Rate of reactions...

[Fifty 3rds]

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I need some help with my mental imagery...

Assume we have a reaction with two different gases at different partial pressures. 2AB(g) + 2C(g) --> A2(g) + 2BC (g) Rate = k[AB]^3[C]

If volume is decreased by decreasing the size of the container, the rate of the reaction increases. Now this is due to increases in the probability of successful molecular collisions allowing the reaction to go forward, but it is not because of the difference in molarities on the two sides of the equation? Are gasses immune to lechatelier's principles? Forgive me, I have not had general chemistry in a few years.

Now, if the pressure is increased by adding a nonreactive gas, what would you expect the rate to do?

I expected it to once again increase because of the molar disparity between products and reactant. Are gases only affected by pressure or actual addition of products or reactants? If so, then I can see that even though the new gas has been added to the container increasing the pressure, it also will interfere with the possibility of successful collisions occurring, and there will be no effect.

 

[loveoforganic]

Don't confuse thermodynamic equilibrium and reaction rate. You can have thermodynamic equilibrium lying tremendously in favor of one direction, but be unable to get there because of activation energy. Gases do obey LeChatlier's - if you increase pressure, (in this particular case) the reaction would be driven to the products because there are overall less moles of gas (reducing pressure). However, this is only thermodynamic equilibrium. Reaction rate increases with increased pressure on a gas phase reaction because there are more opportunities per unit time for the necessary collisions to occur.

If the pressure increase was achieved by adding a nonreactive gas, I'd expect the reaction to still be driven toward the products by LeChatlier's (thermodynamic), but I'd expect the reaction to decrease because you're effectively setting up numerous molecules that can be involved in unreactive collisions, rather than the collision necessary to produce a reaction.

 

[Fifty 3rds]

When an unreactive gas is added thereby increasing pressure, there is no change in rate. That is what is stated in TPR. Does everyone agree?

 

[rocketbooster]

I don't really understand your question, yet alone understand how you got that rate law. How did you get the powers of the rate law? 3 and 1? where did those come from? should be 2 and 2 from what I can tell in your post...

maybe post another sample question?

 

[amikhchi]

I'm assumign it was derived *experimentally* because unless he/she was given that it was an elementary step you can't assume that the rate law would just be reactant^coefficient right?

Or it's just completely a hypothetical example

AFAIK, the rate law is different for each T (or P if you're working with gases): If you decrease the volume, you increase rate of collisions and therefore increase the rate of the reaction.

Lowering the volume, increases the overall pressure, thus due to le chatliers principle you will stress the side with less moles of gas (products in this case). I think the fact you're confused with is that le chatliers principle will tell you which way the rxn will go to reach equilibirum, whereas the rate is just about how fast it goes(am i right?)

Adding an unreactive gas, I would assume since you're increasing pressure you would again stress product side for equilibrium because it's 3M gas vs 4M for reactant side. As for what changes it would have on rate, based on my previous logic it would seem that it would alter the rate by making it a bit more difficult for the reactants to "find eachother"

 

[rocketbooster]

I'm assumign it was derived *experimentally* because unless he/she was given that it was an elementary step you can't assume that the rate law would just be reactant^coefficient right?

Or it's just completely a hypothetical example

AFAIK, the rate law is different for each T (or P if you're working with gases): If you decrease the volume, you increase rate of collisions and therefore increase the rate of the reaction.

Lowering the volume, increases the overall pressure, thus due to le chatliers principle you will stress the side with less moles of gas (products in this case). I think the fact you're confused with is that le chatliers principle will tell you which way the rxn will go to reach equilibirum, whereas the rate is just about how fast it goes(am i right?)

Adding an unreactive gas, I would assume since you're increasing pressure you would again stress product side for equilibrium because it's 3M gas vs 4M for reactant side. As for what changes it would have on rate, based on my previous logic it would seem that it would alter the rate by making it a bit more difficult for the reactants to "find eachother"

yep, must have been done experimentally for the rate powers.

when you add a nonreactive gas, it's not going to change the rate because the nonreactive gas is not a part of the rate equation. don't overthink it. it is nonreactive, so it will have no effect on the rate of reaction.

in terms which direction it flips, you can't really determine that by looking at the number of moles. I think the OP made a typo on the products. he wrote "A2" when I'm pretty sure he meant "2A." the reactant and product ratios are then 2:2 and 2:2. when you increase pressure, the reaction wants to flip to the side with less moles of gas. neither side has less moles of gas...I don't think it changes.

also, when using lechatlier's principle, the only things that change by lechat are the concentrations of the reactants and products and thus the rate. the equilibrium does not change. also, the only variable that can change the the equilibrium constant is temperature.

