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For the Reaction CuS(s)+ H2(g) H2S(g)+ Cu(s) Δ\Deltaf (CuS)= -53

question 24

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For the reaction CuS(s)+ H2(g)  For the reaction CuS(s)+ H<sub>2</sub>(g)   H<sub>2</sub>S(g)+ Cu(s),  \Delta G°<sub>f</sub> (CuS)= -53.6 kJ/mol  \Delta G°<sub>f</sub> (H<sub>2</sub>S)= -33.6 kJ/mol  \Delta H°<sub>f</sub> (CuS)= -53.1 kJ/mol  \Delta H°<sub>f</sub> (H<sub>2</sub>S)= -20.6 kJ/mol Calculate the value of the equilibrium constant (K<sub>p</sub>)at 798 K and 1 atm pressure. H2S(g)+ Cu(s),
Δ\Deltaf (CuS)= -53.6 kJ/mol
Δ\Deltaf (H2S)= -33.6 kJ/mol
Δ\Deltaf (CuS)= -53.1 kJ/mol
Δ\Deltaf (H2S)= -20.6 kJ/mol
Calculate the value of the equilibrium constant (Kp)at 798 K and 1 atm pressure.


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