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 4.1. A Y + Z Initial amounts: mol Amounts at equilibrium: mol Concentrations at equilibrium: mol dm 3 K c (1.5/5) (3.0/5) (1/5) mol dm 3 (1.5) (3) (5).7 mol dm 3 mol dm A + B Y + Z Initial amounts: mol Amounts at equilibrium: 1 mol 0.1; ( ) Thus, initially there must be 4 mol of A A + B Z Initial amounts: 4 0 mol Amounts at equilibrium: 1 1 mol Concentrations at equilibrium: mol dm 3

2 CHEMICAL EQUILIBRIUM n 73 1/5 [( 1)/5] (/5) ( 1) Thus, initially there must be 6 mol of A Two moles of SO 3 (g) produce 3 mol of product; thus Σν +1 mol. Then, from Eq. 4.6, K P K c (RT) +1 (0.071 mol dm 3 ) ( J K 1 mol K) 47.8 J dm J m Pa.48 bar 4.5 I I Initial amounts: mol Equilibrium amounts: ( ) mol 4 ( / 05.) K c / mol mol dm mol dm 3 K P mol dm 3 ( J K 1 mol K) 0.8 Pa bar 4.6. Addition of N at constant volume and temperature necessarily requires the equilibrium to shift to the right. [NH 3 ] n NH3 V K c [N ][H ] 3 n N nh 3 If n N 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 requires that V is increased. If the proportional increase in V is greater than the increase in n N the equilibrium will shift to the left when N is added. The volume V is proportional to n N + n H + n NH3, and V is proportional to (n N + n H + n NH3 ) If n N is very much larger than n H + n NH3, V will increase approimately with n N and therefore increases more strongly than n N. If n N is not much larger than n H + n NH3, an increase in n N will have a relatively smaller effect on V. The increase in ammonia dissociation when N is added is therefore epected when N is in ecess, but not otherwise.

3 74 n CHAPTER 4 On the other hand, n 3 H appears in the equilibrium epression; this varies more strongly than V, and added H therefore cannot lead to the dissociation of ammonia Suppose that in 1 dm 3 there are mol of N O 4 y mol of NO Pressure bar ( + y)rt/v ( + y) mol bar dm K mol K 3 1 dm Therefore, + y (1) Density g dm 3 ( y) g ( + y) g 1 dm 3 1 dm 3 Therefore, + y () From Eqs. (1) and (), we get ; y N O 4 (g) š NO (g) P(1 α) Pα Since partial pressures are proportional to the number of moles of each species present, P(1 α) and y Pα, which means that y from which we get α K c ( ) Pα P(1 α), y ( + y) mol dm mol dm 3 K P K c RT (from Eq. 4.6, with Σν 1) 0.03 mol dm dm 3 bar mol bar K K P P 1 (from Eq. 4.3) 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 ) ( mol dm 3 ) ( mol dm 3 ) mol dm 3 K P K c RT (from Eq. 4.6) mol dm dm 3 atm mol bar K K P (0.355 bar) 1 (from Eq. 4.3) Suppose that, if there were no dissociation, the partial pressure of COCl was P; then the actual partial pressures are COCl (g) CO(g) + Cl (g) P(1 α) Pα Pα P + Pα bar With α P bar 1 + ( ) bar K P ( ) [1 ( )] ( ) bar bar K c K P (RT) 1 (from Eq. 4.6) bar ( dm 3 bar mol 1 ) mol dm 3 K K P P 1 (from Eq. 4.3) bar ( bar)

5 76 n CHAPTER H + I HI Initially: mol At equilibrium: 1 3 mol After addition of mol H : 3 3 mol K 3/ (3 ) K 4(3/) (3/) g iodine 0.05 mol I When all of the solid iodine has just gone, the iodine pressure is 0.10 atm. The consumption of 0.05 mol I leads to the formation of 0.10 mol HI, which eerts a pressure of P HI 0.1 mol dm3 bar K 1 mol K 10 dm bar Then, if P H is the partial pressure of H after equilibrium is established, (0.60 bar) 0 P 0.10 bar H P H bar bar 10 dm n H bar dm 3 mol mol Thus, mol of H is produced in the equilibrium miture, and 0.05 mol of H is required to remove the 0.05 mol of I. Therefore, mol of H must be added Assuming that we start with one mole of N O 4 (g), N O 4 (g) š NO (g) 1 α α We obtain (1+α) moles of gas at equilibrium. Therefore, equilibrium partial pressures are 1 α α PNO P; P 4 NO P, 1+ α 1+ α and

6 CHEMICAL EQUILIBRIUM n 77 4α 1 P c P 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 1.81, and at 6.18 bar, K P 1.45 bar, K c mol dm 3, K Rewriting the reaction in terms of one mole of HCl, we get HCl(g) O (g) 1 Cl (g) + 1 H O(g) (1 y) P O y/ y/ From eamining the equation above, it is possible to establish the following relationships: Cl HCl y ; (1 y) HO Cl 1. It is important to remember that the same ratios hold for partial pressures also. Now, 1/ Assuming ideal behavior, the partial pressure of oygen is 0.51 bar. Therefore, Since no reactants are initially present, we write PCl P H P Cl y 1 K P. 1/ 4 1/ 4 1/ 4 P P P P (1 y) P K P HCl 1/ O O HCl.76 1 (1 0.76) 0.51 O 0 1/ bar 0.5 O H (g) + I (g) HI(g) n. where n is the initial amount of HI: n 10.0 g/(17.91 g mol 1 ) mol. ( n ) n Since KP 65.0, or 65, we can solve for. The solution is mol. Therefore, the equilibrium mole fractions are H I /n ; HI (n )/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) C 6 H 5 COOH Initially: mol dm 3 At equilibrium: 0.1 mol dm 3 (0.1 ) ± ± 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) 0. (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 ( 16.63) 0 (3 0) 33.6 kj mol 1 K P ep( G /RT) bar b. G kj mol 1 K P bar c. G kj mol 1 K P bar 1