 

[amikhchi]

I thought the A2 was referring to a diatomic gas

but good to know that unreactive gas wouldn't alter rate, I'm taking MCAT this saturday so I'm trying to soak up every little bit of info I can until then, and make sure I have my the concepts right.

is it safe to assume that temperature is the only property that affects rate and equilibrium?

 

[Fifty 3rds]

I did not flip anything. The two is meant to be subscript but I have yet to find that font on here. The reaction was a simple single-replacement reaction. A became a bimolecular gas in the reaction.

The rate was experimentally determined based on partial pressure concentrations and rates.
Rocketbooster, I fear you are under-thinking it, but don't take that as an attack. You are right, the non-reactive gas wouldn't interact with the reactants or products, but it would increase overall pressure. That is what is bothering me. I guess the non-reactive gas's physical interference with the reacting gases would nullify any pressure that it adds.

amikchi,
Yes you were right, I was confusing rates and equilibriums. Apparently only concentrations and temperatures affect rates. That is my understanding so far, but the pressure in the first would increase rate, yet the second doesn't. SO.... the only thing I can think of is that the increase in molecules in the container combined with an increase in pressure, results in no change, a canceling affect. What do you think?

I would still love further verification of my understanding of why the second example does not lead to formation of products more quickly. (because that was the answer in TPR)

 

[rocketbooster]

I did not flip anything. The two is meant to be subscript but I have yet to find that font on here. The reaction was a simple single-replacement reaction. A became a bimolecular gas in the reaction.

The rate was experimentally determined based on partial pressure concentrations and rates.
Rocketbooster, I fear you are under-thinking it, but don't take that as an attack. You are right, the non-reactive gas wouldn't interact with the reactants or products, but it would increase overall pressure. That is what is bothering me. I guess the non-reactive gas's physical interference with the reacting gases would nullify any pressure that it adds.

amikchi,
Yes you were right, I was confusing rates and equilibriums. Apparently only concentrations and temperatures affect rates. That is my understanding so far, but the pressure in the first would increase rate, yet the second doesn't. SO.... the only thing I can think of is that the increase in molecules in the container combined with an increase in pressure, results in no change, a canceling affect. What do you think?

I would still love further verification of my understanding of why the second example does not lead to formation of more products. (because that was the answer in TPR)

I just explained that. the pressure increase does cancel out but not for the reasons you mentioned.

like I said, it is nonreactive. it does not react with anything, so does not affect the equilibrium.

YES, it does increase the overall pressure, but guess what, that increases the pressure for both the reactants and the products. so the increase cancels out.

I'll write it out this way:

2AB + C +Inertgas ----> A2 + 2AC + Inertgas

Happy? The inert gas is on both sides of the equations. See, it cancels out. The shifts by Lechat. prin. due to pressure change is due to pressure change between the reactants and products. There is no pressure change between reactants and products. Both the reactants and products have the same increase in pressure, so the effects do nothing. It "cancels," per se.

 

[Fifty 3rds]

okay. Why does the first example result in an increase in more products? We decrease the size of the container, increasing pressure.

 

[amikhchi]

confused, but i think rocketbooster is right

 

[rocketbooster]

confused, but i think rocketbooster is right

I'm definitely right. Here, go read this: http://www.ilpi.com/genchem/demo/n2o4pressure/index.html Scroll down to about half way through the page where it takes about adding an inert gas. It's the exact same explanation I just gave you.

 

[amikhchi]

I'm definitely right. Here, go read this: http://www.ilpi.com/genchem/demo/n2o4pressure/index.html Scroll down to about half way through the page where it takes about adding an inert gas. It's the exact same explanation I just gave you.

yup, makes perfect sense now

thanks

 

[Fifty 3rds]

Thank you. Yes, that is how le chatelier's principle works.

Now in regards to rates (not equilibrium), why would a pressure increase by changing the volume of the container, result in an increase in rate?

