For the reaction CuS(s) + H₂(g) $\leftrightharpoons$ H₂S(g) + Cu(s) ,
$\Delta$ G^$\circ$_f (CuS) = -53.6 kJ/mol
$\Delta$ G^$\circ$_f (H₂S) = -33.6 kJ/mol
$\Delta$ H^$\circ$_f (CuS) = -53.1 kJ/mol
$\Delta$ H^$\circ$_f (H₂S) = - 20.6 kJ/mol
Calculate the value of the equilibrium constant (K_p) for this reaction at 298 K.

Question

For the reaction CuS(s) + H₂(g)

\leftrightharpoons

H₂S(g) + Cu(s) ,

\Delta

G^$\circ$_f (CuS) = -53.6 kJ/mol

\Delta

G^$\circ$_f (H₂S) = -33.6 kJ/mol

\Delta

H^$\circ$_f (CuS) = -53.1 kJ/mol

\Delta

H^$\circ$_f (H₂S) = - 20.6 kJ/mol
Calculate the value of the equilibrium constant (K_p) for this reaction at 298 K.

Quizplus · Accepted Answer

The equilibrium constant (Kp) can be calculated using the formula:

Kp = (P_HS*S/Cu)*(P_CuS/P_HS2S)

where P_HS*S, P_CuS and P_HS2S represent the partial pressures of HS2S, CuS and HS*S respectively at equilibrium.

Using the values given in the question, we get:

ΔG° = -RT ln (Kp)
Kp = e^(-ΔG°/RT)

where R is the gas constant (8.314 J/mol K) and T is the temperature (298 K).

Substituting the values, we get:

Kp = e^(-(-33.6 + 53.6 - 20.6 + 53.1)/(8.314*298)) 
Kp = 3.11 x 10^-4

Converting to scientific notation, we get:

Kp = 3.11 x 10^3

Therefore, the correct answer is C.

For the Reaction CuS(s) + H2(g) ⇋\leftrightharpoons⇋ H2S(g) + Cu(s) Δ\DeltaΔ G ∘\circ∘