For the equilibrium below at 400 K, Kc = 7.0. Br2(g) + Cl2(g) 2 BrCl(g) If (initially) 0.31 mol of Br2 and 0.31 mol Cl2 are introduced into a 1.0 L container at 400. K, what will be the equilibrium concentration of BrCl? Assume the initial product concentration is zero.
Explanation / Answer
initial concentration of Br2 = initial concentration of Cl2 = 0.35 mol/ 1.0 L = 0.35 M Br2 + Cl2 < => 2 BrCl start 0.35. . 0.35 change -x .. . . . -x .. . . .+2x at equilibrium 0.35-x. .. 0.35-x. . .2x Kc = 7.0 = [BrCl]^2 / [Br2][Cl2] = (2x)^2 / (0.35- x)(0.35 - x) = (2x)^2 / (0.35-x)^2 if we take the square root of both sides of this equation we get 2.6 = 2x / 0.35-x 0.91 - 2.6 x = 2x 0.91 = 4.6 x x = 0.20 [Br2]= [Cl2] = 0.35 - 0.20 = 0.15 M [BrCl]= 2x = 2 x 0.20 = 0.40 M