At 800 K,the equilibrium constant,K_p,for the following reaction is 3.2 $\times$ 10^-7.2 H₂S(g) 2 H₂(g)+ S₂(g) A reaction vessel at 800 K initially contains 3.00 atm of H₂S.If the reaction is allowed to equilibrate,what is the equilibrium pressure of S₂?

Question

Quizplus · Accepted Answer

The balanced equation for the reaction is:

2HS(g) ⇌ H2S(g) + S(g)

The equilibrium constant, K, is given as:
K = [H2S][S]/[HS]^2
where the square brackets denote the concentration in units of atm.

Initially, the pressure of HS is 3 atm, and let the equilibrium pressure of H2S and S be P.

At equilibrium, we have:
K = [H2S][S]/[HS]^2
3.2x10^-10 = P^2/(3-P)^2

Solving this equation for P, we get:
P = 9.0x10^-4 atm

Therefore, the equilibrium pressure of S is P = 9.0x10^-4 atm, which is answer choice C.

At 800 K,the Equilibrium Constant,Kp,for the Following Reaction Is ×\times× 10-7 ×\times×

At 800 K,the Equilibrium Constant,K_p,for the Following Reaction Is $\times$ 10^-7 $\times$