Login

Welcome, Guest. Please login or register.

March 28, 2024, 11:49:07 pm

Author Topic: HSC Chemistry Question Thread  (Read 1040548 times)  Share 

0 Members and 3 Guests are viewing this topic.

myopic_owl22

  • Trailblazer
  • *
  • Posts: 29
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3690 on: January 16, 2019, 05:33:14 pm »
+1
So, I thought this dot point was simple but now I'm confused.

analyse examples of non-equilibrium systems in terms of the effect of entropy and enthalpy, for example:
– combustion reactions
– photosynthesis


I thought combustion was a static equilibrium reaction rather than a non-equilibrium one.

Combustion really isn't in an equilibrium at all, since equilibrium reactions must inherently be reversible, and combustion releases too much energy (very low (i.e. negative) enthalpy and high entropy) to be reversible.
Static equilibrium exists in only reversible reactions, where the energy/ pressure requirements are too high to convert reactants to products. Therefore, the rates of each reaction are pretty much zero. The only example I've ever been taught is C(graphite) ⇌ C(diamond)
For the record, photosynthesis will also reach completion, but the higher activation energy required (as per endothermic reaction energy profiles) means it needs a catalyst - UV light.

A lot of people (myself included) will remember respiration to be the opposite of photosynthesis - respiration being a form of combustion. It's important to note that while the reactants and products appear inverted, these are two separate reactions and are not in equilibrium with each other. However, you can use this fact to deduce the relative entropy and enthalpy of each reaction - which is essentially what the dot-point is asking. Photosynthesis is endothermic (+ve enthalpy) reaction with a -ve entropy (given the large molecular size of its products), which is the opposite of all combustion reactions. In either case, ΔG will never be 0, so there'll be no equilibrium involved. Hope this answers your question! :)

david.wang28

  • MOTM: MAR 19
  • Forum Obsessive
  • ***
  • Posts: 223
  • Always do everything equanimously
  • Respect: +29
Re: Chemistry Question Thread
« Reply #3691 on: January 24, 2019, 04:03:26 pm »
0
Hello,
I have trouble with Question 3 in the link below. I have also posted my working out in the link below. Can anyone please help me with this question? Thanks :)
Selection rank: 95.25

2020-2024: Bachelor of engineering and computer science @ UNSW

fun_jirachi

  • MOTM: AUG 18
  • HSC Moderator
  • Part of the furniture
  • *****
  • Posts: 1068
  • All doom and Gloom.
  • Respect: +710
Re: Chemistry Question Thread
« Reply #3692 on: January 24, 2019, 05:13:37 pm »
+4
Hey there!

Your first step is perfect! You are using 0.01mol of phosphorus pentoxide. This also corresponds to 0.02mol of phosphoric acid by molar ratios. Subbing this into the molar ratio of the second equation, you need 0.06mol of sodium hydroxide. Using n=cV, sub in n=0.06, c=0.3 to get that v= 0.2. So the answer is C. Not quite sure where you got the 3 from. :)

Hope this helps :)

EDIT: in red
« Last Edit: January 24, 2019, 05:46:16 pm by fun_jirachi »
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Guide Links:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)
Asking good questions

david.wang28

  • MOTM: MAR 19
  • Forum Obsessive
  • ***
  • Posts: 223
  • Always do everything equanimously
  • Respect: +29
Re: Chemistry Question Thread
« Reply #3693 on: January 24, 2019, 05:29:11 pm »
0
Hey there!

Your first step is perfect! You are using 0.01mol of phosphorus pentoxide. This also corresponds to 0.02mol of phosphoric acid by molar ratios. Subbing this into the molar ratio of the second equation, you need 0.06mol of sodium hydroxide. Using n=c/v, sub in n=0.06, c=0.3 to get that v= 5. So the answer is D. Not quite sure where you got the 3 from. :)

Hope this helps :)
I checked the answer and it said C. Not sure why?
Selection rank: 95.25

2020-2024: Bachelor of engineering and computer science @ UNSW

sweetcheeks

  • Forum Obsessive
  • ***
  • Posts: 496
  • Respect: +83
  • School: ---
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #3694 on: January 24, 2019, 05:39:10 pm »
+2
Hey there!

