Exam 1 p5 - 13 ’A 1.00 flask was filled with 2.00 mol...

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Unformatted text preview: 13. ’A 1.00- flask was filled with 2.00 mol gaseous SO; and 2.00 mol gaseous N02 and heated. e quilibrium was reached, it was found that 1.30 mol gaseous NO was present. Assume that the reaction S02(g) + N02(g) 5 503 (g) + NO (g) occurs under these conditions. Calculate the value of the equilibrium constant, K, for this reaction. (Set up an ICE table.) (Ccolol aléo use L'cvm.> ind» .- 1.00 not” 1.60 me D 0 Slaw. ‘Valume is Eiflul : 2 7 7 (.90 “91 amen away .. 9‘. fi 94 + x + x 9;“ 1212111 Sam ‘- (1‘ 9° " 7‘) (i'oc' x) x j: ragezé—duiis k; [50,] Lug] : (mm 2 L10}; x=1.30' [501'] [“01] (1.00 — «$1 07° 7. 3-H 14. At a particular temperature, K = 1.00 x 102 for the reaction below. In an experiment, 1.00 mol H2, 1.00 mol 12, and 1.00 mol HI are introduced into a 1.00-L container. (a) How do you know in which direction the reaction will proceed? gl‘uce 4“ “WK” V “Minimal Q= ”Mimic-ow“ Vflflihlfl (b) Calculate the concentrations of all species when equilibrium is reached. L00 L H2(g) + 12 (g) * 2H1 (g) <— Q 4- k (Lac 4.100) Thar ore reaction Must Proaed {—9 Hm: FISH—- +0 Tmu’eose numerator LProgluct duct) ' \u:*‘. 1.00 M 1.13054 {.00 g; (anaefl. _. 9; -x + 2x Chgmutse of, 1:121 fah‘o ‘ o co ‘ ‘den'i' S141)“ (Loo’ydfl (l‘°°"7‘)fl (l.oo+1%)fl i“ 249‘ 9) 7. “/i ‘ 19(3 K : Ll.oo+2.ac) OK «0.; :- (Lou— Cwouflz’ (“00’7‘3 \0'0 (:Loc—qb :. (.00 +27c (‘00 +2X CH1]: o‘lsfl “3,0 , max. '4 01.0 __ [2.07; [“111 ; 0.2.5131 x : 9.75 EH1] ; 2.5g ...
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