Question

Consider the reaction. A(aq) - 2B(aq) K-1.14 x 10-6 at 500 K If a 4.20 M sample of A is heated to 500 K, what is the concentration of B at equilibrium? B-

Answer 1

consider ICE table table ne Alag) 2Blag) 1 4. 2 0 с +2x o 4.2- 2x I ke = [B]²/CA] 1.14 x10-6 = (2x)² (4.2-0)

1.14x10-6(4.2-x) = 4x2 0.0000048 - 0.00000114 * = 4*? 4x² to.ooooo 1148 -0.0000048=0 - by 4, we get x² +0.000000285x -0.00000 12 = 0

Using the we quadratic equation , x = -b I √62-4ac get 2a = - 0.000000285 + 1 (.000000285)2 -4*1*-0.000 = -0.000000 285 t o.000000000000081225 + 0.0000048

= -0.000000285 0.0000048 2 = -0.000000 285 + 0.00219 2 -0.000000285 + 0.00219 2

--0-OOUUO0285 +0.00219 Inc -0.030OUO 285-0.00219 2 = 0.00219 = -0.00219 2 2 - Ø11rbreeza pece 0.0011 - 0.0011 x = 0.0011 and -0.0011

i.x=0.0011 because x can't be negative i equilibrium concentration a B = 2x0.0011 = 0.0022 i equilibrium concentration of A=4.2 -0.0011 = 4.1989
[B] = 0.0022 M