I'm not looking for the answer as I already have the answer key. I am looking for some help on how to figure out these two. Any help is greatly appreciated!

Make ICE table for top one

Finishing up eating atm

You mean an ICE table for the bottom one? I know for a fact the bottom one uses one, but the top one as well?

Also the bottom one, I was trying to follow an example like it in my book, but it has the Kp value given, so I don't know how to figure out the problem without it.

Oh, and the pm was just a joke. I didn't realize you could actually help me

Double edit: Looking up a thing for the second question.

N2O4 2NO2
I = 1.50 and 1.00
C = +x -2x
E= 1.50+x 1.00-2x


Why is the change +x for the reactant and -2x for the product?

Triple edit: I figured out the problem, but I don't understand how to figure out that the changes are the +x and -2x. Any help on as to why they are like that is greatly appreciated.

This post was edited by Braxton11 on Oct 29 2013 07:22pm

So for #2, its +x for the reactant and -2x for the product since one n2o4 is formed per 2xno2. so you set x = reactant, and -2x = NO2


also, you can see it goes from 1 atm NO2 to 0.512 atm NO2, so reaction is proceeding towards reactants - hence why reactant side is + and product side is - .

To do #2 with ICE table and no given Kp:

N2O4 2NO2
I = 1.50 and 1.00
C = +x -2x
E= 1.50+x 1.00-2x

You know that at equilibrium though that the partial pressure of NO2 = 0.512. So

1 - 2x = 0.512 -> x = 0.244

partial pressures than being 1.744 (N2O4) and 0.512 (2NO2),

then to find Kp - Kp = [NO2]^2/[N2O4] = [.512]^2/[1.744] ~= 0.150




#1 you can do an ICE table as well - one thing to remember any solid or liquid = 1

for a) Kc = [product]/[reactant] = {[CHCOO-][H3O+]}/{[H2O][CHCOOH]}, But since H2O is liquid and = 1, you can remove it from the equation to get:
={[CHCOO-][H3O+]}/[CHCOOH]

Then for b/c, you can finish up with ICE table as i did above

/e about the pm dont worry about it

This post was edited by cialda on Oct 29 2013 07:58pm

So basically if one of the coefficients is 2 or more, it will have a minus that number and then x? Or if you start with a number and the partial pressure is lower, it will get the minus?

What about halves or something like that?

I will try to figure out the first problem now with that equation.

Edit: Alright so my answer key has this for B.

[CH3COO-] = [H3O+] = 0.0019M

Now for my ICE table I assume I put the 0.0019M for the I part of CH3COO-, but why are they equal? I don't understand that at all. It says in the question "Analysis of the equilibrium mixture shows that the concentration of H30+ is 0.0019 M."

This post was edited by Braxton11 on Oct 29 2013 08:15pm

For the first problem:

you know initially:
0.2 M of CH3COOH is in solution and is allowed to dissociate
Since they dont give any previous concentrations of H3O+ or CH3COO-, you can assume them to be = 0. - H3O+ is not actually zero, but that part is negligible and you probably dont have to worry about that yet.

You know at equilibrium, there is 0.0019 M H3O+

So:
--CH3COOH -> H3O + CH3COO-
I: 0.2 ------->___0_____0
C: -x -------->__+x____+x
E: 0.2-x____>0.0019__+x

from this one, you can see +x = 0.0019, so H3O+ = CH3COO- = 0.0019,

and then CH3COOH = 0.2-0.0019

if you set H3O & CH3COO- to -x and CH3COOH to +x, you get the same result:

-x = CH3COO- = H3O = 0.0019, then +x = -0.0019 (0.2+x = 0.2 +(-.0019)). So regardless of if you designate the reactants +/ products - or reactants -/products +, you should still get the right answer as long as your algebra is correct.


as for which you designate as x, that does not matter either, as long as you keep the same ratio to other reactants/products - i usually just pick the reactant or product that has a coefficient of 1.

This post was edited by cialda on Oct 29 2013 08:36pm

Was editing for a long time while I worked out the problem so just going to copy and paste what I wrote and then read what you said and edit



My ICE table looked like this


CH3COOH CH3COO-
I .2M 0.0019M
C -.0019M Nothing
E .2M - .0019M 0.0019M

This post was edited by Braxton11 on Oct 29 2013 08:40pm

Kc is squared since [H3O+] = [CH3COO-], you could write it as Kc = [H3O+][CH3COO-]/[CH3COOH], and knowing H3O = CH3COO-, you can square it, or you can plug 0.0019 into both.

Wait, how do we know H3O+ and CH3COO- are equal?

Originally/initially, it should be 0 for both CH3COO- and H3O+. then both of them change by +x each, and you know that H3O+ = 0.0019

in this case, it could be easy to do without an ICE table, but im sure they are just using basic ones for now. when you get to more complicated problems, the ICE table will come in handy



both are originally "0" and the reaction proceeds in the direction of producing H3O+ and CH3COO-. when it proceeds, it produces 1 mol of H3O+ / 1 mol of CH3COO- and consumes 1 mol of CH3COOH / 1 mol of H3O & 1 mol of CH3COO- produced

This post was edited by cialda on Oct 29 2013 08:43pm