Chemistry 21

[vivo990]

As far as i know a change in pressure does affect the equilibrium position when there are gases involved.....an increase in pressure favors the side with fewer moles, so in the case of methanol i think the products side was favored.....it is only the equilibrium CONSTANT that is not affected by a change in pressure...

see, what is equilibrium constant; it is the ratio of amt of products to reactants, you know that Kp does not change; and the question asks about the equilibrium yield so at equilibrium the Kp is unchanged so higher pressure will NOT affect the amt of methanol AT EQUILIBRIUM.

 

[vivo990]

obviously increase as it depends on where there are less no. of moles which was in the forward direction i suppose

see, what is equilibrium constant; it is the ratio of amt of products to reactants, you know that Kp does not change; and the question asks about the equilibrium yield so at equilibrium the Kp is unchanged so higher pressure will NOT affect the amt of methanol AT EQUILIBRIUM.

 

[iKhaled]

see, what is equilibrium constant; it is the ratio of amt of products to reactants, you know that Kp does not change; and the question asks about the equilibrium yield so at equilibrium the Kp is unchanged so higher pressure will NOT affect the amt of methanol AT EQUILIBRIUM.

what was the equation exactly i am confused!!!!!!!!!!!

 

[vivo990]

what was the equation exactly i am confused!!!!!!!!!!!

the question stated that if higher pressure is done to the reaction, what is the effect on the equilibrium yield of methanol?Explain your answer.
CO2+3H2 <---------> CH3OH + H2O

 

[vivo990]

that was really tricky!!!! hardly anyone spotted that

 

[iKhaled]

that was really tricky!!!! hardly anyone spotted that

did u do q5 last part?

 

[vivo990]

HO2CCOCO2H

did u do q5 last part?

 

[iKhaled]

HO2CCOCO2H

uk what the hell i did? so freakin stupid really i did the same thing exactly but it was like HO2CCO"C"CO2H...added an extra carbon by mistake arghhh

 

[Jaf]

see, what is equilibrium constant; it is the ratio of amt of products to reactants, you know that Kp does not change; and the question asks about the equilibrium yield so at equilibrium the Kp is unchanged so higher pressure will NOT affect the amt of methanol AT EQUILIBRIUM.

You, sir thought way too much for the question.

Here's the link to a similar reaction and it tells you what happens to the equilibrium yield when you have a higher pressure:
http://webpages.sou.edu/~chapman/ch203/Ch15/15_56.html

 

[Soldier313]

see, what is equilibrium constant; it is the ratio of amt of products to reactants, you know that Kp does not change; and the question asks about the equilibrium yield so at equilibrium the Kp is unchanged so higher pressure will NOT affect the amt of methanol AT EQUILIBRIUM.

bro u r confused...... there is a HUGE diff btwn equilibrium constant and equilibrium yield....but it's okay now chem is over for AS

 

[iKhaled]

whats the mark approx mark for a D in paper 3 ??

 

[Sanis]

higher temp will decrease yield, higher pressure increase yield, adding catalyst does not change yield

 

[mady666]

we had to calculate hydrogen atoms in the last question, what did you guys get? everybody in my school got different answers i got 6 x 10^21

 

[melly713]

if u put -49 in ur working bu didnt put it for d actual ans wud they still giv u ur mark or not ?

 

[iKhaled]

if u put -49 in ur working bu didnt put it for d actual ans wud they still giv u ur mark or not ?

i think you would get ur mark, because it shows in the paper somewhere..it doesnt matter if u write it in the place where u should right ur answer or not but logically u should get ur mark cuz u got it right and that shows the examiner u understood the question, so u deserve the mark...thats what i guess though

 

[vivo990]

bro u r confused...... there is a HUGE diff btwn equilibrium constant and equilibrium yield....but it's okay now chem is over for AS

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 no longer in equilibrium. "hanges in concentrations or pressures result in a perturbation of the equilibrium such that the system is momentarily not at
equilibrium"

 

[vivo990]

You, sir thought way too much for the question.

Here's the link to a similar reaction and it tells you what happens to the equilibrium yield when you have a higher pressure:
http://webpages.sou.edu/~chapman/ch203/Ch15/15_56.html

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. "Changes in concentrations or pressures result in a perturbation of the equilibrium such that the system is momentarily not at
equilibrium."

 

[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