Question

A . EE AU AaBbCctde AaBbCcD pri (B... 11 A A Aa I Uab x x A A REL AaBbcc de No Spacing AaBbCcDdE Heading 2 = = Normal Heading 1 Styles Pana Consider the reaction Alag) + Blag) Claa) + Diag), which has an equilibrium constant, K, equal to 3.4. 1. If one begins a reaction by placing 0.500 moles of A in a 1.0 L container as well as 0.125 moles of B, what will be the equilibrium concentrations of A, B, C, and D? Write your answers in the spaces provided below. a. [A] = b. [B]=_ c. (C)= d. [D] 2. Once the reaction in problem 1 reaches equilibrium, 0.500 moles of additional B is injected into the flask from an outside source. What will the new equilibrium concentrations become? a. [A] = b. [B]= _ c. [C]= d. [D]=_ rds DE English (United States) Focus

please help me out with #2

Answer 1

flags & Blag, - Cags & Diag) Keg = 3.4 Volume = 1 L Ang 0.5 Caps & Dlags at initial & Beag 0.125M - at equilibrium 0.5-1 0.125-n. So, Keg = x.x (0.5-x) (0.125-2) 34 = x² 0.0625-0.625x+x? 3.4(xc² -0.6252 +0.0625) = x² 3.4 x² - 2.1252 +0.2125= x² (3.4-1) x² - 2.125x +0.2125=0 2.4x2 - 2.125 x +0:2125 = 0 So, x = -42.125) + ][-2.125)2 - 4X24X0:2125 2x2.4 - 2.125 I 2.475625 4.8 x = 2.12511:573 4.8 x=0.77 50,0.77 is greater 0.115 than it is greater So, n=0.115. than concentration initial initi

Concentration at equilibrium [A] = 0.5-0.115 = 0.385 M. [B] = 0.125-00115 0.010M [c] = [0] = 0.115M Alag & Blogs - Clags + Diagn o Now, 0.5 mole of B is added to flask So, new concen- tration become. 0.385 0.51 0.115 0.115 O c = 0.115X0.115 - 0.067 0.385 X0.51 So, Qc <Kc Reaction goes in forward direction. At equilibrium 0.385-4 osty 0.listy - 1 0.15ty Keg= (0:115+y)(0.15ty) 10.385-4) (0.51-9) 3.4 = 0.013225 + 0.23y + y² 0.19635-0-895y + ya 34 (0-19635 -0-8954 +42) = 0-013225 +0.234 +y? 0.66759 – 3.0434 + 3.4y2 = 0.013225 +0.23y + y2 3.47- 3.2734 +0.654365 = 0 = 0.039 25 +0.23 O 013225 +0. 23

3:4y? - 3-273y +0.654365 = 0 y = -(-3.273) = }(-3.273)2 + 4x 2.4x0-654365 - 2x2.4 3.273 + 4.430625 4.8 9. 3.2734 2.105 4.8 yo? 3.273+2.105 , 4. 8 9 Š - 3273-2.105 9 = 3.273-2.105 4.8 y = 112 y = 0.24 So, y = 1.12 is greater than initial concentration of B. Hence, we have to take y = 0.24. Now Concentration at nog new equilibrium. [A] = 0.385-y = 0.385-0.24 = 0.145 M (Mamolas = mol L [B] = 0.51-y = 0.51-0.24 = 0.27 M [c] = 0.115 +y = 0:115+0-24 = 0.35519 [D] = 0+15+y = 0-115 +0.24 = 6-355 M )