Chemistry
posted by Jamal .
A mixture of 0.47 mole of H2 and 3.59 moles of HCl is heated to 2800C. Calculate the equilibrium partial pressures of H2 Cl2 and HCl if the total pressure is 2.00 atm. For the reaction Kp is 193 at 2800C.
H2(g) +Cl2(g) = 2HCl (g)
What I've done so far:
found the partial pressures of HCl and H2 by taking moles x / moles total = mole fraction x total pressure (2?)
Got:
1.77 atm for HCl
.233 atm for H2
Plugged into ICE chart but I'm not getting the right answer. These are supposedly the right answers:
HCl: 1.67 atm
H2: .282 atm
Cl2: .051

Your solution is not getting the right answers for two or three reasons.
a. You don't have a value for moles Cl2; therefore, you don't have total moles.
b. The moles given are at the start of the reaction (not at equilibrium) BUT total pressure of 2 atm is at equilibrium.
Here is how you approach the problem.
First, notice that you have two moles of products and two moles of reactants; therefore, Kp = Kc. Also note that you don't have a volume; however, moles/volume will give concn BUT the V term cancels in the Kc expression so we can ignore that (and I have ignored it below). You can assume some volume if it makes you feel any better but it cancels in the end, whatever you choose.
........H2 + Cl2 ==> 2HCl
I....0.47.....0.......3.59
C.....+x.....+x.......2x
E.....0.47+x...x.......3.592x
Kp=Kc=(HCl)^2/(H2)(Cl2)
193 = (3.592x)^2/(0.47+x)(x)
solve for x.
Add moles to find total moles AT EQUILIBRIUM.
Then X*totalP will find partial pressures of each.
I worked the problem and the answers are correct.
Post your work if you get stuck. 
Got it. Thanks man.