Determine KeandK,at 298 K for the reaction:NOBI(g) = % N2(g) + ½ O2(g) + % Br2(g)from the following information at 298 K:NO(g) = % N2(g) + ½ O2ig)Ke = 1.55 x 1015½ NO(9) + % Br2(g) = % NOBR(a)Kp= 2.63Type your answers below.Note: If answering in scientific notation, type your answer this way:1.23 x 1045 → type 1.23e45ANSWERS:Kc =Kp =

Answered: Image /qna-images/answer/9f379e84-ccb7-4201-9002-ed61b22ec0f8.jpg

candkpe NOBR(g) = % N2(g) + O2(g) + % Br2(g) from the following information at 298 K: NO(9) = % N2jg) + % O2ig) % NO(g) + % Br2(g) = % NOBR(ga) Ke = 1.55 x 1015 Kp = 2.63 Type your answers below. Note: If answering in scientific notation, type your answer this way: 1.23 x 1045 → type 1.23e45 ANSWERS: Kc = Kp =

candkpe NOBR(g) = % N2(g) + O2(g) + % Br2(g) from the following information at 298 K: NO(9) = % N2jg) + % O2ig) % NO(g) + % Br2(g) = % NOBR(ga) Ke = 1.55 x 1015 Kp = 2.63 Type your answers below. Note: If answering in scientific notation, type your answer this way: 1.23 x 1045 → type 1.23e45 ANSWERS: Kc = Kp =

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 89AP

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