For the reaction CuS(s)+ H₂(g) $\rarr$ 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.

Question

For the reaction CuS(s)+ H₂(g)

\rarr

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.

Quizplus · Accepted Answer

The Answer of For the reaction CuS(s)+ H₂(g) $\rarr$ 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) →\rarr→ H2S(g)+ Cu(s) Δ\DeltaΔ G°f (CuS)= -53 Δ\DeltaΔ

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