For the reaction Sb2S3(s) + 3H2(g) => 2Sb(s) + 3H2S(g) the equilibrium constant

Question

For the reaction
Sb2S3(s) + 3H2(g) => 2Sb(s) + 3H2S(g)
the equilibrium constant is 4.27 at 600 K. Calculate the amount ofH2S gas that will be produced when 1 mole of H2 gas is added to aclosed vessel containing excess Sb2S3 at 600 K. Sb2S3(s) + 3H2(g) => 2Sb(s) + 3H2S(g)

Explanation / Answer

K = (H2S)3/(H2)3        Sb2S3(s) + 3H2(g) => 2Sb(s) +3H2S(g) initial                                     1 change                                  -x                         +x equil                               1-x                              x K = (x/[1-x])^3 = 4.27 x/(1-x) = 1.6223428 x = 0.619 moles

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