............2IBr ==> I2 + Br2
I........ 0.0124......0....0
C...........-2x.......x....x
E........0.0124-2x....x....x
Keq = 4.13E-5 = (I2)(Br)2/(IBr)^2
Substitute the E line into Keq expression and solve for x = (Br)2 = (I2)
At a certain temperature, Keq is 4.13 x 10^-5 for the equilibrium:
2IBr(g) ↔ I2(g) + Br2(g)
Assume that the equilibrium is established at the above temperature by adding only the
reactant to the reaction flask. What are the concentrations of I2 and Br2 in equilibrium with
0.0124 mol/L of IBr
1 answer
1 answer