Question

courses/1379638/quizzes/2202917 YULJUNTO) 1 pts Question 7 We use an ICE table to solve for equilibrium concentrations when given initial concentrations and the equilibrium constant, K. For each case below, determine the best approach to solve for x using the value of K and the expression of K with x. "X" represent the smallest positive change. K expression Kvalue Solving approach Case KE (0.5-2)2 0.64 [Select] Case K= 0.80 [Select] K = 0.86 [Select] 1.3 x K = 0.5-1 [Select] 107

Answer 1

8 We use an ICE table to solve for equilibrium concentrations when given Intial enneentrations and the equilibrium constant, K. for each ease below, determine the best approach to solve son & using the value of Kand. the expression of k with x. " represent the positive change. x" represent the smallest Kwalue Kexppession I solving approach Case L K = 15 0.54 ease 2 0.80 case 3 0.86 0.5-n ku? K- 0.5-20 case 4 / 1.3x10-7 Ams) Case 1 K = 6.5-x)2 -> (0.3x)² x = x² -) (0.25 + x²-x) k = x² =) x² (1-K) + ka. 0.25k=0 Now, k=0.64 x² (1-0.66) + 0.64x- 0.25 x 0.66 = 0 =) 0.36x² + 0.64n-0.16 = 0 - --0-64+/0.44 +0.23 0.72 = -0.64 & 0.8 Positive 0.72 change considen -0.72 0.16 0.22

Case 2 k- a che = o.sk-ek erre-ost-o & ( 1 ) - 0.5% - na 054 1 * * Now 0.2 0.540.2 - 0.22 Case 3 - & + 2x - 4005)=0 - 2 + vr – 0.5x=0 Now, 14 = 0.26 +0.867.-16.50.44) = 0 x +0.264 - 0.13 = 0 = -0.467 0944 172 0.16 12.45 0.71 = 0.355

Case 4 0.5-X -) + kr-0.5% =0 Now, k-1.3 2107 = x + 1.3x10x - 8.5x105 -0. > *=-1-3-10-254-693 1075 +2.6*10-7 -) = -1-3x102 6x-7 --1-3 *18*45.1x10-5 Only positive taken value 2 = 5-037*10-4 2 - 2.55x100 The ake of x for different case are - for cose , k = 0.64 and x = 0.22 son case 2 x=0.80 and x=0.22 son case 3, 4= 086 and n=0.355 foo vse 4 x = 13x107 and a-2.55x10-4