Your first step is perfect! You are using 0.01mol of phosphorus pentoxide. This also corresponds to 0.02mol of phosphoric acid by molar ratios. Subbing this into the molar ratio of the second equation, you need 0.06mol of sodium hydroxide. Using n=c/v, sub in n=0.06, c=0.3 to get that v= 5. So the answer is D. Not quite sure where you got the 3 from. :)

Hope this helps :)

Great explanation. You've made a little error in the formula. It should be volume = moles/concentration, which would give 0.2 L rather than 5 L.

fun_jirachi

  • MOTM: AUG 18
  • HSC Moderator
  • Part of the furniture
  • *****
  • Posts: 1068
  • All doom and Gloom.
  • Respect: +710
Re: Chemistry Question Thread
« Reply #3695 on: January 24, 2019, 05:44:19 pm »
0
Oooh yes, my bad. I will edit that now. :)
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Guide Links:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)
Asking good questions

myopic_owl22

  • Trailblazer
  • *
  • Posts: 29
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3696 on: January 28, 2019, 03:55:04 pm »
0
Hi there, a quick textbook question:
"Explain why we cannot compare ksp values to determine the relative solubility between two compounds."
Is this because ksp is dimensionless? I always thought that we'd use ksp to judge if something was good at dissolving or not...
Thanks!

szetotess

  • Adventurer
  • *
  • Posts: 5
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3697 on: January 28, 2019, 04:39:53 pm »
0
helloo,

i was wondering why insoluable and sparingly soluable salts reach equilibrium and not soluable salts??

 :)

fun_jirachi

  • MOTM: AUG 18
  • HSC Moderator
  • Part of the furniture
  • *****
  • Posts: 1068
  • All doom and Gloom.
  • Respect: +710
Re: Chemistry Question Thread
« Reply #3698 on: January 28, 2019, 04:47:30 pm »
+4
Hey there!

Hi there, a quick textbook question:
"Explain why we cannot compare ksp values to determine the relative solubility between two compounds."
Is this because ksp is dimensionless? I always thought that we'd use ksp to judge if something was good at dissolving or not...
Thanks!

You can actually use Ksp to compare the relative solubility between compounds, but if and only if they dissociate to produce the same ratio of ions.  For example, you can compare the relative solubility of potassium chloride and sodium hydroxide, or even calcium carbonate, because they dissociate to produce the same ratio of ions as can be seen here:

KCl(aq)⇌K+(aq)+Cl-(aq)
NaOH(aq)⇌Na+(aq)+OH-(aq)
CaCO3(aq)⇌Ca2+(aq)+CO32-(aq)

Another example of this would be say magnesium hydroxide and calcium chloride:

Mg(OH)2(aq)⇌Mg2+(aq)+2OH-(aq)
CaCl2(aq)⇌Ca2+(aq)+2Cl-(aq)

What you can't do is compare two compounds that when ionised, form different ratios of ions. This is fundamentally because of the mathematical definition of Ksp, because for example in the compound AB⇌A++B-

If we now consider the compound AB2⇌A++2B-, our Ksp becomes


One of the reasons we can't is that when considering the units of the former Ksp, it is mol L-1, while for the latter it is mol 2 L-2, but also because you have an extra value when considering the solubility product. However, in addition to this, in the former case, when you're looking at the actual solubility, you get that Ksp=s2, while in the latter case you get that Ksp=4s3 simply because of ion ratios. Thus, even if they had really similar Ksp values, the actual solubility is heavily skewed. What you need to compare is not the Ksp, but the actual solubility s. Note that this doesn't occur with the compounds with the same ion ratios, since it'll just be s2 for both, or 4s3 or something or other. Hence, this is why you (usually) can't compare Ksp to compare relative solubility between two compounds.

