When the reaction 2H₂S(g) ⇌ 2H₂(g) + S₂(g) is carried out at 1065^$\circ$C, K_p = 0.012. Starting with pure H₂S at 1065^$\circ$, what must the initial pressure of H₂S be if the equilibrated mixture at this temperature is to contain 0.250 atm of H₂(g)

Question

Quizplus · Accepted Answer

To find the initial pressure of H₂S, use the equilibrium expression Kp = (P_H2)^2 * (P_S2) / (P_H2S)^2. Given Kp = 0.012 and P_H2 = 0.250 atm, solve for P_H2S. Assume P_S2 = 0.125 atm (since it forms in a 1:2 ratio with H₂). Solving gives P_H2S_initial = 1.06 atm.

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

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