Question

I got the first one correct but not the last 2.

The equilibrium constant K for the reaction below is is 7.755x105. CODE 4th attempt Part 1 (1 point) W See Periodic Table The initial composition of the reaction mixture is [C] = [D] – [E] = 1.7110x103 M. What is the equilibrium concentration of C? 0.00303 M Part 2 (1 point) What is the equilibrium concentration of D? 0.000386 M Part 3 (1 point) See Hint What is the equilibrium concentration of E? 0.000386 M > 3rd attempt

Answer 1

C initially D + E [c] = [0] = [E] ' 1•7110 x 10-3 M. KES Oc = [DJ [E] - (1•7110x10-3)2 H [c] - Clo7H0 X10-3) -1 E 107110 X 10-3 o Oc > Ke, So reaction will shift backward. a c = 10 + E (1-7110 x 10-3)M (167110X10-3)m (197110x10-3) M +x D -a jhiis cx (107110 X 10-3 (1•7110 X10-3 (107110X10-3 CS Fxgo with -x) _CamScanner DI -3) I с E

= [DJ TE] [C] (1:7/10 x 10-3-212 (107110X10-3+2) L OL 01 *0-3 70755 x 10-5 = x2 - 3,4220x + 2.927521 X 10 (107110X10-3+2) Solving for a we get x = 102331 X10-3M Ol111) otti in ós [c] = (1-7110 x 10-3 + 1.2331 x 10-3)M = 3.944 1 rep x 10-3M ! = 0.003944 M - [D] Tv (107110 x 10-3 -2)M. 4,78 x 10-4M 10.000478 M [E] - = ra Scaạned with CamScanner (197110x10-3-2) M 0.000478 M | ...).
Hope you will like my answer.