Answer:
P = 0.557 atm
Explanation:
Let's write the equation again:
H₂S(g) <---------> H₂(g) + S(g) Kp = 0.784
We know that only H2S is present at the beggining of reaction, and we want to know 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
