Chemist in a pickle: solving system of eqns for equilibrium reaction

dcf

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Hello Internet...

I'm trying to solve an equilibrium problem, so far I have tried to employ algebra to calculate the two unknowns. The two equations I know to be true are in red (n.b. there is a typo in the graphic: x-y=1.5, not 15 as written).



Hydrogen and iodine react together and establish equilibrium.

The value of Kp is 64.

Equal amounts of hydrogen and iodine were mixed together. The eqm mixture was found to contain 1.5 moles of iodine.

etc...



H2+ I2\(\displaystyle \rightleftharpoons\, \) 2HI
initial
xx0
change-y
-y
+2y
eqm.x - yx - y+2y

. . .x - y = 1.5

. . . . .\(\displaystyle \textrm{K}_p\, =\, \dfrac{\left[\textrm{HI}\right]^2}{\left[\textrm{I}_2\right]\, \left[\textrm{H}_2\right]}\)

. . . . .\(\displaystyle \color{red}{64\, =\, \dfrac{(2y)^2}{(x\, -\, y)(x\, -\, y)}}\)



Can this be solved with some crazy simultaneous equations?

It's been a long day, maybe I'm overthinking the problem or have completely lost the plot. Either way, help me Maths Wizards :)
 

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That's cool thanks, doing that, I get y=10, x=11.5

Anybody else concur?
 
… I get y=10, x=11.5
Those values satisfy only the equation x-y=1.5

They do not satisfy the other equation:

64 = (2*10)^2/(1.5)^2

64 = 1600/9

Please show your work. :cool:
 
. . .x - y = 1.5

. . . . .\(\displaystyle \textrm{K}_p\, =\, \dfrac{\left[\textrm{HI}\right]^2}{\left[\textrm{I}_2\right]\, \left[\textrm{H}_2\right]}\)

. . . . .\(\displaystyle \color{red}{64\, =\, \dfrac{(2y)^2}{(x\, -\, y)(x\, -\, y)}}\)



Can this be solved with some crazy simultaneous equations?
Why not just plug in the given value? Since x - y = 1.5, then:

. . . . .\(\displaystyle 64\, =\, \dfrac{(2y)^2}{(1.5)^2}\, =\, \dfrac{4y^2}{2.25}\)

. . . . .\(\displaystyle \dfrac{(64)(2.25)}{4}\, =\, y^2\)

...and so forth. ;)
 
Thanks for typing this out stapel...

x = 7.5 and y = 6, those values work out for the rest of the problem (calculating the mole fractions and partial pressures etc)

Thank you all in the maths world, also I learnt a bit along the way :)
 
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