Login | Register

Welcome, Guest. Please login or register.

August 12, 2020, 03:56:21 pm

Author Topic: Rod's Chemistry 3/4 Questions Thread  (Read 13313 times)  Share 

0 Members and 1 Guest are viewing this topic.

nhmn0301

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 379
  • Respect: +15
  • School: The University of Melboure
  • School Grad Year: 2017
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #135 on: June 11, 2014, 10:48:36 pm »
0
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 :D), 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.
2015-2017: Bachelor of Biomedicine

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #136 on: June 11, 2014, 11:38:44 pm »
0
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 :D), 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? :P. 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.
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #137 on: June 11, 2014, 11:41:26 pm »
0
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
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #138 on: June 11, 2014, 11:42:53 pm »
0
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
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

nhmn0301

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 379
  • Respect: +15
  • School: The University of Melboure
  • School Grad Year: 2017
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #139 on: June 12, 2014, 06:40:20 am »
+1
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.
2015-2017: Bachelor of Biomedicine

nhmn0301

  • Victorian
  • Forum Obsessive
  • ***
  • Posts: 379
  • Respect: +15
  • School: The University of Melboure
  • School Grad Year: 2017
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #140 on: June 12, 2014, 06:42:58 am »
+1
Oops sorry, I accidentally hit the bold button on my phone.... Didn't mean holding anything in my explanation.
2015-2017: Bachelor of Biomedicine

lzxnl

  • Victorian
  • ATAR Notes Legend
  • *******
  • Posts: 3430
  • Respect: +210
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #141 on: June 12, 2014, 10:17:24 am »
+2
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.
2012
Mathematical Methods (50) Chinese SL (45~52)

2013
English Language (50) Chemistry (50) Specialist Mathematics (49~54.9) Physics (49) UMEP Physics (96%) ATAR 99.95

2014-2016: University of Melbourne, Bachelor of Science, Diploma in Mathematical Sciences (Applied Maths)

2017-2018: Master of Science (Applied Mathematics)

2019-: ???

Accepting students for  VCE tutoring in Maths Methods, Specialist Maths and Physics! PM for more details

thushan

  • ATAR Notes Lecturer
  • Honorary Moderator
  • ATAR Notes Legend
  • *******
  • Posts: 4978
  • Respect: +623
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #142 on: June 12, 2014, 11:15:28 am »
0
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).
Basic Physician Trainee 1 - Monash Health (2019)
Medical Intern - Alfred Hospital (2018)
MBBS (Hons.) - Monash Uni
BMedSci (Hons.) - Monash Uni


Former ATARNotes Lecturer for Chemistry, Biology

ExamPro Study Guide Author in
- VCE Chemistry
- VCE Biology
- HSC Chemistry

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #143 on: June 12, 2014, 04:35:31 pm »
0
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 ! :)
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #144 on: June 12, 2014, 05:22:58 pm »
0
Thanks Lxnl and nhmnhm :)
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #145 on: June 14, 2014, 05:07:02 pm »
0
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
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

keltingmeith

  • Honorary Moderator
  • Great Wonder of ATAR Notes
  • *******
  • Posts: 5094
  • Respect: +968
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #146 on: June 14, 2014, 05:16:40 pm »
+1
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.
Currently Undertaking: Doctor of Philosophy (PhD) in Supramolecular Photochemistry (things that don't bond but they do and glow pretty colours)

Previous Study:
Bachelor of Science Advanced (Research) - Monash University, majoring in Mathematical Statistics and Chemistry
Something in VCE, shit was too long ago to remember

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #147 on: June 14, 2014, 05:19:45 pm »
0
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 :)
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #148 on: June 15, 2014, 06:33:59 pm »
0
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
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy

Rod

  • Victorian
  • Part of the furniture
  • *****
  • Posts: 1738
  • The harder the battle, the sweeter the victory
  • Respect: +99
Re: Rod's Chemistry 3/4 Questions Thread
« Reply #149 on: June 15, 2014, 06:42:32 pm »
0
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.
2013-2014:| VCE
2015-2018:| Bachelor of Science (Neuroscience) @ UoM
2019-X:| Doctor of Dental Surgery (discontinued)
2019 -2021:| Master of Physiotherapy