So I should start with a topical TYS and get a yearly TYS for cross referencing of answer, right? My concern for the answer sheet is about the answering method, to ensure i get the marks. The current Topical TYS I'm using is from 2003-2012. No problem with that right? How about the CS Toh practice book?
Originally posted by senga:So I should start with a topical TYS and get a yearly TYS for cross referencing of answer, right? My concern for the answer sheet is about the answering method, to ensure i get the marks. The current Topical TYS I'm using is from 2003-2012. No problem with that right? How about the CS Toh practice book?
In addition, if you have Android handphone with internet data access and/or can read Google Play ebooks on your phone or PC, and you can use your parents' or your own credit or debit cards, I strongly recommend you also buy CS Toh's TYS solutions (only soft copy available, but not pdf formal, you need to login to access the Google server each time you want to read CS Toh's ebook), to cross reference against the other 2 or 3 hardcopy TYS you're gonna buy from Popular.
Trust me and use this approach : I myself have bought and use 4 different TYS solutions every year, and have noted that answers vary widely across the different TYSes (just as they would across different Singapore JCs for Prelim papers and TYS papers, since every H2 Chem teacher in every JC has a different opinion of what the best answer should be). Your best bet is to cross reference the different answers across the TYSes, pick 2 of the best answers, and combine their points to write the best possible answer that would secure your marks in the 2016 A levels.
Lastly, be sure to post here on this forum, whenever you come across any such different answers and you're not sure which is the best answer. There are hundreds of JC students who lurk on the forums here, and they'll also benefit from your questions here. I'll advise all of you on which are the best answers to write for your A levels.
Purchase CS Toh's H2 Chem TYS ebooks on Google Play here :
http://www.post-1.com/step-by-step/?a=book_al_soln_chem
PS. Senga, for Physics & Math questions and advice on preparing for A levels (especially from Eagle, the H2 Physics & H2 Math expert), post in this Physics thread here :
http://sgforums.com/forums/2297/topics/492259
So I should read through the study guide once then start doing the yearly TYS for my study approach. For paper 5, just need to diligently practise the planning questions from paper 2 right? thanks for the advice. Will post my queries here.
Originally posted by senga:So I should read through the study guide once then start doing the yearly TYS for my study approach. For paper 5, just need to diligently practise the planning questions from paper 2 right? thanks for the advice. Will post my queries here.
Yes, no worries about Planning and Paper 5. For Paper 5, just follow the given instructions carefully. For Planning Qn (Paper 2), skip it first. Complete the rest of the Paper, then go back to it after you're done with the rest of the Paper, and write whatever you can.
If you *really* want to prepare for Planning Qn (ie. if you need A grade for H2 Chem because of your desired Uni course), you should buy and study Chan Kim Seng & Jeanne Tan's H2 Chem Planning Book, available from Popular Bookstore (the larger outlets, not the smaller outlets) :
Originally posted by senga:So I should start with a topical TYS and get a yearly TYS for cross referencing of answer, right? My concern for the answer sheet is about the answering method, to ensure i get the marks. The current Topical TYS I'm using is from 2003-2012. No problem with that right? How about the CS Toh practice book?
for Math and Physics, i recommend you to use cstoh's books as well.
the standard enthalpy change of combustion, is the enthalpy change(or heat evolved) when one mol of a substance is completely burnt in (excess) oxygen under standard conditions, of 298K and 1 atm.
how can combustion occur at 25°C i.e. 298K?
also, my teacher noted that it's better to specify and include in writing '298K , 1 atm' in addition to writing standard conditions.
for Boiling points and thermal stabilities of hydrogen halides, HX, both involves the increase of temperature(and energy)
is it that in general, the BP(in temperature units K or °C) is lower than the Decomposition temperature.
HI, HBr, HCl, HF are all gaseous state at room temp.
at temperature higher than room temp, then HX will decompose to form H2(g) and X2(g) .
Originally posted by Flying grenade:the standard enthalpy change of combustion, is the enthalpy change(or heat evolved) when one mol of a substance is completely burnt in (excess) oxygen under standard conditions, of 298K and 1 atm.
how can combustion occur at 25°C i.e. 298K?
also, my teacher noted that it's better to specify and include in writing '298K , 1 atm' in addition to writing standard conditions.
Common sense stuff don't need your teacher to tell you, right?
Originally posted by Flying grenade:for Boiling points and thernal stabilities of hydrogen halides, HX, both involves the increase of temperature(and energy)
is it that in general, the BP(in temperature units K or °C) is lower than the Decomposition temperature.
HI, HBr, HCl, HF are all gaseous state at room temp.
at temperature higher than room temp, then HX will decompose to form H2(g) and X2(g) .
Originally posted by UltimaOnline:
The ambient temperature is at 298K, not the temperature of the combustion itself.Common sense stuff don't need your teacher to tell you, right?
as in, can optionally write or omit 298k, 1atm becos when u write definition of standard enthalpy changes, u write under standard conditions, Cambridge should accept, but its better to be precise and just dont take risk and just include 'under standard conditions, 298K ,1atm' , cos Cambridge examiner may think the student may forget the exact standard conditions, and so just write under standard cond without specifying exactly
OOOHHH WTF, THANKS ULTIMAAAA god!!!!
just simply carry out the combustion at an ambient temp of 25°C !!!!
lol, no one thought of that at all!!!! including my teachers, and me *damn embarassing all*
Explain the significance of the value of temperature T, at which ΔS°total = 0
It represents the critical value of temperature, at which any temperature above it ,the reaction becomes thermodynamically feasible.
