When the reaction 2h2s(g) 2h2(g) + s2(g) is carried out at 1,065°c, kp = 0.012. starting with pure h2s at 1,065°c, what must the initial pressure of h2s be if the equilibrated mixture at this temperature is to contain 0.250 atm of h2(g)?

Kp is the equilibrium constant of the equilibrium reaction in terms of partial pressure. It is the pressure ratio of the products to the reactants raised to the power of their stoichiometric coefficients.

So, we let x be the partial pressure of H₂S initially. Let y be the partial pressure that reacted to produce the products. The ICE(Initial-Change-Equilibrium) analysis is as follows:

                 2 H₂S → 2 H₂ + S₂
Initial            x            0       0
Change       -2y       +2y     y

Equilibrium  x-2y       2y    y

Thus, the equilibrium pressure of H₂S is: x-2y = 0.25. Then, using the definition of Kp,

0.012 = (2y)²(y)/(x-2y)²

Substitute 0.25 to (x-2y)

0.012= (2y)²(y)/(0.25)²
y = 0.0572

So, we use the value of y to the first equation to determine x:
x-2(0.0572)= 0.25
x=0.364 atm