Question

Suppose a 250. mL flask is filled with 2.0 mol of Cl, 0.20 mol of CHCl, and 1.7 mol of CCI4. The following reaction becomes possible: Cl2(g) + CHCI, (8) HCI(g) +CCI (8) The equilibrium constant K for this reaction is 0.701 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places. OM 1 x s ?

Answer 1

concentration (M = mol volumell volume of solution = 250 mL = 0.250L 1000 mL = IL) initial concentration of Cl2 = 2.0 mol 0.250L = 8.0M initial concentration of CHCl3 = 0.20mol = 0.250L 0.8M. initial concentration of coly = 1.7mol d. 250L = 6.8 M. ICE Table concentrationcm C1219) Hallgl & 8.0 & CHCl3 19) 0.8 - se 0.8-x coly 19) 6.8 +x 6.8+0 C - x 8.0-x 30 tx x At equilibrium CEO K = equilibrium constant = 0.701 - [HC] [ccly] [cle) (CHC13] - 0.701 = x (6.8+8) (8.0-x) (0,8-2) = 0.299x2 + 12.96880 – 4,4864=0 Solving this quadratic equation we get value of x and x cannot have negahve value. Here, a= 0.299 for ax²+bx+c=0 b= 12.9688 x= -b ± √b2-490 = 20 = 0.34 M c= -4.4864 29 CHCI] = x = 0.34M Hence, the equilibrium mularity of Hel is 0.34 M Please Rate.