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A mixture of 0.01341 mol of CH4, 0.01170 mol of H2S, 0.02118 mol of CS2, and 0.02835 mol of H2 is placed in a 1.0-L stee...

A mixture of 0.01341 mol of CH4, 0.01170 mol of H2S, 0.02118 mol of CS2, and 0.02835 mol of H2 is placed in a 1.0-L steel pressure vessel at 3416 K. The following equilibrium is established:

1 CH4(g) + 2 H2S(g) revrxnarrow.gif 1 CS2(g) + 4 H2(g)

At equilibrium 0.003198 mol of H2S is found in the reaction mixture.
- Calculate the equilibrium partial pressures of CH4, H2S, CS2, and H2.

- Calculate KP for this reaction.

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Answer #1

Write the reacti on as foll ows: 2 Cs ()4H, (s CH, (3)+2H2S() volume of the flask, V = 1.0 L Initially moles of CH= 0.01341mo

(a) Calculate the equilibrium parti al pressures as follows: PECSRT PcS = V (0.02118+x) RT (0.02118 0.004251) mol x0.08206 L

(ь) Write the relati on between K and K as follows: к, 3 к (RT)* (2) gas constant, R= 0.08206L atm/molK temperature, T = 3416

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