Chemistry 21

[Jaf]

see, look at this site http://www.newton.dep.anl.gov/askasci/chem03/chem03747.htm, at the end u will see that when the pressure is altered the
system is nolongerin equilibrium. "hanges in concentrations or pressures result in a perturbation of the equilibrium such that the system is momentarily not at
equilibrium."

So, what's your point? :S
IF we consider the system to NOT be in equilibrium it fails your Kp argument too (which was not a real argument to begin with...).
Any how, I think you need to pay closer attention to what the link says: 'However, the Le Chatelier Principle expresses the system will reacquire equilibrium. This means that, in this case, there will be an increase in the forward rate of reaction so that more product will be formed. This will continue until the pressures are such that the ratio is once again equal to Kp.'

I think what you don't realize here is that the reactants don't necessarily have to increase for the Kp to remain constant. Remember, we're using partial pressures and not the moles themselves.

 

[vivo990]

So, what's your point? :S
IF we consider the system to NOT be in equilibrium it fails your Kp argument too (which was not a real argument to begin with...).
Any how, I think you need to pay closer attention to what the link says: 'However, the Le Chatelier Principle expresses the system will reacquire equilibrium. This means that, in this case, there will be an increase in the forward rate of reaction so that more product will be formed. This will continue until the pressures are such that the ratio is once again equal to Kp.'

what u are saying is totally correct, the question asked about equilibrium yield they even made it bold in the exam ( never asked why?), Kp happens at equilibrium & and u know that change of pressure will give higher yield but the system is NOT at equilibrium. At equilibrium the yield is same if Kp is not affected ( ONLY can be affected by temp.)

I think what you don't realize here is that the reactants don't necessarily have to increase for the Kp to remain constant. Remember, we're using partial pressures and not the moles themselves.

Remeber also that pressure depends on the number of moles... thats why the position of equilibrium shifts in the first place.

 

[geek101]

So, what's your point? :S
IF we consider the system to NOT be in equilibrium it fails your Kp argument too (which was not a real argument to begin with...).
Any how, I think you need to pay closer attention to what the link says: 'However, the Le Chatelier Principle expresses the system will reacquire equilibrium. This means that, in this case, there will be an increase in the forward rate of reaction so that more product will be formed. This will continue until the pressures are such that the ratio is once again equal to Kp.'

I think what you don't realize here is that the reactants don't necessarily have to increase for the Kp to remain constant. Remember, we're using partial pressures and not the moles themselves.

hmm, true. but ull calculate the Kp when the reaction is in equilibrium. when you change the temp or pressure the equilibrium shifts...right? at this point you cant even calculate the Kp. though the change is reversed later on. anyhoo, smart thinking. to all those who thought so

 

[iKhaled]

guys u r confusing my ass...whats the answer of this question? -_______________________________- !!

 

[vivo990]

hmm, true. but ull calculate the Kp when the reaction is in equilibrium. when you change the temp or pressure the equilibrium shifts...right? at this point you cant even calculate the Kp. though the change is reversed later on. anyhoo, smart thinking. to all those who thought so

but at the end, the question asks equilibrium yield, which is at Kp, so it does not change!

 

[geek101]

but at the end, the question asks equilibrium yield, which is at Kp, so it does not change!

if thats true....thats a whole lot ov marks man. -_-

 

[geek101]

but at the end, the question asks equilibrium yield, which is at Kp, so it does not change!

what i meant to say was...that Kp is calculated only when the reaction is equilibrium. When you change any of the factors the equilibrium doesnt remain anymore so how will you calculate Kp. your not seeing the affect on Kp but on the equilibrium :/ hope so atleast.

 

[vivo990]

what i meant to say was...that Kp is calculated only when the reaction is equilibrium. When you change any of the factors the equilibrium doesnt remain anymore so how will you calculate Kp. your not seeing the affect on Kp but on the equilibrium :/ hope so atleast.

u r right, BUT & again the question asked the equilibrium yield of methanol, at when the equilibrium is re-established again after higher pressure, will the amount of methanol will be higher (after equilibrium is re-established)?
of course not! because the temp. didn't change so Kp is same! thus no effect on incresing pressure on the equilibrium yield of methanol

 

[iKhaled]

u r right, BUT & again the question asked the equilibrium yield of methanol, at when the equilibrium is re-established again after higher pressure, will the amount of methanol will be higher (after equilibrium is re-established)?
of course not! because the temp. didn't change so Kp is same! thus no effect on incresing pressure on the equilibrium yield of methanol

:O:O what about the temperature and catalyst? :S:S

 

[Mclovin]

u r right, BUT & again the question asked the equilibrium yield of methanol, at when the equilibrium is re-established again after higher pressure, will the amount of methanol will be higher (after equilibrium is re-established)?
of course not! because the temp. didn't change so Kp is same! thus no effect on incresing pressure on the equilibrium yield of methanol

The amount of methanol DOES increase, but the reactants decrease by the same amount/proportion, so Kp stays the same. It doesen't stay the same with temperature because the products and reactants don't increase/decrease proportionally.

 

[Mclovin]

so increase in pressure increases equilibrium yield of methanol.

 

[geek101]

u r right, BUT & again the question asked the equilibrium yield of methanol, at when the equilibrium is re-established again after higher pressure, will the amount of methanol will be higher (after equilibrium is re-established)?
of course not! because the temp. didn't change so Kp is same! thus no effect on incresing pressure on the equilibrium yield of methanol

hmmm....thats probably true bout the Kp, but surely for the change in temperature the Kc value changes, and this is becuz the eq. yeild chnages. oh forget it brotha...wait till the mark schemes come out XD

 

[vivo990]

temp is fine, normal answer, less methanol,
catlyst no effect

:O:O what about the temperature and catalyst? :S:S

 

[Jaf]

Son, vivo - nobody has any idea what you're talking about.

I'm sure you'd be pretty successful in life but right now fact is you thought too much for the question.

 

[vivo990]

so increase in pressure increases equilibrium yield of methanol.

it is right, but it is not the answer of question as far as i know, the answer is no effect on equilibrium yield of methanol

 

[iKhaled]

Son, vivo - nobody has any idea what you're talking about.

I'm sure you'd be pretty successful in life but right now fact is you thought too much for the question.

he sounds as Einstein though but i think an examiner wont ask a question which needs that much of thinkin at that age right? maybe he thought of it too much cuz me also i have no idea what he is talkin about but he sounds as a prof lol..

 

[vivo990]

Anyway, this is as far as i can approach...

 

[mady666]

yes ofcourse you will, but you'll have 1 mark deducted for not writing it!

 

[roadtrip9o9]

I know Jaf, but the exam is Advanced Subsidary Level of Chemistry, so u have to think advanced, plus any correct chemistry is accepted! thats clearly stated in mark schemes. & thanks anyway.

trust me on this bro ur talking about jaf , and u have no idea about his capablities

 

[vivo990]

Dude, it's a discussion & i could be wrong!