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CH9_Page_15

# CH9_Page_15 - 9.515 The initial concentrations of'Nz and H...

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Unformatted text preview: 9.515 The initial concentrations of'Nz and H; are [N,] = [1-1,] = MD m”! = noose mol+L'1. ZSJElL At equilibrium, 5.12} “In of the N2 had reacted, so 954‘.) % of the N2 remains: [N2] = {GBSGHRGHEU inol - L'l} = {1.01376 niol - L‘1 If it) “is reacted, then H . nose x 9.2m mol N, 3-: m5- = nose mol NH, formed. Inol NE The concentration of NH, formed = % = 8,0 x ICIA1 mol -L'1. The amount of 3 mol H, H2 reacted = {MSG x {1.200 mol N, x mol N, = [MEI] mol H, used. Concentration of H2 present at equilibrium __ UJOD mol — {11.0311} mo] _ 25.0 L = ﬂﬂﬂﬁs mol - L". [NH,]2 _ (SDXIIT‘F r: =——3-—i=3r”192 [N,][H,] (streamlines) 9.53 Note: The volume of the System is not used because we are given pressures and K. Pressures {har}N2[g) + 31mg) ,—_i smug) initial ems sols ii change —x —3 3: +23: ﬁnal 0.025 —x sills — 3x +2x P 2 2 K “"3 = \$— = nose 2 P“, is; [0.025 — Jemima — ix)? Solving this Expllﬂltlj" will lead to a high—order equation, so ﬁrst check to see if the assumytion that 3x {5: I101 5 can be used to simplify the math: 2T1 ...
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