Question

Suppose a 500. ml flask is filled with 1.7 mol of Cl, and 2.0 mol of HCl. The following reaction becomes possible: H2(g) +C12(g) + 2HCl (8) The equilibrium constant K for this reaction is 7.05 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places. x o ?

Answer 1

H₂(g) + Cl21g) ₃ 2011 (9) volume of flask= soom2 (0.500L) initial mol of cl2 = 107 mol initial mol of Hel= 2.0mor molarity of Cl2 = 1.7 mol -= 0.500L 3.4m molarity of HCl = zoomol = um 0.5OOL using ICE table I(M) Haug) + Cl2(g) 22H019, o 3.4 4.0 x x tax (-x) (3.4-X) (ut2x) +2x ccm Elmo cm) ke= [HI] 2 . - tos - [Hz] [2] (4+2x) (-x) (3-4-x) - 7:05 CS Scanned with CamScanner

CW)+(2x)*+ 2(1)(2x) = 7:05 (3.4-X) (-X). 16+ 4x²+ 16x = -23.97 + 7.0502 4x² 70054² + 16x + 16 + 23.97 =0 -3.05X? +1GX +39.97-0 X-5-24- 13.10 :0 it is a quadratic equation (ax²t bx tc=0 a=1, b = -5.24 C= -13.10 -btJ6zual -(-5.24) 1 JL-5-24) Eu (1) (-13.00) 29 2X1 s.24 + /27.45 +52-4 Xe 5.24 + 8.93 2 x= 5.24-8.93 = -1.8um so, equilibrium molarity of Hel = (4+ 2x) m 4721-6.84) - (4-3.68) m = 0.32m a scasold equilibrium CamScanner molarity of Hel is 0.32m