...................Br2(g) ⇌ 2 Br(g)
I.....................2.25........0
C.......................-x.........2x
E.................2.25-x.........2x
Kc = (Br)^2/(Br)
Plug the E line into Kc expression and solve for (Br).
Then % dissoc = [(Br)/2.25]*100
At 2,000 K, the equilibrium constant for the reaction below is Kc = 3.92.
Br2(g) ⇌ 2 Br(g)
What is the percent dissociation of Br2 at 2,000 K, if the initial concentration of Br2 is 2.25 mol L−1 ?
1 answer
1 answer