Stankovic123's chem q's

[lzxnl]

Yeah i suppose i can see how its works through the mathemaics of it all, but not so much in a practical sense. The concentrations of reactants and products in whatever equilibrium system, as you guys said, change by the same factor. So then, why does the equilibrium position change? Because the frequency of successful collisions for both back and forward reactions should then be changed by the same factor and remain the same if there was a change in volume no?

I really don't like discussing equilibrium in terms of reaction rates, as reaction rate expressions can be VERY messy depending on the mechanisms involved. Besides, chemical kinetics (the study of reaction rates) and equilibrium (thermodynamics) are separate, VCE has just blurred them together which can be very confusing (like you can increase both reaction rates with temperature increase, but one rate happens to increase more?). Just think of it in terms of reaction quotients; you'll never go wrong that way.

[zvezda]

I really don't like discussing equilibrium in terms of reaction rates, as reaction rate expressions can be VERY messy depending on the mechanisms involved. Besides, chemical kinetics (the study of reaction rates) and equilibrium (thermodynamics) are separate, VCE has just blurred them together which can be very confusing (like you can increase both reaction rates with temperature increase, but one rate happens to increase more?). Just think of it in terms of reaction quotients; you'll never go wrong that way.

So in fact, any question one crosses for equilibrium will simply require the Le Chatlier reference and thats it?


By the way. Was just reading the solutions for a checkpoints question, and it says that adding an inert gas "decreases the number of collisions per second for reactant molecules"

[lzxnl]

So in fact, any question one crosses for equilibrium will simply require the Le Chatlier reference and thats it?


By the way. Was just reading the solutions for a checkpoints question, and it says that adding an inert gas "decreases the number of collisions per second for reactant molecules"

Generally, yeah. Le Chatelier is a decent guide to work out the effect on the equilibrium but not the underlying mechanism.

Adding an inert gas does result in less collisions because then the reactant molecules run into the inert gas molecules. However, the overall equilibrium is still unaffected as the reaction quotient stays the same.

[jgoudie]

Adding an inert gas does not affect the rate at all.

1) It is the partial pressure that deals with collisions, partial pressure does not change
2) even if the reactant molecule runs into the inert molecule the collision is elastic and will bounce back to hit another reactant molecule.  While you are not reducing the number of pathways the two reactant molecules can reach each other you are also increasing the probability that the the two molecules will hit each other.  the two cancel each other out.

Generally, yeah. Le Chatelier is a decent guide to work out the effect on the equilibrium but not the underlying mechanism.

Adding an inert gas does result in less collisions because then the reactant molecules run into the inert gas molecules. However, the overall equilibrium is still unaffected as the reaction quotient stays the same.

[lzxnl]

Adding an inert gas does not affect the rate at all.

1) It is the partial pressure that deals with collisions, partial pressure does not change
2) even if the reactant molecule runs into the inert molecule the collision is elastic and will bounce back to hit another reactant molecule.  While you are not reducing the number of pathways the two reactant molecules can reach each other you are also increasing the probability that the the two molecules will hit each other.  the two cancel each other out.

I would advise you to read http://en.wikipedia.org/wiki/Collision_theory
You are right in saying that partial pressures deal with collisions. However, hidden in the rate constant is a steric factor which deals with the orientation of the collisions. Although I cannot say with certainty that adding an inert gas would directly affect this steric factor, it seems to me that you cannot conclusively say "adding an inert gas has no effect at all because the partial pressures are not affected".

Also, when I say that the reactant molecules run into inert gas molecules, I am not implying any reaction there. I am merely stating that instead of colliding with another reactant molecule, the first reactant molecule collides with something else. It is like having a large pool of people with different coloured shirts running around and you want two people wearing the same coloured shirt to run into each other. Sure, if we put springs on everyone so that collisions are elastic, then ideally no energy is lost. However, it takes longer for a person to run into someone else with the same colour shirt if you suddenly add a lot more people who run around too but are otherwise obstacles. A guy in a coloured shirt could run into these new obstacles instead of someone else in the same shirt.

[jgoudie]

Yes but say there, but collision theory is to do with probability also, and the likely hood of you running into someone.  Yes you run into someone with the wrong coloured shirt, but then there is equal chance that you turn around and run into someone with the correct coloured shirt quicker than if you were in a straight line.

This is all based on ideal gases, of cause when you go to extremes things change, but when dealing with kinetic molecular theory there is not change in the rate of the reaction.

