For the reaction 2HI (g) <-- --> H2 (g) + I2 (g) Kp = .02 at 445 degrees C
Parital pressures are .05, .01 and .01 respectively. Perform a calc to see which way it goes then figure out the equilibrium partial pressures of the gasses
The Attempt at a Solution
Qp = (.01)(.01)/(.05^2) = .0001/.0025 = .04
Since Qp>Kp reaction goes left
I .05 .01 .01
C +2x -x -x
E .05+2x .01-x .01-x
.02 = (.01-x)^2/(.05+2x)
0 = -x^2 + .06x + .0009
Quadratic equation yields -.0264 and -.0336... neither of which make sense as the final answers are given and they should be:
.0546, .0077 and .0077 atm
Which makes x = .0023...
That was my third time running through this problem and each time I got a different answer... no idea how I screwed this one up. it's only the first problem on an assignment due tomorrow... any help would be greatly appreciated.