Exception to Le Chatlier's

[jlt0530]

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Hey guys,
this concept is driving me nuts for couple of days, and I still don't have a clear picture in my mind about this. I hope some of you guys can help me out.

So, i know there are two exceptions to the le chatlier's law. First one is about the pressure change when non-reactive gas, like helium is added, which I understand. But, the second concept is quite a paradox (at least to me) and it's confusing. Second one says that solubility of salts generally increases with increasing temperature, even when the rxn is exothermic, b/c increase in entropy becomes more significant with increasing temperature? I think i kinda understand where this coming from in terms of thermodynamics (delta G = delta H - T*delta S). But, when it comes to solving actual problems, it's so hard for me to distinguish when to apply thermodynamics vs le chatlier.

for example,
Q1.

H2(g) + Br2(g) -><- 2HBr(g)
delta H = -30KJ

in this case, since its enthalpy is negative, meaning its exothermic rxn, when the temp is incrreased, then the equilibrium will be shifted to the left according to le chatlier's principle. Easily done. But,

Q2.

2SO2(g) + O2(g) -><- 2SO3(g)
delta H = -200KJ

in this case, when the temp is increased, equilibrium doesn't shift to the left as expected according to le chatlier. Instead, the rates of both the forward and reverse rxn will increase according to the explanation.

Do you guys see what I'm struggling with?
how do we know when to apply le chatlier over the thermodynamic and vice versa? am i missing something?

i know it's a long post, but i will be appreciated if anyone can help me out here!! Thanks in advance!

 

[UCfan]

Hey guys,
this concept is driving me nuts for couple of days, and I still don't have a clear picture in my mind about this. I hope some of you guys can help me out.

So, i know there are two exceptions to the le chatlier's law. First one is about the pressure change when non-reactive gas, like helium is added, which I understand. But, the second concept is quite a paradox (at least to me) and it's confusing. Second one says that solubility of salts generally increases with increasing temperature, even when the rxn is exothermic, b/c increase in entropy becomes more significant with increasing temperature? I think i kinda understand where this coming from in terms of thermodynamics (delta G = delta H - T*delta S). But, when it comes to solving actual problems, it's so hard for me to distinguish when to apply thermodynamics vs le chatlier.

for example,
Q1.

H2(g) + Br2(g) -><- 2HBr(g)
delta H = -30KJ

in this case, since its enthalpy is negative, meaning its exothermic rxn, when the temp is incrreased, then the equilibrium will be shifted to the left according to le chatlier's principle. Easily done. But,

Q2.

2SO2(g) + O2(g) -><- 2SO3(g)
delta H = -200KJ

in this case, when the temp is increased, equilibrium doesn't shift to the left as expected according to le chatlier. Instead, the rates of both the forward and reverse rxn will increase according to the explanation.

Do you guys see what I'm struggling with?
how do we know when to apply le chatlier over the thermodynamic and vice versa? am i missing something?

i know it's a long post, but i will be appreciated if anyone can help me out here!! Thanks in advance!

n(mole) is directly proportional to T(temperature). If you increase T, you increase reactants mole (3 moles vs. 2 moles) in equlibrium. Since reactant's mole increases, reactions shifts to right.

 

[GiTsticker]

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[GiTsticker]

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[GiTsticker]

First of all, if I saw an equation on the DAT that had a -deltaH (solubilities or no) and they were adding heat, my OPINION would be that the answer would be that it shifts to the left.

However, to use a real life example:

When you're making syrup, you are dissolving massive quantities of sugar into water. In order to achieve this, you have to raise the temperature. This is an innate quality of solutes, that they dissolve more readily at higher temperatures. This will increase the rate of the reactants being made. However if this sugar so happens to release heat when it dissolves (-deltaH) then turning up the heat will increase the rate of the reactants becoming products. It so happens in this case that the heat increases the forward reaction rate more than the reverse reaction rate. Now if it released a TON of heat when it dissolved, who knows, maybe it would be different.

 

[nze82]

Hey guys,
this concept is driving me nuts for couple of days, and I still don't have a clear picture in my mind about this. I hope some of you guys can help me out.

So, i know there are two exceptions to the le chatlier's law. First one is about the pressure change when non-reactive gas, like helium is added, which I understand. But, the second concept is quite a paradox (at least to me) and it's confusing. Second one says that solubility of salts generally increases with increasing temperature, even when the rxn is exothermic, b/c increase in entropy becomes more significant with increasing temperature? I think i kinda understand where this coming from in terms of thermodynamics (delta G = delta H - T*delta S). But, when it comes to solving actual problems, it's so hard for me to distinguish when to apply thermodynamics vs le chatlier.

for example,
Q1.

