Partial pressures and equilibrium question

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Discussion Overview

The discussion revolves around a homework problem involving the equilibrium of the reaction 2HI (g) ⇌ H2 (g) + I2 (g) at a given temperature, focusing on calculating the reaction quotient (Qp) and determining the equilibrium partial pressures of the gases involved.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant calculates Qp and concludes that since Qp > Kp, the reaction should shift to the left.
  • The same participant sets up an expression for equilibrium partial pressures but encounters difficulties with the quadratic equation derived from the equilibrium expression.
  • Another participant points out a potential error in the quadratic equation's formulation and suggests checking the math.
  • A subsequent reply corrects the quadratic equation but still yields roots that do not match the expected equilibrium partial pressures.
  • There is a humorous suggestion to check the math again, indicating ongoing confusion or frustration with the calculations.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the mathematical steps taken, with some suggesting errors while others attempt to clarify or correct the calculations. The discussion remains unresolved regarding the correct equilibrium partial pressures.

Contextual Notes

Participants have not reached a consensus on the correct values for x or the equilibrium partial pressures, and there are indications of potential errors in calculations that have not been fully clarified.

veitch
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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.
 
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veitch said:
.02 = (.01-x)^2/(.05+2x)

This is OK.

Quadratic equation yields -.0264 and -.0336

This is not. Just check your math.
 
Borek said:
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 :(
 
while (x != 0.0023) check_your_math;
 

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