[NYM]

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?

[Borek]

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

I don't think I get it...

[NYM]

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

[NYM]

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?

[Donaldson Tan]

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 P_H2 is 0.20bar, then P_CS2 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.

P_CH4 + P_CS2 = 0.92 - 0.20 - 0.05 = 0.67 bar
P_CH4 = n_CH4RT/V = (2 - X)RT/V
P_CS2 = n_CS2RT/V = (1 - 2X)RT/V
=> (3-3X)RT/V = 0.67 bar (Eqn1)

P_H2 = 0.20 bar
P_H2 = n_H2RT/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

n_total = (2 + 1) + (1 + 4 -1 - 2)X = 3 + 2X = 3.3656 moles
n_H2S = 1 - 2X = 0.6344
P_H2S = n_H2S/n_total * P_total = 0.6344/3.3656 * 0.92 = 0.17bar

P_CH4 = 0.92 - 0.17 - 0.20 - 0.05 = 0.50 bar

[NYM]

Thank you!
That was difficult.

[edwinksl]

I had the same answers as geodome.

[sdekivit]

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.