Chemistry

The following equilibrium was studied by analyzing the equilibrium mixture for the amount of H2S produced.

Sb2S3(solid)+3H2(gas)<=>2Sb(solid)+3H2S(solid)

A vessel whose volume is 2.5L is filled with 0.0100 mole of antimony(III)Sulfide, Sb2S3, and 0.0100 mole H2. After the mixture came to equilibrium in a closed vessel at 440Celcius, the gaseous mixture was removed and the H2S dissolved in water. Sufficient lead(II)ion was added to react completely with H2S to precipitate lead(II)Sulfide, PbS. If 1.029g of Pbs was obtained what is the value of K(eq) at 440Celcius

k(eq)=the equilibrium constant

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  1. 1.029g PbS x (1 mol PbS/molar mass PbS)= approximately 0.004 mol but you need to calculate more carefully than that.
    .......Sb2S3 + 3H2 ==> 2Sb + 3H2S
    initial.0.01..0.01......0......0
    change..-x......-3x.....2x....3x
    equil..0.01-x..0.01-3x..2x....0.004

    Therefore, H2 must be 0.01-0.004 = 0.006

    I would convert 0.004 mol H2S and 0.006 mol H2 to M, substitute into the Kc expression and solve for Kc. Don't forget: solids are not part of the Kc expression. Also note the correct spelling of celsius.

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