Consider the reaction: 2 H2S(g)--><-- 2H2(g) + S2 Kp= 2.4 x 10^-4 at 1073k

A reaction mixture contains .112 atm of H2,.055 atm of S2, and .445 atm of H2S. Is the reaction mixture at equlibrium? If not, in what direction will the reaction proceed?

To determine if the reaction mixture is at equilibrium, we compare the given concentrations to the equilibrium constant, Kp.

Step 1: Calculate the reaction quotient, Qp
Qp can be calculated using the given partial pressures of the gases at a certain temperature.
Qp = (PH2)^2 * (PS2) / (PH2S)^2

Step 2: Compare Qp to Kp
If Qp is equal to Kp, then the reaction mixture is at equilibrium. If Qp is greater than Kp, the reaction will proceed in the reverse direction to reach equilibrium. Conversely, if Qp is smaller than Kp, the reaction will proceed in the forward direction to reach equilibrium.

Let's calculate Qp:

Given:
PH2 = 0.112 atm
PS2 = 0.055 atm
PH2S = 0.445 atm

Qp = (0.112)^2 * (0.055) / (0.445)^2

Qp ≈ 0.00109

Step 3: Compare Qp to Kp
Given that Kp = 2.4 x 10^-4

Qp (0.00109) > Kp (2.4 x 10^-4)

Since Qp is greater than Kp, the reaction will proceed in the reverse direction (from right to left) to reach equilibrium. This means that more H2S will be formed, while the concentrations of H2 and S2 will decrease.