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For the Reaction CuS(s)+ H2(g)
H2S(g)+ Cu(s) Δ\Delta

Question 14

Short Answer

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

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