Chemistry Questions (1 Viewer)

[LoveHateSchool]

Re: Le Chatelier Question

Oh wait: So the temperature is increasing and due to this more of the KNO3 is being ionised. To maintain equilibrium, the forward reaction must be endothermic? :s

Disregard "overall", my bad.

Equilibrium can exist at different product to reactant ratios though, it is not when the concentration is equal, but the rate of reactions is equal (both forward and backwards equal). When you change the temperature, the system counteracts to reestablish equilibrium, but this may be "left" or "right" of where it was before.

If we are adding more product(changing concentration) or additional heat, the temperature of the system will decrease due to being endothermic, with postive delta H.

 

[Aysce]

Re: Le Chatelier Question

To produce more soluble potassium nitrate when heat increases, heat has to be on the left. When you add heat, for the system to minimise the disturbance it will shift it to the right producing more soluble KNO3.
Also iirc, heat increases rate not yield :S

Since heat increases rate, more of the KNO3 ionises as opposed to when it's lower, less of it ionises, no?

I don't know why you mentioned the bolded part, unless you misinterpreted what I was saying..

 

[LoveHateSchool]

Re: Le Chatelier Question

Since heat increases rate, more of the KNO3 ionises as opposed to when it's lower, less of it ionises, no?

Yes as heat is on the LHS, more heat means more ions (K and NO3 as products) and less means more solid KNO3!

 

[Trebla]

Re: Le Chatelier Question

Another way to think of it is using the principle that bond breaking absorbs energy and bond making releases energy. This approach assumes you understand its mechanism (which is easy for ionic reactions but not so easy when you deal with organic compounds).

A covalent bond is successfully formed if the electrons shared are able to reach a lower energy configuration than they did compared to their atomic states (otherwise the bond will not form since electrons are more stable in low energy states than high energy states).

Looking at the backward reaction, a covalent bond is formed which suggests that the electrons have lowered their energy levels. By conservation of energy, since the electrons went from a high energy state to a lower energy state, the energy difference must have been released into the surroundings (often as heat).

Looking at the forward reaction, the KNO₃ solid contains covalent bonds which must absorb energy from its surroundings in order to be broken into its constituent ions. This is because you are trying to excite electrons from a low energy level to a higher energy level (i.e. exciting the electron to remove itself from K to NO₃ which gives the K⁺ and NO₃^- ions). Going from low energy state to a high/excited energy state requires energy input.

By this interpretation we can see that the forward reaction is therefore endothermic.

(Of course with more complex systems which have complicated mechanisms beyond the scope of the HSC then you can't really use this somewhat intuitive approach)

 

[Aysce]

Re: Le Chatelier Question

Another way to think of it is using the principle that bond breaking absorbs energy and bond making releases energy. This approach assumes you understand its mechanism (which is easy for ionic reactions but not so easy when you deal with organic compounds).

A covalent bond is successfully formed if the electrons shared are able to reach a lower energy configuration than they did compared to their atomic states (otherwise the bond will not form since electrons are more stable in low energy states than high energy states).

Looking at the backward reaction, a covalent bond is formed which suggests that the electrons have lowered their energy levels. By conservation of energy, since the electrons went from a high energy state to a lower energy state, the energy difference must have been released into the surroundings (often as heat).

Looking at the forward reaction, the KNO₃ solid contains covalent bonds which must absorb energy from its surroundings in order to be broken into its constituent ions. This is because you are trying to excite electrons from a low energy level to a higher energy level (i.e. exciting the electron to remove itself from K to NO₃ which gives the K⁺ and NO₃^- ions).

By this interpretation we can see that the forward reaction is therefore endothermic.

(Of course with more complex systems which have complicated mechanisms beyond the scope of the HSC then you can't really use this somewhat intuitive approach)

Thanks for the comprehensive answer, Trebla! Didn't really know much about these technicalities before but now I sort of have an idea

=========================

Thanks to all who have contributed, has helped me a tonne

 

[Trebla]

Re: Le Chatelier Question

Thanks for the comprehensive answer, Trebla! Didn't really know much about these technicalities before but now I sort of have an idea

You'll learn all about these mechanisms at uni (i.e. quantum atomic theory)

 

[Aysce]

Re: Le Chatelier Question

Looking forward to it!

 

[Omed62]

Re: Quick question

Thanks Omed - haven't started yet haha.

Where did you get this information from? It isn't from the textbook and certainly isn't in the state ranking notes I have.

hey whats up guys, hope the info helped you guys. I simply took some time, to write them for my HSC year 2013 at the start of year 11, as i believe helping one is great thing that we can do. hence, I am willing to help you in any needs.

just ask your question.

wish you all HSC success.

