At a certain temperature, the Kp for the decomposition of H2S is 0.784. H2S(g)↽−−⇀H2(g)+S(g) Initially, only H 2 S H2S is present at a pressure of 0.292 atm 0.292 atm in a closed container. What is the total pressure in the container at equilibrium?

To do this, we need to know the partial pressure of all gases in equilibrium. To do that, we use an ICE chart, and solve for the partial pressure in equilibrium. Once we have that, we just sum the value of the pressure in equilibrium

Doing an ICE chart we have:

H₂S(g) <---------> H₂(g) + S(g) Kp = 0.784

I: 0.292 0 0

C: -x +x +x

E: 0.292-x x x

Writting the expression for Kp:

Kp = [H₂] [S] / [H₂S] --> replacing the values of the chart:

0.784 = x² / 0.292-x solving for x

0.784(0.292-x) = x²

0.2289 - 0.784x = x²

x² + 0.784x - 0.2289 = 0

Using the general formula for x, in a quadratic equation we have:

x = -b ±√b² - 4ac / 2a

From the equation, we replace the values of a, b and c, and solve for x:

a = 1; b = 0.784; c = -0.2289

x = -0.784 ±√(0.784)² - 4 * 1 * (-0.2289) / 2 * 1

x = -0.784 ±√0.6147 + 0.9156 / 2

x = -0.784 ±√1.5303 / 2

x = -0.784 ± 1.237 / 2

x1 = -0.784 + 1.237 / 2 = 0.2265

x2 = -0.784 - 1.237 / 2 = -1.0105

The negative value is neglected, so we use x1 = 0.2265

Therefore the equilibrium pressures are:

PpH₂S = 0.292 - 0.2265 = 0.0655 atm

PpH₂ = PpS = 0.2265 atm

So, the total pressure in the container would be:

P = 0.0655 + 0.2265 + 0.2265

P = 0.557 atm