And the slow process:rate=kES] Then we get rale=k,车Gy-ke时 k Chapter Four 14.28 Question:A 2.5-mole quantity of NOCL was initially in a 1.50-L reaction chamber at 400C .after equilibrium was established,it was found that 28.0 percent of the NOCL had dissociated: 2NOCL(g)2NO(g)+CL2(g) Calculate the equilibrium constant Kc for the reaction. Answer:The concentration of NOCL at the beginning of the Cw0=0.72*167=120C0=028*1.67*号-031Cc =号167*028=0.15 The equilibrium constant of the reaction is given as follow Kc= [ojCl_032*015=0.03 NOCL 1.22 14.40 Question:For the synthesis of ammonia N2(g)+3Hg)=2NH3(g) The equilibrium constant Kc at 375C is 1.2.starting with [H]o=[ESkrate ] And the slow process: = 2 [ ][ ] [ ][ ] SEkSE k k = krate = −1 1 Then we get 2 Chapter Four 14.28 Question : A 2.5-mole quantity of NOCL was initially in a 1.50-L reaction chamber at 400℃ .after equilibrium was established , it was found that 28.0 percent of the NOCL had dissociated: 2NOCL(g) ⇒ 2NO(g) + CL2 (g) Calculate the equilibrium constant KC for the reaction. Answer: The concentration of NOCL at the beginning of the reactions 5.1 5.2 =1.67 mol/l : when the reaction reached equilibrium 3 2 CNOCL = 0.72 * 1.67 = 1.20 CNO = 0.28 * 1.67 * =0.31 CCL2 3 1 = * 1.67 * 0.28 =0.15 The equilibrium constant of the reaction is given as follow KC = [ ][ [ ] ] 2 2 2 NOCL CLNO 2 2 2.1 O 15.0*31. = = 0.03 14.40 Question : For the synthesis of ammonia N2(g) + 3H2(g) ⇔ 2NH3(g) [H2 The equilibrium constant KC at 375 is 1.2.starting with ℃ ] 0 =