I know the reaction will shift to the right to regain equilibrium, but i\'m havi
ID: 495020 • Letter: I
Question
I know the reaction will shift to the right to regain equilibrium, but i'm having trouble setting up an ICE table. Please help.
Some CoBr2 s allowed to dissociate into Co and Br2 at 346 K. Att equilibrium, CCoBr2] 0.115 M, and Ico Br 0.147 M. Additional COBr s added so that COBr new 0.181 M and the system is allowed to once again reach equilibrium. 2 CoBr20g) CO(g) Br2 (g) K 0.190 at 346 K (a) In which direction will the reaction proceed to reach equilibrium? (b) What are the new concentrations of reactants and products after the system reaches equilibrium? [CoBr2] M M CO [Br2]Explanation / Answer
COBr2 (g) <-----------> CO (g) + Br2 (g)
0.181 0.147 0.147 (initial)
0.181-x 0.147+x 0.147+x (at equilibrium)
Kc = [CO] [Br2] / [COBr2]
0.190 = (0.147+x)^2 / (0.181-x)
0.0344 - 0.190*x = 0.0216 + x^2 + 0.294*x
x^2 + 0.484*x - 0.0128 = 0
This is quadratic equation (ax^2+bx+c=0)
a = 1.0
b = 0.484
c = -0.0128
Roots can be found by
x = {-b + sqrt(b^2-4*a*c)}/2a
x = {-b - sqrt(b^2-4*a*c)}/2a
b^2-4*a*c = 0.285
roots are :
x = 0.025 and x = -0.509
but x can't be negative
so,
x = 0.025 M
[COBr2] = 0.181 - x = 0.181 - 0.025 = 0.156 M
[CO] = 0.147 + x = 0.147 + 0.025 = 0.172 M
[Br2] = 0.147 + x = 0.147 + 0.025 = 0.172 M