In an equilibrium mixture of HCl, Cl2 and H2, the partial pressure of H2 is 4.2 mPa and that of Cl2 is 8.3 mPa. What is the partial pressure of HCl at 500 oK given that at that temperature Kp = 4.0 x 1018 for

H2(g)+ Cl2(g)--> 2HCl(g)

6 answers

*** I meant Kp = 4.0 x 10^18
If Kp = 4.0E18 = (HCl)^2/(H2)(Cl2)
Isn't this just a matter of substituting values for (H2) and (Cl2) and solving for (HCl)? Or is it more complicated than that.
yes i did that but the answer i get is weird
True you obtained a huge number for (HCl); however, with a K as large as 10^18 you know (HCl) must be very very large.
Is 8.35 x 10^9mPa the right answer?
@DrBob222: is that the answer?
Similar Questions
  1. All (g)3N2H4 + 4ClF3 <=> 12HF + 3N2 + 2Cl2 The mixture initially consisted of 0.880 atm N2H4 and 0.970 atm CIF3(g). At
    1. answers icon 0 answers
    1. answers icon 1 answer
  2. 3H2(g)+N2(g)=2NH3(g)Suppose the equilibrium constant Kp = 0.003337 for the reaction above. If the equilibrium mixture contains
    1. answers icon 1 answer
  3. At 900 K the following reaction has Kp=0.345; 2 SO2(g) + O2 (g) -> 2 SO3 (g)In an equilibrium mixture the partial pressures of
    1. answers icon 4 answers
more similar questions