Question

The equilibrium constant, K, for the following reaction is 1.20x10-2 at 500 K. PC15(E) =PC13(E) + Cl2(E) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.157 M PCI3, 4.34x102 MPC12 and 4.34x102 M CI, What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.17x10 - mol of Cl2(g) is added to the flask? [PC13] = [PC13] = [Cl] =

Answer 1

ANSWER

[ PCl ₅ ] = 0.169 M

[ PCl ₃ ] = 0.0318 M

[ Cl ₂ ] = 0.0635 M

Consider reaction, Pal (9) = Pelz (9) + Cl (9) For above reaction, K. - [PC ] [CI ] [ Polo Let's use ICE table 2 1.20X1o - conc. (M) PC (9) 0.157 Polz (g) + ch (9) 4.34x10² 7.51×10² bH the 0.157+2 4.34x10²-20 7.51810²_x (4.348102-x) (7.51815²-x) = 1.20x102 (0.157+x) (4.34x10²_x) (7.514 10²_x) = 1.20 x 102 (0.157+2) 3.259x183_ 0.1185% + = 1.884 X10 3 + 1.20x1ó? L X2-0.1305 x + 1.375X103=0 - b ± √b²-4ac -(-0.1305) ± √(-0.1385) 2_ 4 (1) (1.375X103) 201) za X = 0 1989 on = 0.01156. We con choose x = 0.01156 M.

. Eam conc. of Pol = 4.34610 x = 4.34x10² -0.01156 = 0.0318 M. - Eq. conc. of a = 7.51x10²x = 7.51x10² _ 0.01156 = 0.0635 M EsTM cone of pels = 0.157 + x = 0.157+0.01156 – 0:16856 -0.169 M.