HW-Solns-Chap-6-5200

# S62 a p o 30 bar for simplicity pcl 5 a pcl 3 b cl 2

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S6.2 (a) p o = 3.0 bar. For simplicity, PCl 5 = A, PCl 3 = B, Cl 2 = C A B + C Equil. Press: (1-x)p o xp o xp o 2 ( )( ) (1 ) 1 B C o o o A o p p xp xp x p K p x p x p o x 2 + Kx - K = 0 = ax 2 + bx - c 2 2 2 4 ( ) 4.5 (4.5) 4(3.0)( 4.5) 4 2 2 2(3.0) o o K K p K b b ac x a p   4.5 74.25 0.686 6.0 x 5 (1 ) (1 0.686)(3.0) 0.942 0.94 A PCl o p p x p bar bar 3 (0.686)(3.0) 2.058 2.06 B PCl o p p xp bar bar 2 (0.686)(3.0) 2.058 2.06 C Cl o p p xp bar bar

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(b) T = 150 o C = 423 K G o = -RTln(K) = -8.31 J/mol-K(423 K)ln(4.5) = -5290 J 3 2 5 ln( ) ln PCl Cl o o PCl P P G G RT Q G RT P     (2.2)(2.2) ln( ) 5290 (8.31)(423)ln 5290 6330 1040 10.4 0.8 o G G RT Q J kJ           S6.3 p init (A) = p init (B) = 3.0 bar p init (C) = 0.0 bar 2 A + B 2C Equil. Press: 3.0-2x 3-x 2x   2 2 2 2 3 2 2 2 (2 ) 4 4 1 10 (3) (3) 27 3 2 3 C A B p x x x K x p p x x 3 27 (1 10 ) 0.082 4 x x p C = 2x = 0.164 bar 0.16 bar S6.4 K 1 = 337 at T 1 = 100 o C = 373 K K 2 = 7.1x10 -5 at T 2 = 500 o C = 773 K (a) 2 1 2 1 1 1 ln( / ) o H K K R T T   5 2 1 2 1 ln( / ) 8.31ln(7.1 10 / 337) 92,080 / 92.1 / 1 1 1 1 773 373 o R K K x H J mol kJ mol T T     1 1 ln( ) o o H S K RT R   Therefore, 1 1 92,080 ln( ) 8.31 ln(337) (8.31)(373) o o H S R K RT 8.31( 23.9) 198.5 / J mol K  
(b) H o = -92,080 J/mol , K 1 = 337 at T 1 = 100 o C = 373 K , K 3 = ?? at T 3 = 300 o C = 573 K 3 1 3 1 1 1 ln( / ) o H K K R T T   3 92,080 1 1 ln( / 337) 10.37 8.31 573 373 K     10.37 5 3 3.14 10 337 K e x K 3 = 0.0106 0.011 (c) N 2 + 3H 2 2NH 3 Equil. Press: 3.0-x 2.0-2x 2x 3 2 2 2 2 2 2 5 3 3 3 (2 ) 4 4 7.1 10 (3.0 )(2.0 3 ) (3.0)(2.0) 24 NH N H P x x x K x P P x x 5 24 (7.1 10 ) 0.0206 4 x x P NH3 = 2x = 0.0412 bar 0.041 bar (d) T = 500 o C = 773 K K = 7.1x10 -5 5 ln (8.31 / )(773 )ln(7.1 10

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