construct the ICE table
        H2(g)
+Â Â I2(g) <=>Â Â 2 HI(g)
IÂ Â Â Â Â Â Â Â
1.43Â Â Â Â Â
3.15Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0
CÂ Â Â Â Â Â Â Â
-x        Â
-x              Â
+2x
EÂ Â Â Â Â Â Â
1.43-x     Â
3.15-x       +2x
Kp = [HI]2 / [H2][I2]
but he has alrready given equilibrium pressure which is 1.844
atm
[HI] = 2x = 1.844 atm
x = 1.844 / 2 = 0.922 atm
[H2] = 1.43-x = 1.43-0.922 = 0.508 atm
[I2] = 3.15-x = 3.15 - 0.922 = 2.228 atm
now we have all equilibrium concentratiions
Kp = [1.844]2 / [0.508][2.288]
Kp = 3.0043