 

[Omed62]

Re: Quick question

Yeh Omed62's info is same as mine... i didnt get it from a textbook, our chem teacher went through each form of radiation with us in class and clarified with a US documentary...

Hey peter,

whats up bro... yeah my brother help me with the info, since his doing medicine, he has a strong knowledge of the sciences.

btw, you hsc marks are great... I am currently year 12.

to get an hsc mark of above 95, what marks will you need to get throughout the year in assigments and HSC exam peter?

 

[Omed62]

Re: Quick question

Great explanation bro!

Thanks bro!

wish you the best in life!!!!

 

[Omed62]

Re: Le Chatelier Question

I'm a little uncertain about this question since there are no answers in the textbook and just want to confirm my answer.

http://imageshack.us/photo/my-images/820/screenshot20130131at720.png/

Hey lol,

its me omed62,

this is pretty simple lol, your at uni right.

here it is, so the solubility of KNO3 increasing as temperature increases, this would be an exothermic rxn,as the temperature release energy but due to the LeChatelier principle, it will reverse the rxn, so thus it is an exothermic rxn. This staff, I never use any sources, I have my one set of notes that explains all these in a simplest way.

emial me, if you still need help.

 

[Omed62]

Re: Le Chatelier Question

Looking forward to it!

hey bro.

Its me Omed62,

here is some info, I wrote for my classmate, because they could understand i, simple as I am concerned is that inorder to undertand chemistry is to first understand the basics.

you see what I mean, I am trying to help here. maybe, you don't or forget the work in year 12, because this question, is simple related to the year 12 HSC sylabus.

Remember how you studyed about the system of LE CHATELIER’S PRINCIPLE: see below notes. My best advise, memorise these asap becuase thrust me, it will help you in medicine, as I am aiming for medicine too, currently year 12- I really enjoy chemistry and biology, so I really enjoy studying medicine hopefully. Do you think I will be able to make it lol, depending on my knowledge.

Also do you have any umat practice papers, I could please borrow, as I am sitting the Umat this year. let me know, I can come to parra liberray to get them, also I can see you then, If you need further explanation on any thing


Reversible Reaction: is one that can proceed in both directions without reaching completion

Equilibrium: The state at which the rate of forward reaction equals the rate of reverse reaction

Closed System: One in which no energy can get in or get out

LE CHATELIER’S PRINCIPLE

“If a chemical system at equilibrium is subjected to a change in conditions, the system will adjust to re-establish equilibrium in such a way as to partially counteract the imposed change”

• Systems at equilibrium have constant concentrations of reactants and products, and equal/opposing reaction rates
• If changes are made to the conditions, the system may no longer be at equilibrium, and the system will tend to re-establish equilibrium by favouring one side of the equation.
• Used in industry to improve the yield of processes by changing conditions under which the reaction is carried out




Changing temperature:

N_2 O_(4(g))+ 57 kJ2〖NO〗_(2(g))
The forward reaction is endothermic and absorbs heat; reverse reaction releases heat.
E.g. if temperature of the system is increased, system will adjust to re-establish equilibrium in such a way as to decrease temperature. The forward reaction is favoured,
as it absorbs heat
Results in an increase in [NO2] and decrease in [N2O4]
Since N2O4 is a colourless gas while NO2 is a dark brown gas, the mixture is dark brown at higher temperatures, and a lighter colour when cooler

 

[Omed62]

Re: Potential difference in context of EMFs

I've googled what potential difference is in this context but all it is giving me are definitions that I don't understand and analogies. I know that it is synonymous for voltage, but I don't understand what it actually is and how it applies to the electrodes in the Galvanic cell. Can someone please explain this to me?

EMF definition in textbook:

The electromotive force or EMF of a galvanic cell is the potential difference (voltage) across the electrodes of the cell when a negligibly small current is being drawn. It is the maximum voltage that the cell can deliver.

Thank you!

hey Aysce,

its me omed,


bro most of the questions you ask are all in my notes somehow, I understand pretty much the whole thing about chemistry and other sciences.

Chemistry is like a piece of cake for me, this how good I am,

So in terms of galvanic cell,

potiential- means how much electrons are released from the anode to the cathode- remeber that in galvanic cell the anode (+) is where the osidation occures (loss of electrons) and in the cathode (-) where reduction occurs (gain of electrons. simply, the electric potential travels from the anode to the cathode. Hence the battery starts to function.

