Any chemistry majors?

[Dragonmas121]

I'm sitting here trying to do this chemistry homework and keep fucking up. Is there anyone out there who knows how to do chemistry pretty well? I need some help.

[Dragonmas121]

At 2/21/13 08:05 PM, Cherry wrote: do your homework yourself, otherwise you will not gain the knowledge necessary to function in day to day life, young child . . .

-- supercherry64

I'm more than happy to do my homework myself. It's just that I've worked the same answer 5 times and i'm not getting the right answer. I need advice as to what i'm not doing right and which steps I should take.
I'm not a dumb guy. It's AP Chemistry. This is just a difficult subject, I'm not the strongest in math.

[notYert]

At 2/21/13 08:12 PM, Cherry wrote: it starts with "advice" and before you know it NG is turning up at school in your place. . . trust me child, stay off drugs and do your own homework like a good boy and soon you will live a fulfilled life, much like i do

-- super cherry 65

i hope you don't think you're being funny by impersonating supergandhi

[Dragonmas121]

What are you talking about? I'm not inviting anyone to my house, I was going to write the problem out here. Oh well, might as well.

At 1100 K, Kp = .25 for the reaction
2SO2(g) + O2(g) <----> 2SO3(g)

Calculate the equilibrium partial pressures of SO2, O2, and SO3 produced from an initial mixture in which Pso2 = Po2 = .5 atm and Pso3 = 0.

(I think Pso2 AND O2 both have partial pressures of .5 atm. Either that or its a typo.)

Anyways, I know that the first thing I need to do is use x to symbolize the loss or gain in pressure.

2SO2 = (.5atm + 2x)^2 O2 = (.5 atm + x) and 2SO3 = (0 + 2x)^2.

Then I know I need to put products over reactants.

4x^2
-----------------------------------------------------
(.5 atm + 2x)^2 (.5 atm + x)

Then I need to multiply the bottom two numbers together which gets

4x^2
---------------------------------------
4x^3 + 3x^2 + 1.25x + .125

So I set .25 equal to that
4x^2
.25 = -------------------------------
4x^3 + 3x^2 + 1.25x + .125

And then I solve. Does anyone see any error in the way I calculated anything?

[KatMaestro]

Don't rely on this place to help you. Likely they help you roll donut more like it.

Go to Math Club and ask those nerd.

[BumFodder]

At 2/21/13 08:17 PM, notYert wrote: i hope you don't think you're being funny by impersonating supergandhi

Trust me, its hilarious

-- superfodder69

[Dragonmas121]

You guys really are no help.

[draco889]

I was all excited to help. Then I saw it was an equilibrium problem.

[Dragonmas121]

Yeah, these damn equilibrium problems. It was all fine until Zumdahl started making us solve for unknown concentrations.

[Cordyceps]

I hate to be one to say this, but you should be able to do this yourself.

[ohbombuh]

At 2/21/13 08:22 PM, Dragonmas121 wrote: At 1100 K, Kp = .25 for the reaction
2SO2(g) + O2(g) <----> 2SO3(g)
Calculate the equilibrium partial pressures of SO2, O2, and SO3 produced from an initial mixture in which Pso2 = Po2 = .5 atm and Pso3 = 0.
Anyways, I know that the first thing I need to do is use x to symbolize the loss or gain in pressure.

Loss or gain, huh? Seems like a rather important distinction to keep track of...

2SO2 = (.5atm + 2x)^2 O2 = (.5 atm + x) and 2SO3 = (0 + 2x)^2.

Here's where you messed up. Since you start out with only reactant gases, the system shifts towards the right (products). In other words, the amount of each reactant decreases. That means the equilibrium pressure for SO2 will be (0.5 - 2x)^2, and the pressure for O2 will be (0.5 - x). On the bright side, you were right about SO3.

[Piemix]

I'm glad I'm a history major sometimes

[i-am-ghey]

it appears to me that you are not given enough information to do the calculation (or there are some implict assumptions not stated in the question).

one could have easily calculated the equilrbium composition if the actual number of moles of gases are given because you can then add and subtract the compositions directly.

but this is not the case if partial pressure are given.. in general you cannot add and subtract the pressures directly.

this question is wrong.

[i-am-ghey]

At 2/22/13 05:18 AM, i-am-ghey wrote:

stuff

okay. suppose everything is kept fixed, indeed you can do the calculation as follows:

final partial pressures of SO2, O2 and SO3 are 0.5-2x, 0.5-x and 2x respectively.

Kp=(2x)^2/[((0.5-2x)^2)(0.5-x)]=0.25 which yields

32x^3+96x^2+10x-1=0

but then the equation has no nice solution. (the only feasible solution is between x=0.06 and 0.07, which is expressed in radicals. the graph is slightly increasing and there is no real roots beyond x=0.07)

[Cyberdevil]

I actually only had one year of chemistry in school, that's all. To top it off it was a boring a violent year, gutting frogs, playing around with cows eyeballs, breeding banana flies... not anything I'd like to work with the rest of my life. :/

[i-am-ghey]

just want to add a small remark here. it is quite useful sometimes.

occasionally, you might be asked to approximate the root of a given equation f(x)=0, which you don't know how to solve (for instance, x^5+x-1=0). in this case, you should first determine the possible range that the solution (you are interested in) lies.

the quickest way is to use newton's method, which converges to the solution quickly. another way is to use the goal seek function in excel.

but if you don't have these tools/don't know the method, you can always try the following method:

e.g. you want to find a root that lies between 0.5 and 1 with f(x)=x^5+x-1

step 1: determine the sign of f(0.5) and f(1). in this example f(0.5)<0, f(1)>0. so indeed there is a solution between 0.5 and 1.

step 2: set x=(0.5+1)/2=0.75. determine the sign of f(0.75). since f(0.75)<0, you know the solution lies between 0.75 and 1.

step 3: set x=(0.75+1)/2=0.875. determine the sign of f(0.875). since f(0.875)>0, the solution lies between 0.75 and 0.875.

step 4: simliarly, repeat the above procedure until you get a good covergence.