H2(g) + Br2(g) -><- 2HBr(g)
delta H = -30KJ

in this case, since its enthalpy is negative, meaning its exothermic rxn, when the temp is incrreased, then the equilibrium will be shifted to the left according to le chatlier's principle. Easily done. But,

Q2.

2SO2(g) + O2(g) -><- 2SO3(g)
delta H = -200KJ

in this case, when the temp is increased, equilibrium doesn't shift to the left as expected according to le chatlier. Instead, the rates of both the forward and reverse rxn will increase according to the explanation.

Do you guys see what I'm struggling with?
how do we know when to apply le chatlier over the thermodynamic and vice versa? am i missing something?

i know it's a long post, but i will be appreciated if anyone can help me out here!! Thanks in advance!

I think you're making this more complicated than it really is. Just ignore all the nonsense about thermodynamics, because it will only confuse you.
Now, for example-2, you said that the explanation says that the rate of forward and reverse reaction increase. This is nothing new. Every time you increase the temperature, the rate of forward and reverse reaction will increase. But why doesn't this reaction go to the left, when temperature is increased? Read along...

Notice that all species that participate in this reaction are gaseous. Assuming that the volume of the container doesn't change, when we increase the temperature, the pressure increases. Why?

PV = nRT

You can see that pressure is directly proportional to temperature, so increasing the temperature, results in a higher pressure, when volume is constant.

But how does the system compensate for this change?

In order to compensate for this increased pressure, the equilibrium should move in the direction, which provides fewer moles of gas. Why?

Again, PV = nRT

Since, a pressure and number of moles are directly proportional, a drop in number of moles will lower the pressure and restore the initial condition.


2SO2(g) + O2(g) -><- 2SO3(g)

Notice that the left side has total of 3 moles of gas, while the right side has total of 2 moles of gas. Since we mentioned the equilibrium must move in the direction of few moles of gas, then an increase in temperature will cause the equilibrium to move to the right side.

Why didn't we do any of this for your first example?

In the first example, the number of moles of gas are equal at both sides, so none of this would apply.

Hope this helped!

 

[jlt0530]

n(mole) is directly proportional to T(temperature). If you increase T, you increase reactants mole (3 moles vs. 2 moles) in equlibrium. Since reactant's mole increases, reactions shifts to right.

Really? if that's the case, heat will be accumulated in the product side since this rxn is exothermic. Heat accumulation only on the product side without alleviating stress would be unfavorable, isn't it? I thought that equilibrium should be shifted to the left, increasing the concentration of reactants, not the products.

you said, increasing temperature also increases reactant moles? i know increasing temperature will increase reactant molecule's kinetic energy, but i don't think it will increase the number of molecules.

 

[GiTsticker]

Really sorry for the multiple posts...

 

[nze82]

Really? if that's the case, heat will be accumulated in the product side since this rxn is exothermic. Heat accumulation only on the product side without alleviating stress would be unfavorable, isn't it? I thought that equilibrium should be shifted to the left, increasing the concentration of reactants, not the products.

you said, increasing temperature also increases reactant moles? i know increasing temperature will increase reactant molecule's kinetic energy, but i don't think it will increase the number of molecules.

You must realize that this question requires additional though, because you're dealing with gaseous species. So, you can't simply determine which direction the equilibrium shifts, just by looking at Delta-H.
As I mentioned in my post, since you have gaseous species, changes in temperature will change the pressure, and to compensate for this change in pressure the equilibrium will do either one of the following:

a)Temp increase -> Pressure Increase -> Reaction moves to the side with lower number of moles of gas in order to compensate for this increase in pressure.

b)Temp decrease -> Pressure decrease -> Reaction moves to the side with higher number of moles of gas in order to compensate for this drop in pressure.

 

[jlt0530]

You must realize that this question requires additional though, because you're dealing with gaseous species. So, you can't simply determine which direction the equilibrium shifts, just by looking at Delta-H.
As I mentioned in my post, since you have gaseous species, changes in temperature will change the pressure, and to compensate for this change in pressure the equilibrium will do either one of the following:

a)Temp increase -> Pressure Increase -> Reaction moves to the side with lower number of moles of gas in order to compensate for this increase in pressure.

b)Temp decrease -> Pressure decrease -> Reaction moves to the side with higher number of moles of gas in order to compensate for this drop in pressure.

i posted my previous post before i read your post (did i just use post three times in one sentence? lol~). But now, i've read youf explanation, it totally makes sense!! it's amazing how helpful SDNer's explanations are, much better than many books out there! Thanks a lot!!! 😍