For the reaction CuS(s)+ H₂(g) H₂S(g)+ Cu(s), $\Delta$G°_f (CuS)= -53.6 kJ/mol $\Delta$G°_f (H₂S)= -33.6 kJ/mol $\Delta$H°_f (CuS)= -53.1 kJ/mol $\Delta$H°_f (H₂S)= -20.6 kJ/mol Calculate the value of the equilibrium constant (K_p)at 798 K and 1 atm pressure.

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The Answer of For the reaction CuS(s)+ H₂(g) H₂S(g)+ Cu(s), $\Delta$G°_f (CuS)= -53.6 kJ/mol $\Delta$G°_f (H₂S)= -33.6 kJ/mol $\Delta$H°_f (CuS)= -53.1 kJ/mol $\Delta$H°_f (H₂S)= -20.6 kJ/mol Calculate the value of the equilibrium constant (K_p)at 798 K and 1 atm pressure.

For the Reaction CuS(s)+ H2(g) H2S(g)+ Cu(s) Δ\DeltaΔ G°f (CuS)= -53

For the Reaction CuS(s)+ H₂(g) H₂S(g)+ Cu(s) $\Delta$ G°_f (CuS)= -53