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Equilibrium


ironizer

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Give me a clue please. I have no idea what do do:

 

Suppose 1.51 atm of CH4(g), 2.23 atm of C2H6(g), and 15.53 atm of O2(g) are placed in a flask at a given temperature. The reactions are given below.

 

CH4(g) + 2 O2(g) <-> CO2(g) + 2 H2O(g) KP = 1.0 x 10^4

2 C2H6(g) + 7 O2(g) <-> 4 CO2(g) + 6 H2O(g) KP = 1.0 x 10^8

 

Thanks a lot. Anything helps.

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OK I'll tell you now i've never seen a question quite like this...

 

I think ONE way to solve it is to combine the two equilibria into one. You can get the value of Kp for the new equilibrium by multiplying the other two together. the trouble there is that you end up with a Kp expression in terms of x^13, which is going to be tricky to solve. Currently I can't think of another method... perhaps when you expand out the brackets it turns out to be an easy equation (perhaps a perfect 13th power).

 

another good idea might be to assume the reaction goes to completion (not a bad assumption when Kp is huge) and then calculate the small leftward shift from that point. Doing that means that the simplifying assumption that a number plus or minus x is equal to that number since x is very small is more likely to be true.

 

When I get time i'm gonna think some more about this one. If we figure it out i'm gonna show my advanced class... they'll love it.


Merged post follows:

Consecutive posts merged

i'm gonna move this to homework help in the hope we can get some more ideas.

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Yeah I see what you're saying.

 

But even if you don't make the assumption, you can still solve 13th power equations using Newton's method.

 

Try both, then we can see if it's close enough to just assume the reaction goes to completion.

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I'm not sure combining the two equilibria is the right way to go about this. I tried it, wrote out an ICE-table and ended up with a 13th power equation. I then solved this equation graphically and via the goal-seek operation in excel, and both methods gave an answer which makes no sense.

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