So this is the question:
At 1,700 K, the equilibrium constant for the reaction below is Kc=0.506.
Br2(g) --> 2 Br(g)
What is the percent dissociation of Br2 at 1,700 K, if the initial concentration of Br2 is 1.75 mol/L?
This is what I did:0.506 = (2Br)^2 / (Br2)
let X = moles/L of Br2 dissociating
moles of Br = 2X/L
0.506 = (2X)^2)/ (1.75 - X)
0.8855 - 0.506X = 4X^2
4X^2 + 0.506X - 0.8855 = 0
X = 0.5486
0.5486/1.75 =
31.35%But the system keeps telling me that it's wrong!!! why?
Can someone help me here???