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Topic: Equilibrium with pressures  (Read 6878 times)

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Offline NYM

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Equilibrium with pressures
« on: August 21, 2006, 02:03:28 PM »
2 moles of CH4 and 1 mole of H2S is heated to 1000 K where this occures:
CH4 (g) + 2 H2S (g) <-> CS2 (g) + 4 H2 (g)

At equilibrium p(H2) is 20 bar and p(total) is 0,92.

Find the partial pressures of CH4, 2 H2S, CS2 (g).

Ok, so I figured that p(CS2 = 0,05 bar. But what about the other two?

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Re: Equilibrium with pressures
« Reply #1 on: August 21, 2006, 02:31:25 PM »
At equilibrium p(H2) is 20 bar and p(total) is 0,92.

I don't think I get it...
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Offline NYM

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Re: Equilibrium with pressures
« Reply #2 on: August 21, 2006, 02:54:32 PM »
Oops. It was
"... At equilibrium p(H2) is 0,20 bar and p(total) is 0,92"

NOT

"... At equilibrium p(H2) is 20 bar and p(total) is 0,92."

Sorry :)

Offline NYM

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Re: Equilibrium with pressures
« Reply #3 on: August 22, 2006, 11:47:15 AM »
5b is the answer (the attached file).
But I don't get it. Why isn't p(CH4) = (90bar - 0,25bar)/3, while p(H2s) =(90bar - 0,25bar)/3*2?

Offline Donaldson Tan

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Re: Equilibrium with pressures
« Reply #4 on: August 24, 2006, 06:51:51 PM »
Your corrected question:
2 moles of CH4 and 1 mole of H2S is heated to 1000 K at which this occurs:
CH4 (g) + 2 H2S (g) <-> CS2 (g) + 4 H2 (g)

At equilibrium p(H2) is 0.20 bar and p(total) is 0.92 bar.

Find the partial pressures of CH4, H2S, CS2 (g).


Initially, you have 2 moles of methane and 1 mole of hydrogen sulphide. Upon heating to 1000K, the chemical composition of the mixture changed according to the chemical equation above. There was zero mole of CS2 and H2 initially. This means the molar ratio of the amount of CS2 to that hydrogen gas must be the same as the stoichiometric ratio, ie. 1:4.

Since PH2 is 0.20bar, then PCS2 must be (1/4)*0.20 = 0.05bar.

Assuming perfect gas
Let V be volume of the chemical system
let T be the temperature of the chemical system (1000K)
let X be the number of moles of CH4 consumed.

PCH4 + PCS2 = 0.92 - 0.20 - 0.05 = 0.67 bar
PCH4 = nCH4RT/V = (2 - X)RT/V
PCS2 = nCS2RT/V = (1 - 2X)RT/V
=> (3-3X)RT/V = 0.67 bar (Eqn1)

PH2 = 0.20 bar
PH2 = nH2RT/V = 4XRT/V
=> 4XRT/V = 0.20 bar (Eqn2)

Eqn1 divide by Eqn2 => (3-3X)/4X = 0.67/0.20
3/4X - 0.75 = 3.35
3/4X = 3.35 + 0.75 = 4.10
4X =3/4.10 => X = (1/4)*(3/4.10) = 0.18293

ntotal = (2 + 1) + (1 + 4 -1 - 2)X = 3 + 2X = 3.3656 moles
nH2S = 1 - 2X = 0.6344
PH2S = nH2S/ntotal * Ptotal = 0.6344/3.3656 * 0.92 = 0.17bar

PCH4 = 0.92 - 0.17 - 0.20 - 0.05 = 0.50 bar

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Offline NYM

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Re: Equilibrium with pressures
« Reply #5 on: August 26, 2006, 04:05:38 AM »
Thank you!
That was difficult.

Offline edwinksl

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Re: Equilibrium with pressures
« Reply #6 on: August 26, 2006, 06:17:32 AM »
I had the same answers as geodome.

Offline sdekivit

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Re: Equilibrium with pressures
« Reply #7 on: August 26, 2006, 08:17:58 AM »
I used a slightly different method yielding the same answer:

We know that at equilibrium there is 2-x mol CH4 and 1-2x mol H2S.

Now we also know that p(CH4) + p(H2S) = 0.67 bar and p(H2) + p(CS2) = 0.25 bar

We can also say:

p(CH4) = (2-x) * 8314 / V = (16628 - 8314x) / V
p(H2S) = (1-2x) * 8314 / V = (8314 - 16628x) / V

Equivalently we can say:

p(H2) = 8314x / V (yield = x mol and RT = 8314)
pCS2) = 33256x / V (yield = 4x mol)

These equations combined yields 2 equation with 2 unknowns:

(1) 41570x = 0.25V
(2) 24942 - 24942x = 0,67V

Now substitue x = 41570x/0.25 = V = 166280x in equation 2, to yield x:

111407.6x = 24942 - 24942x

--> x = 0,183 mol. Thus V = 166280 * 0.183 = 30429.24 (0,30 m^3 because we are calculating in bar here).

Using the ideal gas law and substituting the values for x and V, this will p(CH4) = 0,634 * 8314 / 30429,24 = 0.17 bar and p(H2S) = 1.817 * 8314 / 30429.24 = 0.50 bar.

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