Problem 49. To test the equation, plot 1/ ν against 1/[A]: 1 ν 1 n + 1 nk s [A] One of the intercepts is 1/n. Alternatively, plot ν against ν /[A]: ν ν n K s [A] If K n is very much larger than K 1, K 2, and so on, the equation obtained in Problem 51 reduces as follows: ν n(k 1 K 2... K n )[A]n 1 + (K 1 K 2... K n )[A] n nk[a] n 1 + K[A] n where K K 1 K 2... K n is the overall equilibrium constant for the binding of n molecules: na + M K MA n The fraction of sites occupied, θ, is ν /n: θ K[A] n 1 + K[A] n or θ 1 θ K[A]n

23 94 n CHAPTER 4 The slope of a plot of ln {θ/(1 θ)} against [A] is thus n. If the sites are identical and independent (Problem 50), the slope is 1. Intermediate behavior can give nonlinear plots; the maximum slope of a Hill plot cannot be greater than n.