Hope this helps :D

EDIT:
helloo,

i was wondering why insoluable and sparingly soluable salts reach equilibrium and not soluable salts??

 :)

Hey :)

Soluble salts will not reach equilibrium generally because they will just dissolve completely and dissociate into their ions, hence leaving none of the compound. The system will just go to completion instead of equilibrium. The opposite is true for insoluble and sparingly soluble salts; they don't have as much compound dissociating, so they will be 'stuck' in equilibrium. To put it in terms of Ksp, we know that if Ksp is small, the substance is not as soluble, and if it is large, it is more soluble. eg. For a soluble salt, more ions will be produced at equilibrium, producing a more 'top-heavy' solubility product which is larger --> more soluble. In your case, this is when the substance is completely soluble, and there are no reactants left, leading to a Ksp value that tends to infinity. For an insoluble or sparingly soluble salt, the Ksp value will be zero or close to zero respectively, indicating that at equilibrium, the solubility product is 'bottom-heavy' with a lot more reactants, and thus that the salt is not as soluble, thus forming an equilibrium.

Hope this explanation makes sense :)
« Last Edit: January 28, 2019, 04:57:58 pm by fun_jirachi »
Spoiler
HSC 2018: Mod Hist [88] | 2U Maths [98]
HSC 2019: Physics [92] | Chemistry [93] | English Adv [87] | 3U Maths [98] | 4U Maths [97]
ATAR: 99.05

UCAT: 3310 - VR [740] | DM [890] | QR [880] | AR [800]
Guide Links:
Subject Acceleration (2018)
UCAT Question Compilation/FAQ (2020)
Asking good questions

myopic_owl22

  • Trailblazer
  • *
  • Posts: 29
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3699 on: January 29, 2019, 01:22:11 pm »
0
Thanks so much fun_jirachi! Fantastic explanation - makes sense now :)

mirakhiralla

  • Trailblazer
  • *
  • Posts: 36
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3700 on: January 31, 2019, 10:45:50 pm »
0
Hey can someone please help me out with this q, thankuuu:

Calculate the volume of CO2 needed to produce 100ml of saturated solution of carbon dioxide at 25 degrees. (assume that the solution contains 100g of water)

szetotess

  • Adventurer
  • *
  • Posts: 5
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3701 on: February 06, 2019, 06:27:44 pm »
0
hi all,

i was wondering what the acid base reaction equation for
1. phosphoric acid + ammonia -->
2. acetic acid + sodium oxide -->
3. calcium oxide + ammonium nitrate -->

thanks  :) :)

sweetcheeks

  • Forum Obsessive
  • ***
  • Posts: 496
  • Respect: +83
  • School: ---
  • School Grad Year: 2016
Re: Chemistry Question Thread
« Reply #3702 on: February 06, 2019, 06:39:22 pm »
+1
hi all,

i was wondering what the acid base reaction equation for
1. phosphoric acid + ammonia -->
2. acetic acid + sodium oxide -->
3. calcium oxide + ammonium nitrate -->

thanks  :) :)

What are your thoughts? Look at the molecules, which one is acidic? which one is basic? What happens during an acid-base reaction?

szetotess

  • Adventurer
  • *
  • Posts: 5
  • Respect: 0
Re: Chemistry Question Thread
« Reply #3703 on: February 06, 2019, 06:51:04 pm »
0
What are your thoughts? Look at the molecules, which one is acidic? which one is basic? What happens during an acid-base reaction?

in the answers the answer for the first one is (NH4)3.PO4
but is there not meant to be a water molecule if it is acid base

david.wang28

  • MOTM: MAR 19
  • Forum Obsessive
  • ***
  • Posts: 223
  • Always do everything equanimously
  • Respect: +29
Re: Chemistry Question Thread
« Reply #3704 on: February 07, 2019, 06:18:57 pm »
0
Hello,
I was wondering what to do for Q6 and 2.1 a). Can anyone please help me out with these questions? Thanks :)
Selection rank: 95.25

2020-2024: Bachelor of engineering and computer science @ UNSW