Question

Suppose a 500. ml flask is filled with 0.10 mol of H, and 1.0 mol of HCl. The following reaction becomes possible: H2(g) +C1, (g) = 2HCl(g) The equilibrium constant K for this reaction is 4.69 at the temperature of the flask. Calculate the equilibrium molarity of HCl. Round your answer to two decimal places. x 6 ?

Answer 1

reaction J. H2(g) with Cl2 H2(g) + clalu) = 2HUID) initial mol g H2 = 0.10 mol volume og flask = 500m ImL = 0.0014 500ml = 0.52 Molarity J H2 = mol volume = 0.lomol = 0.2M 0.5L

mole J Hel=10 molarity = lomol = 2M 0.52 H₂ (g) + (212) 224029) initial 0.2 0 2 change - mol og HU = 1.0 mol Molarity = 10 mol = 2M 05 la since concentration of ne is greater than ha, present so reaction proceed backward and so no or left

2 Hully) = 2.0 H219) +4219) 0.2 o kl= = = 0.213 4.69 inisial (M) Change X+0 2 2 2.0-22 equilibrium K'= [he] [che] [u 0.213 = (2+0.2) (x) (2.0-2x)² 0.213 - + 0.22 4 +482-83 0.852 + 0.854x? – 1704 x = +0:2x 0.1482? + 1.904 X - 0.852 = 0 Compare this equation with ax²+bx+c=0 a= 0.148 b = 1.904 (= -0.852 = -b £ 5b2-4ac 20 = -1.904 — 51-904-4X0.148 X-0-852 2X0.148 = -1.904 I 3.63 +0.504 2X0.148 = -1.904 4.134 2* 0.148

X= 0.43 At equilibrium [nce]= 2.0-20 = 2:0 - 2x0.43 = 1.14 = [nce] =1014m Answer