How is Kp(T) Derived for Dissociation of H2O at High Temperatures?

AI Thread Summary
The discussion focuses on deriving the equilibrium constant Kp(T) for the dissociation of H2O at high temperatures. It starts with the reaction 2H2O -> 2H2 + O2 and defines variables for the fraction of dissociated H2O and total pressure. Participants clarify the relationships between the moles of reactants and products, leading to the expression Kp(T) = p f^3 / ((1-f)^2(2+f)). Key calculations include determining the remaining moles of H2O, H2, and O2 in terms of the initial moles and the dissociation fraction. The conversation concludes with a clear understanding of the mole fractions and total moles involved in the reaction.
Dassinia
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Hello
I'm solving previous exams and there is this exercise that appears in almost every exam :

Homework Statement



Suppose v moles of a gas H2O are enclosed in a constant volume V at a low temperature so the gas is in aN undissociated state. At highest temperatureS, H2O molecules are dissociated
2H2O-> 2H2 + O2

If "f" is the fraction of H2O molecules that are dissociated at a known temperature T, and p the total pressure of the gas at that temperature, show that Kp(T) at this pressure is :
Kp(T)= p f^3/ (( 1-f)^2(2+f) )

Homework Equations


The Attempt at a Solution


Kp(T)=[ (P02)*PH2² ]/(PH20)²

We have Pi=xi*p
Kp(T)= [ x02*p*(xH2*p)² ]/(xH20*p)²
Kp(T)= [ p*x02*xH2² ]/xH20²

xi=ni/ntot

Kp(T)= [ p*n02*nH2² ]/(nH20²*ntot)

nH20²=(1-f)²
but I don't get why ntot=2+f and n02=nH2=f
 
Last edited:
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If you have v moles of H2O to start with, and fv moles H2O dissociated to H2 and O2, how many moles of H2O are left, and how many moles of O2 and H2 are formed (in terms of v and f)? What is the total number of moles of H2O remaining (in terms of v and f)? What is the total number of moles in the container after fv moles of H2O has dissociated? What is the mole fractions of H2O, H2, and O2? What are the partial pressures of these species?

Chet
 
For H2O we have (v-vf) remaining
If we have fv moles of H2O are dissociated we will have fv moles of H2 and fv moles of O2 ?
 
Dassinia said:
For H2O we have (v-vf) remaining
If we have fv moles of H2O are dissociated we will have fv moles of H2 and fv moles of O2 ?
From the stoichiometry, you will have fv/2 moles of water, correct?

Chet
 
My chemistry courses are so far away :rolleyes:
Why fv/2 moles of H2O and not (v-vf)moles ?
 
Dassinia said:
My chemistry courses are so far away :rolleyes:
Why fv/2 moles of H2O and not (v-vf)moles ?

Oops. Sorry. I meant fv/2 moles of oxygen.

Chet
 
Oh so it is fv/2 of O2 and fv of H2

So we have ntot=fv/2+fv-fv+v=v/2(f+2)
nO2=fv/2
nH2=fv
nH2O=v(1-f)

Thank you thank you ! now it is really clear in my head !
 
Last edited:
Good. Nice job.

Chet
 

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