∆Stotal = ∆Ssystem + ∆Ssurroundings
If ∆Stotal for a reaction is positive, the reaction will be feasible, if negative it will not be feasible. extracted from http://www.presentingscience.com/quantumcasino/tutorial/totalentropychange.html
A very well written page on Entropy :
http://chem.libretexts.org/Core/Physical_Chemistry/Thermodynamics/State_Functions/Entropy/Calculating_Entropy_Changes
∆Stotal = ∆Ssystem + ∆Ssurroundings
for ∆Stotal to be positive, ∆Ssystem > ∆Ssurroundings
As a common case, if the surrounding is big enough (e.g. the lab or the universe, with respect to the beaker in which the experiment is being done is too big), one can assume that temperature of surroundings does not change during the course of the reaction
Therefore equation can be rewritten as
∆Stotal = ∆Ssystem
ΔG = ΔH - TΔS
for reaction to be thermodynamically feasible, ΔG <0
ΔH - TΔS < 0
ΔH/ΔS < T
T must be more than the calculated value of ΔH/ΔS (of the system) for reaction to be feasible.
From the conservation of energy, qsurroundings and q system are related :
qsurroundings = - qsystem
for example, heat gained by surroundings = heat flows from the system to the surroundings
https://en.m.wikipedia.org/wiki/Gibbs_free_energy # gibbs_free_energy_of_reactions
To solve the 1st part of the Cambridge question, use the formula
ΔS surroundings = - ΔH / T
(Cambridge may, in a challenging Singapore A level H2 Chem exam question, require you to qualitatively explain this formula, ie.
1. an exothermic reaction (ie. negative ΔH) generates heat which results in an increase in the disorderliness and hence entropy of the surroundings
2. the more negative the value of ΔH, ie. the more exothermic the reaction, the more positive the entropy increase of the surroundings
3. the same amount of energy released into the surroundings will make a bigger difference at a lower ambient temperature, hence the division by T)
To solve the 2nd part of the Cambridge question, use
ΔS total = ΔS system + ΔS surroundings
To solve the 3rd part of the Cambridge question, note that :
ΔS total = ΔS system + ΔS surroundings, hence
ΔS total = ΔS system + (- ΔH / T ), hence
Multiplying through by -T, we get :
- T x ΔS total = ΔH - T x ΔS system
ie. Δ Gibbs free energy = ΔH - T x ΔS system
ie. Hence, when ΔS total = 0 (as specified by the exam question), which means Δ Gibbs free energy = 0, then
0 = ΔH - T x ΔS system, hence
ΔH = T x ΔS system, hence
T = ΔH / ΔS system, which will enable you to solve the 3rd part of the Cambridge question.
To solve the 4th part of the Cambridge question, you need to understand that the reaction will only proceed if the total entropy change of the universe is zero or positive, ie. Gibbs free energy change = zero or negative. Hence to ensure the reaction proceeds, ie. for the reaction to be thermodynamically feasible, you need the temperature to be above the value calculated when you used (as instructed by the question) total entropy change = zero.
Thank you god. i am very thankful for your help. Period. (North American; period=full stop)
Thank you for explaining
ΔS surroundings = - ΔH / T ,
Δ Gibbs free energy
and
reaction will only proceed if the total entropy change of the universe is zero or positive, ie. Gibbs free energy change = zero or negative.
神
https://en.m.wikipedia.org/wiki/Geminal_diol
A geminal diol is any organic compound having two hydroxyl functional groups (-OH) bound to the same carbon atom.
The two hydroxyls in a geminal diol are easily converted to a carbonyl or keto group C=O by loss of one water molecule, thus turning the diol into a ketone. Conversely, ketones tend to combine with water to form the corresponding geminal diols. The equilibrium in water solution may be shifted towards either compound
why stability I- >Cl- > Br- > F- ?
I- less electronegative, and anionic charge density is smaller?
Originally posted by Flying grenade:why stability I- >Cl- > Br- > F- ?
I- less electronegative, and anionic charge density is smaller?
but it's a confirm fact that stability is in the order I- >Cl- > Br- > F-
that's why acidity HI > HCl > HBr
so i guess, small anionic charge density of I- (due to large ionic radius ) allows it to be more stable than the other halides, and this effect also outweighs the low electronegativity of I- (which is destabilising because it will tend to accept proton ah or idk??)?
because if the ion is electronegative then it wont accept a proton to share it's electrons and thus more stable? ?
Cambridge mark scheme answer states
the decrease in number of oxidation states after Mn (i.e. Fe to Zn) , is attributed to the increasing ionisation energy from Fe to Zn
pang peng cheong's hwach book writes after Mn(3d5) , pairing of d-electrons occurs such that there is a decrease in the number of e- available for covalent bond formation. this results in the decrease in the number of O.S. exhibited by elements with more than five d electrons (Fe to Zn)
page 62 pang peng cheong book, Transition metals
Examples of homogeneous catalyst include FeBr3 in the electrophillic substitution of benzene
Benzene and Br2 state symbols (l) , FeBr3 state symbol (l) ah????