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guys u r confusing my ass...whats the answer of this question? -_______________________________- !!
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Chemistry 21
- Thread starter imanmalik
- Start date
but at the end, the question asks equilibrium yield, which is at Kp, so it does not change!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
if thats true....thats a whole lot ov marks man. -_-
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
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:O:O what about the temperature and catalyst? :S:S
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.
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
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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.
I'm sure you'd be pretty successful in life but right now fact is you thought too much for the question.
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
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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..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.
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yes ofcourse you will, but you'll have 1 mark deducted for not writing it!
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trust me on this bro ur talking about jaf , and u have no idea about his capablities
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u should remove the 'could be'
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vivo990
Assalamoalaikum wr wb!
Sorry to say, but you're wrong abt that yield part you're saying!
Check this link: http://ibchem.com/IB/ibnotes/brief/equ-sl.htm It's long, so you could just check the first line under the Haber Process, this is what it says:
There are more moles of gas on the left than the right, so a greater yield will be produced at high pressure. (The equilibrium position will lie further to the right)
^ this was the case for haber process...
For the answers, this might help: [from the same link]
Effect of Temperature
The effect of a change of temperature on a reaction will depend on whether the reaction is exothermic or endothermic. When the temperature increases, Le Chatelier's principle says the reaction will proceed in such a way as to counteract this change, ie lower the temperature. Therefore, endothermic reactions will move forward, and exothermic reactions will move backwards (thus becoming endothermic). The reverse is true for a lowering of temperature.
Effect of Pressure
In reactions where gases are produced (or there are more mols of gas on the left), and increase in pressure will force the reaction to move to the left (in reverse). If pressure is decreased, the reaction will proceed forward to increase pressure. If there are more mols of gas on the left of the equation, this is all reversed.
Effect of catalysts on equilibrium
A catalyst does not effect either Kc or the position of equilibrium, it only effects the rate of reaction. As the rate of forward reaction and reverse reaction is affected equally then the equilibrium cannot be affected.
Assalamoalaikum wr wb!
Sorry to say, but you're wrong abt that yield part you're saying!
Check this link: http://ibchem.com/IB/ibnotes/brief/equ-sl.htm It's long, so you could just check the first line under the Haber Process, this is what it says:
There are more moles of gas on the left than the right, so a greater yield will be produced at high pressure. (The equilibrium position will lie further to the right)
^ this was the case for haber process...
For the answers, this might help: [from the same link]
Effect of Temperature
The effect of a change of temperature on a reaction will depend on whether the reaction is exothermic or endothermic. When the temperature increases, Le Chatelier's principle says the reaction will proceed in such a way as to counteract this change, ie lower the temperature. Therefore, endothermic reactions will move forward, and exothermic reactions will move backwards (thus becoming endothermic). The reverse is true for a lowering of temperature.
Effect of Pressure
In reactions where gases are produced (or there are more mols of gas on the left), and increase in pressure will force the reaction to move to the left (in reverse). If pressure is decreased, the reaction will proceed forward to increase pressure. If there are more mols of gas on the left of the equation, this is all reversed.
Effect of catalysts on equilibrium
A catalyst does not effect either Kc or the position of equilibrium, it only effects the rate of reaction. As the rate of forward reaction and reverse reaction is affected equally then the equilibrium cannot be affected.
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vivo990 And for a detailed explanation as to why does this happen, like Kp stays constant, but conc. thing changes...this gives a detailed explanation, http://www.chemguide.co.uk/physical/equilibria/change.html#top that'd be helpful, inshaAllah