A couple of other references, agreed nothing binding, i don't like to reference wiki's or answers.com but nothing is really spelt out of reaction rates decreasing or increasing anywhere i have found thus far.

1) http://www.adichemistry.com/physical/kinetics/factors/factors-affecting-rate-reaction.html
2) http://en.wikibooks.org/wiki/General_Chemistry/Introduction_to_Kinetics
3) http://chem.answers.com/chemistry-basics/the-five-rate-factors-that-influence-chemical-reaction

I would advise you to read http://en.wikipedia.org/wiki/Collision_theory
You are right in saying that partial pressures deal with collisions. However, hidden in the rate constant is a steric factor which deals with the orientation of the collisions. Although I cannot say with certainty that adding an inert gas would directly affect this steric factor, it seems to me that you cannot conclusively say "adding an inert gas has no effect at all because the partial pressures are not affected".

Also, when I say that the reactant molecules run into inert gas molecules, I am not implying any reaction there. I am merely stating that instead of colliding with another reactant molecule, the first reactant molecule collides with something else. It is like having a large pool of people with different coloured shirts running around and you want two people wearing the same coloured shirt to run into each other. Sure, if we put springs on everyone so that collisions are elastic, then ideally no energy is lost. However, it takes longer for a person to run into someone else with the same colour shirt if you suddenly add a lot more people who run around too but are otherwise obstacles. A guy in a coloured shirt could run into these new obstacles instead of someone else in the same shirt.

[lzxnl]

Yes but say there, but collision theory is to do with probability also, and the likely hood of you running into someone.  Yes you run into someone with the wrong coloured shirt, but then there is equal chance that you turn around and run into someone with the correct coloured shirt quicker than if you were in a straight line.

This is all based on ideal gases, of cause when you go to extremes things change, but when dealing with kinetic molecular theory there is not change in the rate of the reaction.

A couple of other references, agreed nothing binding, i don't like to reference wiki's or answers.com but nothing is really spelt out of reaction rates decreasing or increasing anywhere i have found thus far.

1) http://www.adichemistry.com/physical/kinetics/factors/factors-affecting-rate-reaction.html
2) http://en.wikibooks.org/wiki/General_Chemistry/Introduction_to_Kinetics
3) http://chem.answers.com/chemistry-basics/the-five-rate-factors-that-influence-chemical-reaction

I know what you are getting at, and I think we'd really have to see what experiment says.

The thing is though, once you run into someone with the wrong shirt, you then face a new scenario, which is independent to before the collision assuming a roughly uniform distribution of particles. After the collision, you also face the chance of running into some guy with the wrong coloured shirt again.

The rate equation doesn't immediately suggest any change in rate, that is correct, but I still feel as if the presence of different molecules in the way will decrease the proportion of reactant molecules out of total molecules and hence the chance that a given collision will be with a reactant molecule.

[SocialRhubarb]

... I still feel as if the presence of different molecules in the way will decrease the proportion of reactant molecules out of total molecules and hence the chance that a given collision will be with a reactant molecule.

Wouldn't adding an inert gas increase the frequency of collisions as well?

[lzxnl]

I'm not convinced that the two cancel each other out to be honest.

[Mao]

There seems to be a lot of confusion here.

@stankovic123, re: effect of pressure on equimolar reactions:
Increasing the number of collisions does not necessarily imply an imbalance in rates. Consider A+B <--> 2C, where the equilibrated system has [A]=[B)=1 M, [C]=10 M. That means, the collision C+C is 100 times more frequent than A+B. However, since the system is at equilibrium, this must mean the rates are equal, that implies C+C collisions has much fewer fruitful collision compared to A+B collisions (C+C's success rate is 1% of the success rate of A+B).
Now, if we half the volume, the concentration of everything doubles, with [A]=[B)=2M, [C]=20M. Even though the concentration of [C] dramatically increased, the frequency of collision are still in the same ratio: C+C = 400x, A+B = 4x, the ratio is still 100, the success rate of collision has not changed, so the equilibrium position does not shift.
When we discuss increases/decreases in reaction rates and their effect on equilibrium, we must talk about their relative increase compared to each other, and more importantly, the ratio of the new reaction rates. So long as the ratio of rates does not change, equilibrium position does not shift.

@nliu1995, re: kinetics vs thermodynamics:
The two are very related. A description of equilibrium generally does not give enough information to deduce kinetics, but a description of the total kinetics can give all the information about thermodynamics. Kinetics is the most fundamental manifestation of thermodynamics. If we choose to ignore kinetics because it's messy, then we are limiting ourselves to a new framework (equilibrium/thermodynamics) without any proper foundation.

