How am I doing this equilibrium constant question wrong?

General Questions
#1

Right, I am aware that the SDMB is not a homework club, and I am asking for advice only with my technique and mathematics, because it’s the weekend, and I’ve no-one else to ask!

I’m on an assignment titled “Assignment:Kc & Kp” at the moment, and I can’t work out what I’m doing wrong with this question;

Basically, it’s about this reaction:
CO[sub]2(g)[/sub] + H[sub]2(g)[/sub] <==> COsub[/sub] + H[sub]2[/sub]Osub[/sub]
which has a value of Kc of 1.60

It says that 2.00moles of CO[sub]2[/sub] and H[sub]2[/sub] are mixed in a closed container at a constant temp. The question is; how much CO[sub]2[/sub] will remain at equilbrium?

ok, Kc = ([CO] x [H2O]) / ([CO2] x [H2])
where is the concentration of x

Since we are given no total volume, but know it remains constant, we can disregard because it would cancel out on the top and bottom of the formula.

Also, according to the initial equation, we can call the equilibrium concentrations (2.00 - x) for CO2 and H2, and x for CO and H20. We should then rearrange to a quadratic, which can be solved to give x. From this, we can find 2.00 - x, which is our answer.

The assignment sheet states that I should get an answer of .88mol for the amount of CO2, but no matter how many different ways I re-arrange the equation, I cannot get this.

Is there anyone out there who can help me? My maths is rather poor, and I think it must be an elementary mistake I’ve made!

Thankyou in advance,

Harry

#2

It’s been a LOOOOOOONG time since I did anything like this and I don’t have a gen chem book handy, but

Are you sure you should be using 2.00-x for H2? I seem to remember somewhere that the natural state of an element should be given a value of 1. Or am I mixing up some other rule with what you are doing. Anyway, try looking that tidbit up in your book and recalculate it that way.

#3

Your method looks like it’s entirely correct, but check your arithmetic. A solution to the equation x^2/(2-x)^2 = 1.6 is x = 1.117, so 2 - x is about 0.883, as desired.

If you need help solving the quadratic equation, try looking here. Remember that in general, one of the solutions will be physically impossible; and make sure you’re not forgetting that extra factor of 2 in the denominator (which I did the first time I tried to solve this. Oops.)

#4

To second MikeS, your method is correct; you are just making a math error.

Here’s a hint for future problems: look to see if you can simplify the equation. In this case, you can avoid the quadratic by simply taking the square root of both sides of the equation.

(i.e. You have x[sup]2[/sup] in the numerator, and (2.00-x)[sup]2[/sup] in the denominator.) This is set equal to 1.60. Simply take the square root of both sides of the equation!)

Yes, you are mixing this up with something else. For equilibrium problems, you simply leave non-aqueous and non-gas species out of the equilibrium expression. It has nothing to do with species’ standard states.

Standard states come up when working with chemical thermodynamics, such as calculating heats of reaction.

#5

Hmm, I had this big reply worked out, but it got lost somewhere in the depths of the SDMB. Hampsters hate me for some reason.

Anyway, it transpires that my maths had been correct, but as soon as I’d got the ‘wrong’ answer for x, (ie, not 0.88), I’d given up and tried another method.

MikeS and robby, you made me realise that I should have had confidence in my working for once, but you also made me kick myself when you showed me that it’s not x that should be equal to 0.88, but x-2 that should instead. Thanks.

Infact, looking back to my OP, I’d even written that x-2 is the answer, not x. I dont quite know how I’d got the whole question so hideously wrong.

Right, I’ve done it now! Fantastic

Thanks for your help

Cheers, Harry