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CH9_Page_16

# CH9_Page_16 - (2x(0.025{0.015}3 “I-xI = 3.04 x 10"3...

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Unformatted text preview: (2x): (0.025}{0.015}3 “I-xI = 3 .04 x 10"3 x = 2.8 x 1075 = 0036 Comparing x to 0015, we see that the approximation was justiﬁed. At eqmlibnum, Pm! = 2 x 2.3 x10'5 bar = 5.6 x10"5 bar; the pm of N2 and H1 remain essentially,r unchanged. P 9.61} K = ”3'5 Panama 3 5x104 _1.3>-<102 ‘ (9.56)}3 1.3x102 P =—=3.9 10“]: Ct (9.sa}(3.5xlo‘;u x M 9.62 We use the reaction stoichioinetly to calculate the amounts of substances present at equilibrimn: Amounts(n1ol) {30(3) + Hzﬂtg} ;“ C02{g} + H2(g) initial 1.000 1.000 0 0 change —x +3: +x . +x ﬁnal 1.000 — .x 1.000 — x 0.665 +x (a) Because x = 0.665 mol, there will he (1.000 — 0.665}rno1 = 0.335 mol CC); 0.335 mol H20; 0.665 mol H2. The concentrations are easy to calculate because V= 10.00 L: [CD}=[H201= c.0335 moi - L"; [woo2 ] = {H2} = 0.0665 mol+L" (h) K =[co21[nz]=(n.oae.5}= =33; [COJU-IEO] {0.0335}: 9.64 The initial concentration of H25 = W = 0.0100 Incl - L" 10.0L 272 ...
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