Say for example you started off with two containers, both filled with the same amount of N2 and H2 in the same 1:3 ratio.
Now let's assume one container is heated to 200 C and one to 400 C.
The container at 400 C will obviously have the reactants forming products at a faster initial rate so therefore will reach equilibrium in a smaller time span.
However, when both containers have reached equilibrium and you compare the equilibrium compositions, will you notice a greater amount of NH3 in the container at 200 C Or 400 C?
I know this seems like a pretty basic question with an obvious answer.....but i just wanted to clarify (since every chemistry textbook i've read doesn't seem to) that the initial temperatures and other conditions affect the eventual equilibrium composition and that they're not just referring to what changing the conditions at equilibrium will do to the composition.
p.s: the answer i have to this question already: The lower temperature will favour the forward exothermic reaction so therefore the NH3 in the 200 C container will be at a higher amount at equilibrium than in the container at 400 C. However, the container at 400 C will obviously have reached equilibrium faster but the NH3 yield will be smaller.
Inflaton,
I am not sure I understand your question. The answer you give seems correct to me. You may be over thinking this.
Regards,
Caleb
I am not sure I understand your question. The answer you give seems correct to me. You may be over thinking this.
Regards,
Caleb
Sorry, what i was basically asking is 'do you consider the initial conditions to have an effect on the eventual equilibrium or do you only refer to conditions having an effect on the equilibrium content once the reaction has reached equilibrium?'
I also have some other queries about equilibrium:
1.If rate of reaction can be defined as the change in concentration of a reactant or product over time then how, at equilibrium, can the rates be anything over than zero since technically there's no further change in concentration of reactants or products?
2. If you have a reversible reaction X<->2Y then to maintain equilibrium the same conc of X decrease per second must be equivalent to the increase in X caused by the Y being used up. However, in this reaction to produce, say 5 mol of X per second would require 10 mol of Y to be used per second which would mean the rate of decrease of X per second is not the same as Y. This isn't really a question just a problem.
3.When the position of equilibrium shifts to cancel any changes introduced to the system at equilibrium does the rate of one reaction increase for this to occur?
This isn't made clear in any textbook I've read....it just says "shifts"
4.A seemingly paradoxical statement i need help understanding
"If the reaction quotient is less/more than the value of Kc then the system is not at equilibrium. Therefore the position of equilibrium shifts until the system reaches equilibrium."
This doesn't make sense since the position of equilibrium is meant to be the ratio of reactants to products at equilibrium.
I also have some other queries about equilibrium:
1.If rate of reaction can be defined as the change in concentration of a reactant or product over time then how, at equilibrium, can the rates be anything over than zero since technically there's no further change in concentration of reactants or products?
2. If you have a reversible reaction X<->2Y then to maintain equilibrium the same conc of X decrease per second must be equivalent to the increase in X caused by the Y being used up. However, in this reaction to produce, say 5 mol of X per second would require 10 mol of Y to be used per second which would mean the rate of decrease of X per second is not the same as Y. This isn't really a question just a problem.
3.When the position of equilibrium shifts to cancel any changes introduced to the system at equilibrium does the rate of one reaction increase for this to occur?
This isn't made clear in any textbook I've read....it just says "shifts"
4.A seemingly paradoxical statement i need help understanding
"If the reaction quotient is less/more than the value of Kc then the system is not at equilibrium. Therefore the position of equilibrium shifts until the system reaches equilibrium."
This doesn't make sense since the position of equilibrium is meant to be the ratio of reactants to products at equilibrium.
QUOTE (Inflaton+Feb 2 2009, 06:02 AM)
Sorry, what i was basically asking is 'do you consider the initial conditions to have an effect on the eventual equilibrium or do you only refer to conditions having an effect on the equilibrium content once the reaction has reached equilibrium?
Generally speaking if you're talking about an equilibrium reaction, and an equilibrium constant, then it's the equilibrium conditions you should be paying attention to, not the initial conditions.
Generally speaking if you're talking about an equilibrium reaction, and an equilibrium constant, then it's the equilibrium conditions you should be paying attention to, not the initial conditions.
QUOTE (Inflaton+Feb 2 2009, 06:02 AM)
I also have some other queries about equilibrium:
1.If rate of reaction can be defined as the change in concentration of a reactant or product over time then how, at equilibrium, can the rates be anything over than zero since technically there's no further change in concentration of reactants or products?
Equilibrium constant is different from reaction rate, the equilibirum constant is simply the ratio of the product concnetion and the reactant concentration.
