Unit 4 Questions MEGATHREAD!

[kenhung123]

Our discussion is limited to ideal gases

[vea]

You need to understand that it is the particles themselves who create the pressure in the first place. Less particles=release some pressure.

but doesnt the size of the particle have any part to play?

Even though the products side (3C) has less particles, the particles are bigger due to the law of conservation of mass. The size of the particles does not affect the number of collisions or rate of reaction.

[kenhung123]

Actually conservation of mass doesn't have to mean products and reactants have same mass because remember the infamous E=mc^2

[nacho]

You need to understand that it is the particles themselves who create the pressure in the first place. Less particles=release some pressure.

but doesnt the size of the particle have any part to play?

Even though the products side (3C) has less particles, the particles are bigger due to the law of conservation of mass. The size of the particles does not affect the number of collisions or rate of reaction.

hm after some thinking, i've come to the conclusion that the difference is accounted by the change in surface area. Am i correct?


i asked a friend, and he said he wasnt completely sure about what i was confused about.
I want to clearly re-state what i am confused about.
You have this scenario:

Box 1          Box 2

1 particle, weighing 5 grams, in the first box, and  in the second box
there are two particles, which weigh 4.99999999 grams and 0.00000001 grams (total 5 grams)
The environment in box 1 is EXACTLY the same as that of box 2.

pressure in each of the boxes = different or the same?

it is different right? despite the fact the mass in each is the same.
i am wondering why

[REBORN]

Two particles will collide with the box more often then one particle? :S

[nacho]

Two particles will collide with the box more often then one particle? :S

okay, but wouldnt the forces exerted by the two particles be less than that of the one particle, given the weight difference? And this difference, account for the frequency of collisions, and thus the pressure is the same?

[REBORN]

I'm sorry I have nfi, I was just using the kinetic molecular theory of gases based on my [u2] knowledge.

[Mao]

When we get down to the molecular level, almost all of our calculations are performed at the 'mean field' level, that is we no longer count the individual particle's masses/velocities, and instead take an average of everything over a long time. Why can we do this? Because firstly we have many many particles, secondly things at the molecular level happen very fast (like, less than nanosecond fast), and so everything do average out on the scale we observe things.

What is the significance of that? Based on those assumptions, we can go through a lot of statistical mechanics to find the below equation. We also find that almost all gases at low density behave according to this equation:

pV = nRT

Which means, gases behave as if they do not have a mass. This is because we know the temperature. Temperature is a measure of the mean-field kinetic energy (actually, RT=kinetic energy), and that kinetic energy includes the mass (KE=mv^2/2). From this relationship, we can see that for fixed number of moles, decreasing volume increases pressure. We can also see that if we keep volume constant, decreasing the number of particles decreases pressure.

Basically, you ought to trust the ideal gas equation. In the derivation of this equation, people have included all the things that you are worried about, and it turns out at the molecular level it doesn't matter.

[liuetenant]

pressure is essentially another word for 'concentration' that's used in relation to gases. If the pressure is high, it means you've got MORE gas particles in a given amount of space..... And leChat will want to oppose this INCREASE in 'concentration' of gas particles...So in order to decrease, the system will to in the direction where it produces the least amount of particles....hence lowering concentration....(aka pressure).... That's how i think of it...

the size of the particles don't matter. its just the amount hope that helps.

[nacho]

When we get down to the molecular level, almost all of our calculations are performed at the 'mean field' level, that is we no longer count the individual particle's masses/velocities, and instead take an average of everything over a long time. Why can we do this? Because firstly we have many many particles, secondly things at the molecular level happen very fast (like, less than nanosecond fast), and so everything do average out on the scale we observe things.

What is the significance of that? Based on those assumptions, we can go through a lot of statistical mechanics to find the below equation. We also find that almost all gases at low density behave according to this equation:

pV = nRT

Which means, gases behave as if they do not have a mass. This is because we know the temperature. Temperature is a measure of the mean-field kinetic energy (actually, RT=kinetic energy), and that kinetic energy includes the mass (KE=mv^2/2). From this relationship, we can see that for fixed number of moles, decreasing volume increases pressure. We can also see that if we keep volume constant, decreasing the number of particles decreases pressure.

