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

question 57

Short Answer

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 $\Delta$ G at 798 K and 1 atm pressure (assume $\Delta$ S° and $\Delta$ H° do not change with temperature).

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