Question

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14. At a certain temperature the equilibrium constant, Kc equals 0.11 for the reaction: 2 ICl(g) = 12(g) + Cl2(g). What is the equilibrium concentration of ICl if 0.45 mol of 12 and 0.45 mol of Cl2 are initially mixed in a 2.0-L flask? A) 0.34 M B) 0.14 M. C) 0.27 M D) 0.17 M

Answer 1

moles of Iz Given , ke = 0.11 Initial concentration of Iz = [[2] = at = 0, 45 mol 2 l = 0.225M Initial concentration of che a moles of cl volume 0.45 mol - 2 l 0.225M 2 Ic lg - Initial 0 = Iz (9) tel. 0.22 0.225 concentration Gredel - a reallag - 2 0.225-x 0.225-x change +20 Equillitrium an concentration Kc = [I] [ch ] _ (0.225-2) Co. 225-2) (2012 (0.225 - & - Ke = 0.11 K – 0.11 (2x2 01225-x – Joill = 0.33166 2x

- 0.225-x - 0.331.66 = = 0.225 - x 0,225 - x = 0,33166 X 2n 0.663 32 x + x = 0.225 0.66 332 x 1.66332X = 0.225 a = 0.225 1.66332 x = 0.13 527 concentration of ICI at equilibrium = 2x = 2x 0. 13527 -0. 27054.M. = 0.27 m ( by rounding nearest i. The correct option is (c) figures)
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