The equilibrium constant for the reaction I2(g) + Br2(g) --> 2 IBr(g) is 120.0. If 1.40 moles of I2and 1.40 moles of Br2 react in a 3.50 L container, calculate the equilibrium concentrations of I2, Br2 and IBr.
(I2) = mols/L = 1.40 moles/3.50 L = 0.400 = (Br2)
.....................I2(g) + Br2(g) --> 2 IBr(g) .......K = 120.0
I...................0.4...........0.4.............0
C...................-x.............-x............+2x
E................0.4-x........0.4-x.............2x
K = 120.0 = (IBr)^2/(I2)(Br2)
Substitute the E line into the Keq expression and solve for x. Then evaluate 2x and 0.4-x tp determine (ICl), (Br2) and (I2) at equilibrium. Post your work if you get stuck.