Rod's Chemistry 3/4 Questions Thread

[nhmn0301]

Crap, last post 50 days ago! Doesn't vce go fast!!

Having a hard time getting my head around the effects of changing pressure in a reaction;

So apparently, with any change in pressure, the reaction will oppose it by moving to the side with less gas particles. Not sure what it means by 'moving to the side with less gas particles'. Does it mean just move to the side of the reaction with less coefficients?

Don't understand the whole theory of pressure vs reaction

So for the pressure factor (it's related to volume factor as well): according to Le..... (bleh, I never know how to spell his name, but whatever).
For eg, if I have PCl3 + Cl2 <=> PCl5
assume constant temperature, if we double the pressure, we halve the volume and hence we have increase the concentration of each gas (amount/volume).  Think of it like the negative feed back system (since you did Bio ), the system will try to partially negate the "stimulus" by partially decrease the concentration of each gas/unit (i.e decrease the pressure). For this particular equation, the RHS only has 1 molecule and the LHS and 2 molecules. As a result, the system responds by favouring the forward reaction in an attempt to partially reduce the concentration of the LHS species until equilibrium is regained.
You can also think of it by using CF value:
CF = [PCl5] / ( [PCl3] x [Cl2] ) = Kc
when you introduce a change (halve volume => double concentration)
CF = 2[PCl5] / ( 2 [PCl3] x 2 [Cl2] ) = 1/2   [PCl5]/[PCl3]x[Cl2] => which is half of the original Kc value, hence CF < Kc. Hence, the system will try to equilibrate by increasing the CF value, by producing more PCl5 at the expense of PCl3 and Cl2. Hence, forward reaction is favoured.

[Rod]

So for the pressure factor (it's related to volume factor as well): according to Le..... (bleh, I never know how to spell his name, but whatever).
For eg, if I have PCl3 + Cl2 <=> PCl5
assume constant temperature, if we double the pressure, we halve the volume and hence we have increase the concentration of each gas (amount/volume).  Think of it like the negative feed back system (since you did Bio ), the system will try to partially negate the "stimulus" by partially decrease the concentration of each gas/unit (i.e decrease the pressure). For this particular equation, the RHS only has 1 molecule and the LHS and 2 molecules. As a result, the system responds by favouring the forward reaction in an attempt to partially reduce the concentration of the LHS species until equilibrium is regained.
You can also think of it by using CF value:
CF = [PCl5] / ( [PCl3] x [Cl2] ) = Kc
when you introduce a change (halve volume => double concentration)
CF = 2[PCl5] / ( 2 [PCl3] x 2 [Cl2] ) = 1/2   [PCl5]/[PCl3]x[Cl2] => which is half of the original Kc value, hence CF < Kc. Hence, the system will try to equilibrate by increasing the CF value, by producing more PCl5 at the expense of PCl3 and Cl2. Hence, forward reaction is favoured.

Ahhhh, makes much more sense nhmn. Just one more thing, how does decreasing the volume decrease increase concentration for gases?

Thanks

Oh also, where is your school up to for chem? . We finished unit 3 last week, and we're barely moving through unit 4, getting lazy. Sensing we are going to rush through it all in term 3.

[Rod]

Oh silly me, we increase the volume, hence dilute it so concentration decreases, whereas if we take out some volume concentration increase. Is that right or am I still wrong? Need to get the basics right

[Rod]

Or is it that decreasing the pressure results in less collisions so the reaction rate decreases and so does the concentration? But wouldn't everything decrese consequently?

far out aren't I confusing myself?? ahh

[nhmn0301]

Or is it that decreasing the pressure results in less collisions so the reaction rate decreases and so does the concentration? But wouldn't everything decrese consequently?

far out aren't I confusing myself?? ahh

For this example, you are changing the pressure of GAS PARTICLES. So there's no dilution here. If you have a container with gas particles inside, when you increase the volume of the container, the concentration of gas particle (I.e the amount of gas particle per volume) decreases, hence the PARTIAL pressure of each gas will decrease (they exert less force on the surface of the container). When I say increase or decrease pressure, I mean partial pressure of each gas here. To disrupt equilibrium, the change in pressure must effect the partial pressure of the gases in the reaction equation UNEQUALLY to necessitate a response, not total pressure of the whole thing. For this particular example, there is 2 molecule of gas on LHS and 1 on the RHS, hence, when I use CF calculation from above, you see that the 2 on the numerator and 4 on the denominator end up doesn't cancel down but just simplify each other? Hence, you got 1/2 of Kc. If I have something like
A + B <=> C + D
For this particular example, even if you change the pressure, the equilibrium position won't change since the partial pressure of each gas all increase EQUALLY.
CF= 2 [C] x 2 [D] / 2 [A] x 2 = Kc since 4 and 4 cancels out. Hence, the system won't trigger a response.
Alternatively, think about P= nRT/V. Pressure and volume is inversely proportional.
In short, decrease volume => increase partial pressure => system tries to decrease partial pressure => nett forward reaction is favored in this first example.
Increase volume => decrease partial pressure => system tries to increase partial pressure => nett backward reaction is favored in the first example.
Let me know of there is any errors in this explanation though.
Err.... My school is in the middle of nowhere now -,-. We are having aspirin SAC next week and only about to finish unit 3 right before term 2 ends. Meaning we only have term 3 for unit 4....comparing to where other schools are up to, my school goes the slowest lol.