In other words:
for:
aA + bB <-> cC + dD
Kc = ( [C]^c . [D]^d ) / ( [A]^a . [B ]^b )
The bigger the number, the more strongly it favours the products. The smaller the number, the more strongly it favours the reactants, and it is strongly temperature dependant.
1.If rate of reaction can be defined as the change in concentration of a reactant or product over time then how, at equilibrium, can the rates be anything over than zero since technically there's no further change in concentration of reactants or products?
Equilibrium constant is different from reaction rate, the equilibirum constant is simply the ratio of the product concnetion and the reactant concentration.
In other words:
for:
aA + bB <-> cC + dD
Kc = ( [C]^c . [D]^d ) / ( [A]^a . [B ]^b )
The bigger the number, the more strongly it favours the products. The smaller the number, the more strongly it favours the reactants, and it is strongly temperature dependant.
QUOTE (Inflaton+Feb 2 2009, 06:02 AM)
2. If you have a reversible reaction X<->2Y then to maintain equilibrium the same conc of X decrease per second must be equivalent to the increase in X caused by the Y being used up. However, in this reaction to produce, say 5 mol of X per second would require 10 mol of Y to be used per second which would mean the rate of decrease of X per second is not the same as Y. This isn't really a question just a problem.
Think of it this way.
what you have is two competing reactions
X -> 2Y
2Y -> X
All you're doing when you change the conditions, is your making one reaction more favoured, and the other less favoured, so when you look at it at any occasion, you see the products of one reaction more than you see the products of the other.
Think of it this way.
what you have is two competing reactions
X -> 2Y
2Y -> X
All you're doing when you change the conditions, is your making one reaction more favoured, and the other less favoured, so when you look at it at any occasion, you see the products of one reaction more than you see the products of the other.
QUOTE (Inflaton+Feb 2 2009, 06:02 AM)
3.When the position of equilibrium shifts to cancel any changes introduced to the system at equilibrium does the rate of one reaction increase for this to occur?
From what I recall, pretty much, yeah.
From what I recall, pretty much, yeah.
QUOTE (Inflaton+Feb 2 2009, 06:02 AM)
4.A seemingly paradoxical statement i need help understanding
"If the reaction quotient is less/more than the value of Kc then the system is not at equilibrium. Therefore the position of equilibrium shifts until the system reaches equilibrium."
This doesn't make sense since the position of equilibrium is meant to be the ratio of reactants to products at equilibrium.
The equilibrium constant is taken ONLY at equilibrium.
The reaction quotient is taken at any time.
Reaction quotient and Equilibrium constant are calculated using the same formula, so if they're not equal, then the reaction isn't at equilibrium.
If the system isn't at equilibrium, then the reaction that will bring it back into equilibrium will be favoured.
For example, if your concentration of reactants is to high, relative to the concentration of products, then the reaction which consumes the reactants will be favoured.
"If the reaction quotient is less/more than the value of Kc then the system is not at equilibrium. Therefore the position of equilibrium shifts until the system reaches equilibrium."
This doesn't make sense since the position of equilibrium is meant to be the ratio of reactants to products at equilibrium.
The equilibrium constant is taken ONLY at equilibrium.
The reaction quotient is taken at any time.
Reaction quotient and Equilibrium constant are calculated using the same formula, so if they're not equal, then the reaction isn't at equilibrium.
If the system isn't at equilibrium, then the reaction that will bring it back into equilibrium will be favoured.
For example, if your concentration of reactants is to high, relative to the concentration of products, then the reaction which consumes the reactants will be favoured.
A) But what if you have two separate containers heated to different temperatures initially with the same reactants......what would the difference in equilibrium composition be?
A modified question 1, "if, at equilibrium, the rates of the forward and back reaction are indeed the same because there's no further change in concentration of reactant or product then how does that work if you have a reaction x<->2y since in one second the amount of decrease in x will be different than the amount of decrease in y since for example 5 mol of x being depleted per second would require 10 mol of y being depleted per second?
The rate of the forward reaction would be 5mol per dm per second whereas the rate of the back reaction is 10 mol per dm per sec.
A modified question 1, "if, at equilibrium, the rates of the forward and back reaction are indeed the same because there's no further change in concentration of reactant or product then how does that work if you have a reaction x<->2y since in one second the amount of decrease in x will be different than the amount of decrease in y since for example 5 mol of x being depleted per second would require 10 mol of y being depleted per second?The rate of the forward reaction would be 5mol per dm per second whereas the rate of the back reaction is 10 mol per dm per sec.
This stuff makes my head hurt now lol..
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