 

[rocketbooster]

okay. Why does the first example result in an increase in more products? We decrease the size of the container, increasing pressure.

If you decrease volume, pressure goes up. Letchat wants to offset that by decreasing the pressure. What is pressure? Pressure is the force/area=force of collisions with container wall/area of container wall. So, we need to decrease pressure.

You decrease P by either decreasing F or increasing area. Lechat can't change the area of the container, so we have to decrease F. How do we decrease the total force of gas molecules hitting the container wall? We decrease the # of collisions with the wall. We decrease the # of collisions by decreasing the # of molecules. Less molecules means less collisions with the wall which means less force. Less force means less pressure on wall. Voila!

How do we decrease the # of gas molecules in the reaction? We shift to the side of the reaction produces the smaller # of gas molecules. The left side of the equation has 2 moles + 2 moles of gas molecules while the right side has 1 mole and 2 moles of gas molcules. 4>3. We want the 3 because 3<4! So, we shift to the 3! That means we shift to the right, or toward the products. 👍

 

[Fifty 3rds]

And that increases the rate of the reaction? Not just the Keq correct?

I'm sorry I didn't mean to say more products, I meant to say a quicker rate of formation of products. Geez, as if I haven't made this confusing enough.

 

[rocketbooster]

Thank you. Yes, that is how le chatelier's principle works.

Now in regards to rates (not equilibrium), why would a pressure increase by changing the volume of the container, result in an increase in rate?

pressure increases by decreasing the volume of the container. the molecules have less free space to move around in, so they are colliding with the wall more and colliding with each other more (collisions occur at a greater frequency). more collisions=more excitation of molecules=more kinetic energy=increase in rate.

 

[rocketbooster]

And that increases the rate of the reaction? Not just the Keq correct?

I'm sorry I didn't mean to say more products, I meant to say a quicker rate of formation of products. Geez, as if I haven't made this confusing enough.

it's based on more collisions of molecules. whenever you have more collisions, you have more KE, thus increased rate. there's an equation that actually relates that but you don't need to memorize it.

don't you know the effect of temperature on rate? temp always increases right? why? because it increases collision frequency, more KE, more rate etc. same idea with decreasing volume because there's less free space so the collision frequency goes up.

just to make sure you know...increased KE increases rate because it decreases the activation energy needed to start the reaction. i'm sure you know that part

 

[Fifty 3rds]

I apologize if this seems redundant. But by that same logic, shouldn't all pressure increases result in an increased rate of reaction? Apparently inert gases do not push the reaction forward, so while they do increase the pressure, there is something else nullifying that affect on the rate, because the rate stays the same. Why? (and thanks for the answers so far)

 

[rocketbooster]

I apologize if this seems redundant. But by that same logic, shouldn't all pressure increases result in an increased rate of reaction? Apparently inert gases do not push the reaction forward, so while they do increase the pressure, there is something else nullifying that affect on the rate, because the rate stays the same. Why? (and thanks for the answers so far)

If it is an increase in pressure due to a reactive gas, then yes. With a nonreactive gas, though, no.

You are only getting an increase in rate of reaction if you are increasing the collision frequency of the reactive molecules with other reactive molecules. When you add in inert gas, you increase collision frequency between reactive and nonreactive gases. No energy is transferred from a reactive to a nonreactive gas because they do not react. If no energy is being transferred, there is no increase in kinetic energy, thus no increase in rate. It does nothing.

Like before, nonreactive effects do nothing. They increase total pressure of everything in the container, but only increases in pressure of the REACTIVE molecules affects the rate and equilibrium.

 

[Fifty 3rds]

thanks Rb, that was a big help.

 

[canswiss]

I apologize if this seems redundant. But by that same logic, shouldn't all pressure increases result in an increased rate of reaction? Apparently inert gases do not push the reaction forward, so while they do increase the pressure, there is something else nullifying that affect on the rate, because the rate stays the same. Why? (and thanks for the answers so far)

I think it might be helpful to think of the increase in pressure of a gas as an increase in concentration. That is if you reduce the volume of the container you are increasing the concentrations of the gasesinvolved in the reaction. If you add an inert gas you increase the total pressure but you do not increase the concentrations of the gases involved in the reaction and , therefore do not increase the rate of the given reaction.