Basically, you ought to trust the ideal gas equation. In the derivation of this equation, people have included all the things that you are worried about, and it turns out at the molecular level it doesn't matter.

(thanks)^1000 mao, completely forgot about the gas equation LOL!
it makes sense now

edit: just wondering though, (i do not have my textbook with me)
lechatlier's principle in regards to this doesnt just apply to gases, does it?
if not, and it also applies to solids and liquids, what accounts for the change in pressure then? since we are not dealing with gases anymore

[Mao]

Another way to think about it:

At the same temperature (kinetics energy), a heavy molecule would move slowly, and a light molecule would move very fast. However, a heavy molecule moving slowly exerts the same pressure as a light molecule moving very fast. Therefore you see the same pressure regardless of how heavy it actually is.

[nacho]

Another way to think about it:

At the same temperature (kinetics energy), a heavy molecule would move slowly, and a light molecule would move very fast. However, a heavy molecule moving slowly exerts the same pressure as a light molecule moving very fast. Therefore you see the same pressure regardless of how heavy it actually is.

oh sweet
okay this makes perfect sense now , thanks 1 million times

[Mao]

edit: just wondering though, (i do not have my textbook with me)
lechatlier's principle in regards to this doesnt just apply to gases, does it?
if not, and it also applies to solids and liquids, what accounts for the change in pressure then? since we are not dealing with gases anymore

VCE-level theory doesn't deal with solids and pure liquids. We assume that they don't participate (or participate at a constant rate), thus we don't include them in the concentration quotient (e.g. Kw for water, Ka for acids, Ksp for sparingly soluble salt [you deal with this at uni]). Most of the time this is fine.

But VCE-level theory does include solutions. Consider this:

A(aq) + 2B(aq) --> C(aq)

K=C/(A*B^2)

If we dilute the mixture by a factor of 2, the concentration of each species is halved. Our concentration quotient in effect has quadrupled: Q = 0.5C/(0.5A * (0.5B)^2) = 4 C/(A*B^2). Therefore, the system will move to the left. (high concentration quotient means too many products, not enough reactants)

We can think about this as 'pseudo-pressure', that is A, B or C each exert some kind of gas-like 'pressure' within the liquid on each other. When diluted, the pressure decreases, and so we must create more particles. This is only an aid to understand the concept. In reality there is no 'pseudo-pressure', it goes that way because the maths say so.

You can apply the gas equation to A(g) + 2B(g) <--> C(g), you will find that pV=nRT can be rearranged to give p/RT = n/V, where n/V is the concentration. Thus c=p/RT. This linear proportionality between c and p means we can use p wherever we use c. Apply the same mathematical argument as above (with the concentration quotient), you will see that the normal Le Chatelier description is simply a way to express the mathematics in words.

In conclusion, Le Chatelier's principle is a mathematical principle. We explain it qualitatively in chemistry because we cannot teach high level mathematics to every chemistry student (it's not always necessary).


From another perspective, when was the last time you were able to compress a solid or pure liquid?
Most of the time, these are essentially incompressible. So pressure doesn't really affect how a solid/pure liquid behave that much. Therefore we can't use a pressure argument.

[ruchika5]

Thermochemistry question for you,

A student mixes 50mL of 1.00M HCl with 50mL of 1.00M of NaOH and allows the following reaction to occur

H+ (aq) + OH (aq) ---> H2O (l); enthalpy= - 57.2 kJ mol-1
the student observes the temperature of the mixture rising from 25.0 degrees Celsius to 31.2 degrees Celsius.

Calculate the specific heat capacity of water.

Thanks!

[Graphite]

The trick to this question is to realise your finding the amount of energy released per mole of H+ or OH- and not the combined effect of both reactants. Hope it helps