Just wondering if anyone can give any hints as to how to go about solving the following questions:

 

 

 

 

 

 

 

A reaction mixture, CO2 + H2 <----> CO + H2O when at equilibrium was found to contain 0.8 mol of CO2, 0.3mol of H2, 0.6mol of CO and 0.5mol of H2O in a 1.0L container. How many moles of CO2 would have to be removed from the system in order to reduce the amount of CO to 0.3mol?

 

 

 

 

 

 

 

For this question, I'm thinking I need to set up Keq = [P]/[R], which would give me 50 = [0.3][0.5-x]/[0.8-x][0.3-x]

 

 

 

Does that sound like it makes sense?

 

 

 

 

 

 

 

Also,

 

 

 

 

 

 

 

A reaction mixture, 2NO + O2 <---> 2NO2 was found to contain 0.24 mol of NO, 0.08 mol of O2 and 1.2mol of NO2 when at equilibrium in a 2.0L bulb. How many moles of O2 would have to be added to the mixture to increase the number of moles of NO2 to 1.28 when equilibrium is re-established?

 

 

 

 

 

 

 

If I do it the same way as before, I get x^3 which I'm unsure how to solve. Perhaps it's because I'm doing it the wrong way...?

 

 

 

 

 

 

 

Thanks for the help.

I don't know but what i do know is...

 

 

 

 

 

 

 

 

 

 

 

nitroglycerin + gunpowder + oil + gasoline + match = fun

I don't know but what i do know is...

 

 

 

 

 

 

 

 

 

 

 

nitroglycerin + gunpowder + oil + gasoline + match = fun

 

 

 

 

 

 

 

From how far are you throwing the match? I hope you survive.

That looks right but I havent finished cover that in chem yet. The second one im not sure.

 

 

 

 

 

 

 

EDIT: I got an answer for the first problem. remove .2 mole of CO2. Im confident its right but no garentees.

ehhh... too much nerd talk here

ehhh... too much nerd talk here

 

 

 

 

 

 

 

Abyssal, if you don't have anything constructive to contribute, then...?

Meh? :-s

 

 

 

 

 

 

 

I'm in Chemistry...but not that far yet. Good Luck.

We just finished this in chem, but since I don't pay attention, I can't think of how to do it off the top of my head. School starts tommorow (doing your homework on the last day of holidays eh? That makes two of us :P ), so I guess if when I have my chem book open tommorow and I happen upon this, I might see what's up.

 

 

 

 

 

 

 

But like I said, school is tommorow so you can probably just get help there ;).

Heh, it's an assignment that's been killing me over the break. I couldn't seem to solve it, but I finallyl believe I figured it out this evening. Thanks anyhow!

Heh, it's an assignment that's been killing me over the break. I couldn't seem to solve it, but I finallyl believe I figured it out this evening. Thanks anyhow!

 

 

 

 

 

 

 

Just to be sure, I thought I'd dust off my chemistry skills from year 12 and see if I can help (in a colaborative sense).

 

 

 

 

 

 

 

Question 1:

 

 

 

Firstly, I worked out the numerical value for Keq, given the unaltered equilibrium scenario, as described in the reaction given. This value I calculated as 1.25. Secondly, I used this value in the altered equilibrium scenario (with half the moles of CO) and designated the concentration of CO2 as 'x' as it is the variable in this insance. After some simple algebra, I came up with the new concentration of CO2 to be 0.4 molar, which seems to make sense as it is half of the original amount.

 

 

 

 

 

 

 

Question 2:

 

 

 

Again, I worked out the numerical value of Keq which gave me 313.04. I then substituted the altered values (i.e. the concentration of NO2 at 1.28 instead of 1.2) and designated the concentration of oxygen as 'x'. After algebra not dissimilar to the previous question, I worked out that the concentration of O2 in the altered equilibrium was 0.091, an increase of .011.

 

 

 

 

 

 

 

Hope this helped, had to say it was interesting seeing if I remembered senior chemistry. i think i did ok, but dont take my word for it. :wink:

the first i think you set up right

 

 

 

 

 

 

 

the second, i remember theres a "five percent rule", but... that was last year, and i dont know where my chem notes are >_< maybe i'll look tomorrow, since its... late

Isn't it strange how a child can come onto OT and ask for help on a certain "maths/English/science question" and be flamed to do his/her own homework by themselves or to work it out themselves, but when someone comes onto OT and asks for help on a harder question (which only someone who is older and has learnt what their talking about can answer) still revolving around home/school/university work in maths/English/science etc. we happily answer the question. :roll:

Not really. A lot of younger people tend to come across as really pushy and almost demand help when they've done bugger all themself.

 

 

 

 

 

 

 

Bmw on the other hand actually made an effort himself before asking for help - in truth; he was a fair way along before tripping up and getting stuck.

Not really. A lot of younger people tend to come across as really pushy and almost demand help when they've done bugger all themself.

 

 

 

 

 

 

 

Bmw on the other hand actually made an effort himself before asking for help - in truth; he was a fair way along before tripping up and getting stuck.

 

 

 

 

 

 

 

Quite True. But its not always like that. Some of the time the younger people post saying that they are stuck with a few questions and need help, and can get a flame or two from it. Anyway you have made me aware now and I recall that it is true that they are a lot of the time pushy about it asking for help.

Hmm, I'm not too sure about that warrior. What I did was set up I.C.E., using 0.8-y as the initial conc. for CO2. The change will be 0.3 all across the board, since the mole ratios are the same. So now I factored in the 0.3 and replugged my numbers into the equation. I got 1.02, which seems impossible since the initial conc is 0.8. However, its an equilibrium, so although it says 0.8 we actually have a lot more to work with, due to CO shifting sides :)

 

 

 

 

 

 

 

Question 2 I solved in the exact same manner, using 0.12 + y instead of -y. I ended up with 0.1648.

 

 

 

 

 

 

 

It was two somewhat uncommon questions, so the usual methods didn't work. My earleir attemps gave much more reasonable answers, but had a few flaws in the method. Thanks for the attempt though :P

Your probably right, I didnt even think to use I.C.E. :roll:

Gahhhh. General chem.. still gives me shivers thinking about that STUPID class.

 

 

 

 

 

 

 

The fun is in organic chem :)