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# Kp Chemistry Question

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1. "2 moles of X was heated to a temperature of 2000°C until equilibrium was established under a total pressure of 8 x 10^7 Nm^-2. At equilibrium, X was found tohave undergone 20% dissociation. Calculate Kp.
X(g) +Y(g) + 2 Z(g)"
I have attempted this question but I think I may have gotten the calculating the moles at equilibrium part incorrect because the 20% dissociation confused me a bit.
Can anyone explain how I could do this correctly? Much appreciated.
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3. (Original post by CryingWaffle)
"2 moles of X was heated to a temperature of 2000°C until equilibrium was established under a total pressure of 8 x 10^7 Nm^-2. At equilibrium, X was found tohave undergone 20% dissociation. Calculate Kp.
X(g) +Y(g) + 2 Z(g)"
I have attempted this question but I think I may have gotten the calculating the moles at equilibrium part incorrect because the 20% dissociation confused me a bit.
Can anyone explain how I could do this correctly? Much appreciated.
The term dissociation suggests that one compound has broken apart. This is not reflected by the equation given. I will show you a method based on:

X(g) <==> Y(g) + 2Z(g)

20% dissociation simply means that 20/100 of the original sample has decomposed (dissociated) at that temperature.

You are told that you start with 2 moles therefore 2 x 20/100 moles have dissociated = 0.4 moles

So at equilibrium you have a mole ratio of:

X = 2.0 - 0.4 = 1.6 mol
Y = 1 x 0.4 = 0.4 mol
Z = 2 x 0.4 = 0.8 mol

Gas partial pressure = total pressure x mol fraction

Mol fractions:

X= 1.6/2.8 = 0.571
Y = 0.4/2.8 = 0.143
Z = 0.8/2.8 = 0.286

Partial pressures:

X = 0.571 x 8 x 107 = 4.57 x 107
Y = 0.143 x 8 x 107 = 1.14 x 107
Z = 0.286 x 8 x 107 = 2.29 x 107

now substitute values into kp = p(Y) x p(Z)2/p(X)

kp = 1.30 x 1014
4. Thanks a lot for your help!

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