[nhmn0301]

Oops sorry, I accidentally hit the bold button on my phone.... Didn't mean holding anything in my explanation.

[lzxnl]

For this example, you are changing the pressure of GAS PARTICLES. So there's no dilution here. If you have a container with gas particles inside, when you increase the volume of the container, the concentration of gas particle (I.e the amount of gas particle per volume) decreases, hence the PARTIAL pressure of each gas will decrease (they exert less force on the surface of the container). When I say increase or decrease pressure, I mean partial pressure of each gas here. To disrupt equilibrium, the change in pressure must effect the partial pressure of the gases in the reaction equation UNEQUALLY to necessitate a response, not total pressure of the whole thing. For this particular example, there is 2 molecule of gas on LHS and 1 on the RHS, hence, when I use CF calculation from above, you see that the 2 on the numerator and 4 on the denominator end up doesn't cancel down but just simplify each other? Hence, you got 1/2 of Kc. If I have something like
A + B <=> C + D
For this particular example, even if you change the pressure, the equilibrium position won't change since the partial pressure of each gas all increase EQUALLY.
CF= 2 [C] x 2 [D] / 2 [A] x 2 = Kc since 4 and 4 cancels out. Hence, the system won't trigger a response.
Alternatively, think about P= nRT/V. Pressure and volume is inversely proportional.
In short, decrease volume => increase partial pressure => system tries to decrease partial pressure => nett forward reaction is favored in this first example.
Increase volume => decrease partial pressure => system tries to increase partial pressure => nett backward reaction is favored in the first example.
Let me know of there is any errors in this explanation though.
Err.... My school is in the middle of nowhere now -,-. We are having aspirin SAC next week and only about to finish unit 3 right before term 2 ends. Meaning we only have term 3 for unit 4....comparing to where other schools are up to, my school goes the slowest lol.

The explanation is pretty good, although the system isn't trying to increase the partial pressure; the change is the number of gas particles per unit volume and the system is trying to increase that. Also, increasing the volume is a dilution; your first line is slightly off.

Or is it that decreasing the pressure results in less collisions so the reaction rate decreases and so does the concentration? But wouldn't everything decrese consequently?

far out aren't I confusing myself?? ahh

The reason why you're confusing yourself here is that yes, increasing the volume and decreasing the pressure will decrease reaction rates. However, it affects BOTH the forward and backward reactions; it affects them different in different scenarios.

[thushan]

http://www.vtextbook.com/?p=watch&subject=5&unit=4&area=15&topic=39&video=1544069

Rod - for further understanding, check the video above; if you think you have a general grasp of equilibrium and Le Chatelier's Principle, then watch from 10.25 to help answer your question, but if you think you don't understand the concept at all, then watch the whole video (~30 minutes).

[Rod]

http://www.vtextbook.com/?p=watch&subject=5&unit=4&area=15&topic=39&video=1544069

Rod - for further understanding, check the video above; if you think you have a general grasp of equilibrium and Le Chatelier's Principle, then watch from 10.25 to help answer your question, but if you think you don't understand the concept at all, then watch the whole video (~30 minutes).

Thanks so much Thushan!

Big fan of your and edward21's chemistry videos by the way !

[Rod]

Thanks Lxnl and nhmnhm

[Rod]

Hey everyone,

Why do we assume, in acid base equilibria, that the concentration of hydronium ions is the same as the concentration of the conjugate base produced in all acid-base reactions?

thanks

[keltingmeith]

n = CV
Volume is the same, so,
n = C
So, you're actually using the mole ratio. If the stoichiometry is 1:1, you assume the concentration is the same. In all of those acid-water reactions the stoichiometry will be 1:1, so you can assume the concentrations of the hydronium with the conjugate base will be the same.

[Rod]

n = CV
Volume is the same, so,
n = C
So, you're actually using the mole ratio. If the stoichiometry is 1:1, you assume the concentration is the same. In all of those acid-water reactions the stoichiometry will be 1:1, so you can assume the concentrations of the hydronium with the conjugate base will be the same.

Ahh I see, thanks buddy

[Rod]

Hey everyone, just want to confirm this, I've been hearing different things from different people

application of equilibrium and rate principles to the industrial production of one of ammonia,
sulfuric acid, nitric acid:
– factors affecting the production of the selected chemical
– waste management including generation, treatment and reduction
– health and safety considerations
– uses of the selected chemical.

That's a point from the study design. Do we need to know this for just our SAC? Or both SAC and exam?

It's not explicit in the study design and my teacher has said we don't need to know it for the exam

[Rod]

Oh also another question here; can someone please correct me:

An increase in pressure results in a decrease in volume. Hence there is less space for the particles, resulting in more collisions, increasing the rate of reaction. On the other hand, a decrease in pressure results in an increase in volume, hence there is more space for particles, resulting in less collisions. This decreases the overall reaction rate.