Question

Suppose a 500. ml flask is filled with 0.90 mol of NO, and 0.80 mol of NO. The following reaction becomes possible: NO2(g) + NO(g) + 2NO, (g) The equilibrium constant K for this reaction is 8.41 at the temperature of the flask. Calculate the equilibrium molarity of NO. Round your answer to two decimal places. IM x 3 ?

Answer 1

k = 8.41 volume of flook - 500 ml = 0.52 moles of No₂ = 0.90 mol. molarity of No = 0.gonal - 108M O.SL mobs of No = 0.80 mol. molarity of no= 0.8o me a loom. ongl now - NO₂ (g) + No 1g) = 2 No, cg): K= 844 1,8 16 0 Initially: 4+ equilibrium (1.800) (1.60) 20 [No 72 [HQ] x [ro] (2x)2 (118d) (1.60) 28.9= =) (8.91) (1.80) (1.6-2) 2902 2 24.22 - 28.600 & 8.910 2.902 :) 9.1122 - 28.604+ 24.22 20 a=548 or asl & Connet be greaks than 1.

I ho = of to lep uspe Lens, ugnmbo hoo equilibrium molarity of No₂ = 28