KEY idea is to remember: Oil Rig means Oxidation is loss and Rductaion is gain.

Again here is some other information about galvanic cell.

 The above cell can also be represented as: [ Mg(S) | Mg2+ || Pb2+ | Pb(S) ]:

Where Mg | Mg2+ represents a metal/metal ion couple.
The double line || represents the salt bridge.

Electrode/electrolyte couples have a FIXED voltage, no matter how many moles of each substance is present.

The electromotive force or EMF of a galvanic cell is the potential difference (voltage) across the electrodes of the cell when a negligibly small current is being drawn. It is the maximum voltage that the cell can deliver.

Yes, it is the maximum voltage ( in other terms to make it simple for you.... it is the largest electric current that travels from the anode electrode to the cathode electrode in a galvanic cell), see you do know it, well done....

let me know if it helped you lol,

Maybe, come to parra liberray, maybe this sat or sun day if you really need an understand I will bring my notes to help you out. let me know by PM, if your coming... also Pm me your phone number.

 

[Aysce]

Determining acidity/alkalinity of a salt

Hey guys.

I'm a little stuck on these 2 questions.

Q. For each of the salts below, give the formula and state whether you would expect 0.1 mol/L aqueous solutions to have a pH of about 7, less than 7 or greater than 7. Explain why, giving equations where necessary.

f. KBr

K+ won't hydrolyse but stay in ionic form

Br^- + H2O <===> HBr + OH-

I've checked on Google that KBr is neutral but I'm not sure why :/

h. K2SO4 - just need confirmation here

2K+ won't hydrolyse but rather stay in ionic form

SO42^- + H2O <===> HSO4^- + OH^-

Hence pH of aqueous solution is > 7.

Thanks guys!

 

[madharris]

Re: Determining acidity/alkalinity of a salt

I was always taught this way:
K+ + H2O <--> KOH + H+ (Doesn't Work as KOH is a strong base, shifting the equilibrium to the left)
Br- + H2O <--> HBr + OH- (DOesn't work as HBR is a strong acid, shifting the equilibrium to the left)
.: since neither H+ ions or OH- ions are produced, it is a neutral salt

K+ + H2O <--> KOH + H+ (Doesn't work as KOH is a strong base, shifting the equilibrium to the left)
SO42- + 2H2O --> H2SO4 + 2OH- (Doesn't work as H2SO4 is a strong acid, shifting the equilibrium to the left)
.: since neither H+ ions or OH- ions are produced, it is a neutral salt

 

[nightweaver066]

Re: Determining acidity/alkalinity of a salt

f) All halogen halides are strong acids.

Since K+ is a very weak acid (since KOH is a strong base), it won't cause hydrolysis but will stay in ionic form as you said.

HBr -> H+ + Br-

Br- is a very weak base (as HBr is a strong acid) as it goes to completion and so won't cause hydrolysis but will stay in ionic form.

Therefore, KBr is a neutral salt.

Another way of seeing this is that a strong acid + strong base -> neutral salt, so KBr would be a neutral salt.

h) HSO4- *equilibrium* H+ + SO42-

HSO4- is a weak acid, so SO42- will be slightly basic so will cause hydrolysis.

Similar reasoning to f), K+ won't cause hydrolysis.

Therefore, it will be slightly basic and pH of aqueous solution is >7.

Another way to look at this is that HSO4- is a weak acid and KOH is a strong base, therefore the product must be a basic salt => pH > 7.

 

[Aysce]

Re: Determining acidity/alkalinity of a salt

Alright, thanks a lot guys

And madharris, can I just ask why the equilibrium will shift left if KOH is a strong base? Is it because that it has a great tendency to attract a proton, hence it does this and forms the reactants, causing the equib to shift left?

 

[Dylanamali]

Re: Determining acidity/alkalinity of a salt

Studying whole HSC Chem syllabus by yourself in prep for uni??
love dat commitment.

 

[Aysce]

Re: Determining acidity/alkalinity of a salt

Studying whole HSC Chem syllabus by yourself in prep for uni??
love dat commitment.

Kekeke

Thanks Dylan

Means a lot coming from you

 

[someth1ng]

Re: Determining acidity/alkalinity of a salt

K2SO4 is actually a basic salt.

Even though H2SO4 is a strong acid, only the first ionisation is strong.

H2SO4 --> HSO4- + H+
HSO4- <--> SO42- + H+

Therefore:
K2SO4 --> 2K+ + SO42-
SO42- + H2O <--> HSO42- + OH-