(the above method uses intermediate value theorem, which asserts that for a continous f, if f(a)<0 f(b)>0, then you can find c such that f(c)=0)

[ohbombuh]

At 2/22/13 07:01 AM, i-am-ghey wrote: Kp=(2x)^2/[((0.5-2x)^2)(0.5-x)]=0.25 which yields
32x^3+96x^2+10x-1=0

Check your work.
http://www.newgrounds.com/dump/item/94f5b1afa08ec618b6db6215 588c9995

[i-am-ghey]

At 2/22/13 09:26 AM, ohbombuh wrote:

At 2/22/13 07:01 AM, i-am-ghey wrote: Kp=(2x)^2/[((0.5-2x)^2)(0.5-x)]=0.25 which yields
32x^3+96x^2+10x-1=0

Check your work.
http://www.newgrounds.com/dump/item/94f5b1afa08ec618b6db6215 588c9995

and then multiply both sides by 32 to make it simplier QED.

by the way, you should not major in chemistry if you literally think the equilirbium partial pressure involves squares.

[Silverspecks]

Sorry to say this but chemistry used to be my major. When I graduated high school I wanted to become a bio-chemist and advance genetic research so I could one day rid the world of my eye disorder. But the second chemistry course I took in college made me realize my comprehensive limits. Since then my aspiration has been to become a fiction author.

[ohbombuh]

At 2/22/13 09:49 AM, i-am-ghey wrote: by the way, you should not major in chemistry if you literally think the equilirbium partial pressure involves squares.

I don't believe in equilirbiums, but using molar coefficients as exponents relates partial pressures to the equilibrium constant. Solving for the change in concentration with exponents doesn't mean I'd use them for the final concentration of each chemical.

[Dragonmas121]

At 2/22/13 11:26 AM, ohbombuh wrote:

At 2/22/13 09:49 AM, i-am-ghey wrote: by the way, you should not major in chemistry if you literally think the equilirbium partial pressure involves squares.

I don't believe in equilirbiums, but using molar coefficients as exponents relates partial pressures to the equilibrium constant. Solving for the change in concentration with exponents doesn't mean I'd use them for the final concentration of each chemical.

You're right, you don't keep the exponents when solving for the change in concentration. You simplify those exponents and solve for X. Once you find X, you add or subtract that value to the initial concentrations.

And to the other guy who corrected me from using (.5 + x)^2.. Thanks, you're right, it is (.5 - x)^2. I made that mistake earlier. I should have known that.

[Dragonmas121]

At 2/22/13 07:01 AM, i-am-ghey wrote:

At 2/22/13 05:18 AM, i-am-ghey wrote:

stuff

okay. suppose everything is kept fixed, indeed you can do the calculation as follows:

final partial pressures of SO2, O2 and SO3 are 0.5-2x, 0.5-x and 2x respectively.

Kp=(2x)^2/[((0.5-2x)^2)(0.5-x)]=0.25 which yields

32x^3+96x^2+10x-1=0

but then the equation has no nice solution. (the only feasible solution is between x=0.06 and 0.07, which is expressed in radicals. the graph is slightly increasing and there is no real roots beyond x=0.07)

And no, the equation doesn't have a nice solution. In most cases we have to use about 10 significant digits.

[Sevkat]

damn i only majored in meth cooking

[Soviet]

I was going to do chemistry but I decided to do something else (business/economics related.)

God that work looks complex.

[111122223138]

At 2/21/13 08:17 PM, notYert wrote:

At 2/21/13 08:12 PM, Cherry wrote:
-- super cherry 65

i hope you don't think you're being funny by impersonating supergandhi

Shut up supernotyert65
- Super11112222313865

[Ceratisa]

At 2/22/13 04:55 PM, Sevkat wrote: damn i only majored in meth cooking

Why did the breaking bad reference come so late..

[372]

At 2/22/13 08:45 PM, 111122223138 wrote:

At 2/21/13 08:17 PM, notYert wrote:

At 2/21/13 08:12 PM, Cherry wrote:
-- super cherry 65

i hope you don't think you're being funny by impersonating supergandhi

Shut up supernotyert65
- Super11112222313865

Cmon guys I made that joke like a week ago

-supernumbers372

[TracyJacksonTAW]

Not a chem major, but I am a liberal arts person though...specifically a political science major. Best to ask your professor or instructor for help !

[KatMaestro]

At 2/22/13 09:16 PM, TracyJacksonTAW wrote: Not a chem major, but I am a liberal arts person though...specifically a political science major. Best to ask your professor or instructor for help !

Good god, a liberal student!