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# image005 - 9.54 Note The volume of the system is not used...

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Unformatted text preview: 9.54 Note: The volume of the system is not used because we are given pmssur-es and K. Pressures (bat)N2iE} + 3 Hgtg) ""1: 2 NH3{g} initial [1.1325 [LG] 5 i] change -.t: —3 or +2.: final 0.11125 - x t}.ﬂl‘5 - 31' +21: K PNH‘I = ——{2"‘}z—H. = {Loss = PM: PHf [11.1125 — meets a 3x)’ Solving this explicitly will lead to a high-order equation. so ﬁrst check to see if the assumption that 3:: at 13.015 can be used to simplify the math: (ll-ti2 tomsxoor 5f 4::1 = so: x to"’ x = 2.3 x 1o“ = ﬂ.ﬂ3ﬁ Comparingx to [1.015, we see that the approximation wasjustif'ied. At equilibrium, PM: = 2 x 2.8 is ill!"5 bar = 5.6 x iii":r bar; the pressures of N2 and H; remain essentialiy unchanged. ...
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