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   ocr: CHAPTER 15 Solutions to Exercises Chemical Equilibrium 15.40 (a) A(g) 2B(g) initial 0.75 atm 0 atm change -0.25 atm +0.50 atm equil. 0.50 atm 0.50 atm Pr=PA+Pe- 0.50 atm + 0.50 atm = 1.00 atm (b) Kp PA (Pa)?. 0.50 (0.50)2 = 0.50 (c) K- Kp (RT)an , An=+1, T-0C-273-273K 0.50 = 0.022 Kc = (0.0821 X 273)* 15.41 (a) Kp = PNz PNHS XPA, ll 4.34x10-3 PNHa = gRT MV - 17.03 0.753 9 g/mol X 0.0821 Kin L. mol atm X 573K T.000L = 2.08 atm N2(g) + 3H219) G 2NH3g) initial 0 atm 0 atm ? change X 3x -2x equil. X atm 3x atm 2.08 atm (Remember, only the change line reflects the stoichiometry of the reaction.) Kp ...