8 CHEMICAL EQUILIBRIUM n 79 d. G ( 50.7) 68.6 kj mol 1 K P K P Σv K c K P (RT) Σv K 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 O CO + H Initially: 0 0 mol At equilibrium: mol K P K c ( ) (Note that the total volume cancels out and so need not be considered.) ; The amounts are therefore mol (CO); mol (H O) 4.. For reaction (1) mol (CO ); mol (H ) K 1 ep( 3 800/ ) For reaction () dm 3 mol 1 K ep(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 7c 4. Therefore, c (K sp /7) 1/ mol dm 3. n Temperature Dependence of Equilibrium Constants 4.5. 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 10.6 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. H (g) + O (g) H O(g) f G f G H O f G H f G O ( 8.57) (0) kj mol 1 f H f H H O f H H f H O ( 41.8) (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(000 C)/kJ mol 1 f H f S ( ) 81.6 kj mol 1 ln (K P /bar ) K P bar 1

11 8 n CHAPTER r G f G (O 3, g) 3 f G (O, g) (163.) 3(0) 36.4 kj mol 1. K ep( r G /RT) ep[ /( )] a. S H G T 35.1 J K 1 mol b. ln K c K c mol dm 3 c. G (5 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 bar, partial pressure of n-pentane (1 ) bar ; 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 5 C, G RT ln S ( 3.36) J mol 1 H G T J K 1 mol 1 b. The difference between the reciprocals of the absolute temperatures corresponding to 5 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 5 C to 37 C ln K w is thus increased by At 5 C, ln K w is 3.44 and at 37 C it is therefore K w at 37 C is therefore mol dm 6. log 10 (K P /bar) K P.59 bar G /kj mol ln.59 G 4549 J mol kj mol 1 From Figure 4. (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.59 ( ) mol dm 3 G /kj mol ln G J mol kj mol 1 c. I + cyclopentene HI + cyclopentadiene (0.1 ) For a very approimate solution, neglect in comparison with 0.1: (0.1) For a better solution, calculate 4 3 /(0.1 ) at various values: Final concentrations are [I ] M; (0.1 ) [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 799 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.1, d J K 1 mol 1 e 10 3 ( ) J K mol 1 f 10 4 ( ) J K 1 mol 1 Then, from Eq..5, H T /J mol [(T/K) 98.15] [(T/K) ] T/K (T/K) (T/K) /(T/K) H d. ln K dt RT / K 3 ( T / K) T ( T / K) d ( T / K) (T/K) 1.55 ln (T/K) (T/K) (T/K) + I

15 86 n CHAPTER T/K 1.51 ln (T/K) (T/K) + (T/K) + I I is obtained from the fact that at 98.15, ln K I ln K T/K 1.51 ln (T/K) (T/K) + (T/K) e. ln K P at 1000 K K P (1000 K) ln Partial pressures at 1395 K are: CO: atm; CO : ( ) atm; O : atm K P ( ) (0.9999) atm bar At 1443 K, K P ( ) ( ) atm bar At 1498 K, K P ( ) ( ) atm bar Then T/K 10 1 K P /atm 10 4 /(T/K) ln(10 1 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) 09 J K 1 mol 1

16 CHEMICAL EQUILIBRIUM n Suppose that there are present mol of I and y mol of I: + y P At 800 C ( + y) mol RT V atm ( + y) dm3 atm K 1 mol dm K + y y ; y Degree of dissociation, α At 1000 C, + y y ; y ; α 0.0 At 100 C, + y y ; y ; α a. α , 0.0, 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 100 C, ( K c 3 ) mol dm 3

17 88 n CHAPTER 4 c. K P K c (RT) Σν (Eq. 4.6) K c RT since Σν 1 At 800 C, K P mol dm dm 3 m J K 1 mol K 1.16 kpa 9.1 Torr 0.01 bar At 1000 C, K P kpa At 100 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 ep(34 140/ ) The equilibrium constant is equal to unity when G is equal to zero T 170. T K G ln S J mol 1 H G T J K 1 mol 1 At 5 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 6 75 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 (l) Br (g) the vapor pressures are the equilibrium constants. The corresponding G values (standard state: 1 bar) are G ( K) RT ln 1 0 G (8.45 K) (J mol 1 ) ln J mol 1 We can therefore set up two simultaneous equations: 0 H ( K) S J mol 1 H (8.45 K) S Subtraction gives S J mol 1 /48.9 K J K 1 mol 1 At 5 C, H J K 1 mol K J mol 1 G ( ) J mol 1 31 J mol 1 The vapor pressure at 5 C is the corresponding equilibrium constant: P K P ep( 31/ ) 0.74 bar For the reaction, G kj mol 1 Then K P ep( G /RT) ep( / ) bar 1 (The unit arises from the standard state of 1 bar.) Σν for the reaction is 3 1. From Eq. 4.6, K c From Eq. 4.3, K P (RT) Σν bar 1 ( J) Pa bar m 3 mol 1 (since J Pa 1 m 3 ) dm 3 mol 1