Pressure Equilbrium Question

[jeffreyska]

Problem:
Calculate Kp for H2O (g) + 1/2 O2 (g) = H2O2 (g) at 600K, using the following data:
H2 (g) + O2 (g) = H2O2 (g) Kp = 2.3 * 10^6 at 600 K
2H2 (g) + O2 (g) = 2H2O (g) Kp = 1.8 * 10^37 at 600 K

Homework Equations

Kp = [Products]/[Reactants]
Kp= K(RT)^delta N

The Attempt at a Solution

Honestly really lost I have

Kp = [H2O2]/[H2O][O2]^1/2

Kp= K(RT)^delta N

Other than that I'm thoroughly confused, can someone please explain? Thanks!

 

[Iodine]

Problem:
Calculate Kp for H2O (g) + 1/2 O2 (g) = H2O2 (g) at 600K, using the following data:
H2 (g) + O2 (g) = H2O2 (g) Kp = 2.3 * 10^6 at 600 K
2H2 (g) + O2 (g) = 2H2O (g) Kp = 1.8 * 10^37 at 600 K


You don't need these equations
Kp = [Products]/[Reactants]
Kp= K(RT)^delta N

What you need to do is use the reactions that you have to make this reaction:H2O (g) + 1/2 O2 (g) = H2O2 (g)

Start with this one
2H2 (g) + O2 (g) = 2H2O (g) Kp = 1.8 * 10^37

Turn it backwards so that is has H2O in the reactants side just like the equation I wrote in red. When you do this the Kp becomes negative.
2H2O (g)= 2H2 (g) + O2 (g) Kp = -1.8 * 10^37
But the red rxn only has 1 H2O so you must divide by 2
H2O (g)= H2 (g) + (1/2)O2 (g) Kp = -0.9 * 10^37

Now use the other rxn2 (g) + O2 (g) = H2O2 (g) Kp = 2.3 * 10^6

Now add the two reactions together
H2O (g)= [STRIKE]H2 (g)[/STRIKE] + (1/2)O2 (g) Kp = -0.9 * 10^37
[STRIKE]H2 (g)[/STRIKE] + O2 (g) = H2O2 (g) Kp = 2.3 * 10^6
H2O (g) + 1/2 O2 (g) = H2O2 (g) Kp=(-0.9*10^37) +(2.3*10^6)
You get your desired reaction. And to calc Kp add the Kps.