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Partial pressures and equilibrium question

  • Thread starter veitch
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  • #1
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Homework Statement



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.
 

Answers and Replies

  • #2
Borek
Mentor
28,407
2,808
.02 = (.01-x)^2/(.05+2x)
This is OK.

Quadratic equation yields -.0264 and -.0336
This is not. Just check your math.
 
  • #3
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This is not. Just check your math.
You're right.. it should be 0= x^2 -.06x -.0009
Which has roots of 0.072 and -0.012 ...

But .05 + 2(.072) = 0.194 yet it should equal .0546 atm :(
 
  • #4
Borek
Mentor
28,407
2,808
while (x != 0.0023) check_your_math;
 

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