I've got another question. it's similar to this but this time with Kp. Here's the question:
For the equilibrium
2IBr(g) <-> I2(g) + Br2(g)
Kp = 8.5 10-3 at 150.°C. If 0.016 atm of IBr is placed in a 2.5-L container, what is the partial pressure of all substances after equilibrium is reached?
I used I.C.E
2IBr <--> I2 + Br2
I .016 0 0
C x -x -x
E .016+x -x -x
and i got the quadratic equation as:
-0.0085x^2 + .032x + 2.56*10^-4 =0
Please Help
..........2IBr ==> I2 + Br2
initial....0.016.....0....0
change......-2x......x.....x
equil......0.016-2x...x.....x
Kp = 8.5E-3 = (x)(x)/(0.016-2x)^2
8.5E-3(0.016-2x)^2 = x^2
8.5E-3*(2.56E-4 - 0.064x + 4x^2) = x^2
etc.
I think your 8.5E-3 is in the wrong place.
thnx!!!
To solve for the equilibrium partial pressures of all substances, you need to find the value of x in the quadratic equation you obtained. This value of x represents the change in the concentration (or partial pressure) of IBr, I2, and Br2 when they reach equilibrium.
To solve the quadratic equation, you can use the quadratic formula:
x = [-b ± √(b^2 - 4ac)] / 2a
For the equation -0.0085x^2 + .032x + 2.56*10^-4 = 0, the coefficients of a, b, and c are:
a = -0.0085
b = 0.032
c = 2.56*10^-4
Plugging these values into the quadratic formula, you have:
x = [-0.032 ± √(0.032^2 - 4*(-0.0085)*(2.56*10^-4))] / [2*(-0.0085)]
Simplifying the equation further:
x = [-0.032 ± √(0.001024 + 8.768*10^-7)] / -0.017
x = [-0.032 ± √0.0010240008768] / -0.017
Now you can calculate the partial pressures of all substances after equilibrium is reached:
Partial pressure of IBr = 0.016 + x
Partial pressure of I2 = -x
Partial pressure of Br2 = -x
Substituting the value of x into these equations will give you the partial pressures.