@nliu1995, re: effect of inert gas:
What you are claiming here is that the probability of A+B collisions must be lower in the presence of some non-interacting C. However, since we still operate at the same temperature, then every gas molecule follow the correct Boltzmann distribution of energy, so that on average, A, B and C still move with the same speed. A simple application of Brownian motion will show that regardless of C, so long as the assumptions of Brownian motion holds, the frequency of A+B collisions do not change.
When the assumptions of Brownian motion break down, we then start dealing with more complex interactions that should not be treated with simple collision theory. So long as the ideal gas assumption is approximately true (i.e. pressure is not crazy), we can safely assume that addition of inert gas does not affect rates of reaction.
Your reference to stearic factors is to do with the actual reaction mechanism, not inert gases. If you choose to center your frame of reference on A, then you will see that the spatial distribution of collisions from B will not be changed by additional collisions from C (due to Brownian motion and ideal gas assumption). This means, where the ideal gas assumption holds, stearic factors do not change with the addition of inert gas.

[lzxnl]

OK. I stand corrected.

[zvezda]

There seems to be a lot of confusion here.

@stankovic123, re: effect of pressure on equimolar reactions:
Increasing the number of collisions does not necessarily imply an imbalance in rates. Consider A+B <--> 2C, where the equilibrated system has [A]=[B)=1 M, [C]=10 M. That means, the collision C+C is 100 times more frequent than A+B. However, since the system is at equilibrium, this must mean the rates are equal, that implies C+C collisions has much fewer fruitful collision compared to A+B collisions (C+C's success rate is 1% of the success rate of A+B).
Now, if we half the volume, the concentration of everything doubles, with [A]=[B)=2M, [C]=20M. Even though the concentration of [C] dramatically increased, the frequency of collision are still in the same ratio: C+C = 400x, A+B = 4x, the ratio is still 100, the success rate of collision has not changed, so the equilibrium position does not shift.
When we discuss increases/decreases in reaction rates and their effect on equilibrium, we must talk about their relative increase compared to each other, and more importantly, the ratio of the new reaction rates. So long as the ratio of rates does not change, equilibrium position does not shift.

@nliu1995, re: kinetics vs thermodynamics:
The two are very related. A description of equilibrium generally does not give enough information to deduce kinetics, but a description of the total kinetics can give all the information about thermodynamics. Kinetics is the most fundamental manifestation of thermodynamics. If we choose to ignore kinetics because it's messy, then we are limiting ourselves to a new framework (equilibrium/thermodynamics) without any proper foundation.

That does make sense, but then how does the equilibrium position shift for reactions where the mol ratios on each side of the equation are not equal? Whats the mechanism behind this? Your concentrations would still be changed by the same factor with a decrease/increase in volume wouldnt they?

[zvezda]

Also,
There is a question from vcaa 2008 which asks to find "the concentration of a methanoic acid solution that will have the same pH" as one of the acids given. What my question is, how do i know if this means the initial concentration or concentration at equilibrium? Because in the previous parts of the question, they gave concentrations of the acid solutions, some of which were weak acids, and these concentrations were used as initial concentrations.
Am i the only one finding vcaa's wording a but ambiguous?

[lzxnl]

Also,
There is a question from vcaa 2008 which asks to find "the concentration of a methanoic acid solution that will have the same pH" as one of the acids given. What my question is, how do i know if this means the initial concentration or concentration at equilibrium? Because in the previous parts of the question, they gave concentrations of the acid solutions, some of which were weak acids, and these concentrations were used as initial concentrations.
Am i the only one finding vcaa's wording a but ambiguous?

Weak acid => little ionisation => assume initial concentration = equilibrium concentration of methanoic acid. Not ambiguous

That does make sense, but then how does the equilibrium position shift for reactions where the mol ratios on each side of the equation are not equal? Whats the mechanism behind this? Your concentrations would still be changed by the same factor with a decrease/increase in volume wouldnt they?

The mechanism is the altering of the reaction quotient; it's now different to K, so a net reaction takes place.

[zvezda]

Weak acid => little ionisation => assume initial concentration = equilibrium concentration of methanoic acid. Not ambiguous

The mechanism is the altering of the reaction quotient; it's now different to K, so a net reaction takes place.

That actually leads in nicely to my next question lol. When can we not assume that initial conc. and equilibrium conc. are equal?