Question

cs 6. The equilibrium constant for the reaction in question 2c is K. = 721 M . (a) Assuming this reaction is at equilibrium, calculate [I] when [12] = [1:1 = 0.0010 M. (b) Assuming this reaction is at equilibrium, find [I:-) when [12] = 0.0025 M and [1] = 0.010 M. (c) Assuming this reaction is at equilibrium, find [12] when [13"] = 0.00030 M and [1'] = 0.10 M. (d) Before the reaction reaches equilibrium, a solution has [15] = 0.0020 M, and [1"] = [12] = O M. Calculate the concentrations of all three species after the solution reaches equilibrium. (Assume that x M of triiodide dissociates to iodide and iodine, and solve for x.)

Help please! Show work

Answer 1

Iz + Is Ij 12+ I" = 0.0020 M gy 0.0020-% e at equilibrium; L] = [B'] = 0.00101 Ke= [13] 103 (L') [?] . 10-3 *[") 721 M1 = 1 a [1] (BJ (1) - Ke= 721 = (0.0020-2) (2) () (I) = 1 1 2 (17) = 0.00138M 721 = 0.002- 721242 + x -0.002 = 0 an of bet cao x= -b I ✓ 62-49c by at equilibrium; (Iz) - 0.002SM (P) = 0.0LOM Ret (13) . [1] (101.3 0.010) (0.0025) - za -1 12+4*721% 2X103 2x721 7 721 mm x 102x 25x10 4 m2 2 = -14 2.601) 1442 I only positive value acceptable) (33= 0.0180 M) c) x = 0.0011 M at equilibrium (1, 1 = 0.00030 m, (19= 0.10M Ke= [12 (1) (2) . 3x10²4 107 x (12) after equilibrium (1,]: x = 0.0011.my (I ") =2 0.0011 m (23), 0.0020-23 0.0009m 721 = - 3X10 4x10 (2) = 4.16 X106m