If the reaction 2H₂S(g) ⇌ 2H₂(g) + S₂(g) is carried out at 1065^$\circ$C, K_p = 0.0120. Starting from pure H₂S introduced into an evacuated vessel at 1065^$\circ$C, what will the total pressure in the vessel be at equilibrium if the equilibrated mixture contains 0.300 atm of H₂(g)

Question

Quizplus · Accepted Answer

Using the equilibrium constant expression: 
Kp = (PH2S)/(PSS)(PH2S)²
0.0120 = (0.300/PTOTAL)/[(PTOTAL-0.300)/PTOTAL]^2
Solving for PTOTAL, we get PTOTAL = 1.51 atm. Therefore, the correct choice is E.

If the Reaction 2H2S(g) ⇌ 2H2(g) + S2(g) Is Carried ∘\circ∘

If the Reaction 2H₂S(g) ⇌ 2H₂(g) + S₂(g) Is Carried $\circ$