Originally posted by Flying grenade: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? ?
Originally posted by Flying grenade: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)
Originally posted by Flying grenade: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????
Originally posted by UltimaOnline:
Whateva. As long as you don't use (aq) can liao. And if you're kiasu, specify "anhydrous" for all reactants and catalysts.
FeBr3 is solid
Originally posted by UltimaOnline:
Who do you trust?
ahh yes. ok got it liao, thanks ultima.
hmm i think both Cambridge and pangpcheong makes sense
nvm move on alr
2013 A level paper 2, qn 2aiv)
i understand that we need to address the effect of increasing temp, and increasing pressure, as separate cases.
do we need to address the effect of the concurrent increase in temperature and pressure ?
what would be the effect on the system if there were a concurrent increase in temperature and pressure ? is it correct/sufficient to write 'when temp and pressure are increased, there may not be a change in partial pressure of CO at equilibrium' ?
Originally posted by Flying grenade:2013 A level paper 2, qn 2aiv)
i understand that we need to address the effect of increasing temp, and increasing pressure, as separate cases.
do we need to address the effect of the concurrent increase in temperature and pressure ?
what would be the effect on the system if there were a concurrent increase in temperature and pressure ? is it correct/sufficient to write 'when temp and pressure are increased, there may not be a change in partial pressure of CO at equilibrium' ?
Hi, first timer here! i am facing some problems with chem equilibrium/thermochemistry... the front page of my lecture notes states that we are to 'show understanding that the position of equilibrium is dependent on the standard Gibbs free energy change of reaction' but nowhere in the lecture notes mention anything about it... can you please explain how they are related?
Originally posted by Nikkilyx:Hi, first timer here! i am facing some problems with chem equilibrium/thermochemistry... the front page of my lecture notes states that we are to 'show understanding that the position of equilibrium is dependent on the standard Gibbs free energy change of reaction' but nowhere in the lecture notes mention anything about it... can you please explain how they are related?
Tis the Age of the Internet
Google be thy Sword
Wikipedia be thy Shield
- BedokFunland JC
For general Chem conceptual qns like these, there are many websites on the internet that you're encouraged to check out on your own first (you know what they say about Singapore education system spoonfeeding students with notes, until most Sg students don't know or aren't used to taking the initiative to research out and learn for themselves in self-directed lifelong learning).
The best type of questions to ask here, is when you have doubts or disagreements with TYS or Prelim answers, and seek clarification or a 2nd opinion. That's the best way I can help you guys here.
The relevant formula is ΔG = - R T ln Kc
Mathematically, JC students are expected to know that ln 1 = 0. So if position of equilibrium lies on the LHS, Kc < 1, then ΔG > 0, hence the forward reaction is not thermodynamically feasible (under standard conditions and molarities), the backward reaction is.
Conversely, if position of equilibrium lies on the RHS, Kc > 1, then ΔG < 0, hence the forward reaction is thermodynamically feasible (under standard conditions and molarities), the backward reaction isn't.
what flows through the salt bridge? is it electrons and spectator ions ?
thanks for your reply. sorry to bother you so late. i saw this formula on my brother's old uni notes but not on my lecture notes... My lecture notes also mention that 'a quantitative treatment is not required', which i assume we have to explain without numbers/ formulas... and this is where i am having problems with... this is my explanation: when temperature increased for an exothermic reaction, TdeltaS also increases. Hence deltaG becomes more negative since deltaG = deltaH - TdeltaS and deltaH is negative. Hence the reaction occurs more readily and more products are formed. Since K = [Products]/[Reactants], and [Products] increases while [Reactants] decreases, K increases correspondingly, thus the equilibrium shifts to the right... do you think my explanation is correct?
Originally posted by UltimaOnline:
Welcome, and feel free to post your H2 Chem qns here. Nonetheless, do note that :Tis the Age of the Internet
Google be thy Sword
Wikipedia be thy Shield
- BedokFunland JCFor general Chem conceptual qns like these, there are many websites on the internet that you're encouraged to check out on your own first (you know what they say about Singapore education system spoonfeeding students with notes, until most Sg students don't know or aren't used to taking the initiative to research out and learn for themselves in self-directed lifelong learning).
The best type of questions to ask here, is when you have doubts or disagreements with TYS or Prelim answers, and seek clarification or a 2nd opinion. That's the best way I can help you guys here.
The relevant formula is ΔG = - R T ln Kc
Mathematically, JC students are expected to know that ln 1 = 0. So if position of equilibrium lies on the LHS, Kc < 1, then ΔG > 0, hence the forward reaction is not thermodynamically feasible (under standard conditions and molarities), the backward reaction is.
Conversely, if position of equilibrium lies on the RHS, Kc > 1, then ΔG < 0, hence the forward reaction is thermodynamically feasible (under standard conditions and molarities), the backward reaction isn't.
Thanks for another Great post, Ultima
Originally posted by Nikkilyx:thanks for your reply. sorry to bother you so late. i saw this formula on my brother's old uni notes but not on my lecture notes... My lecture notes also mention that 'a quantitative treatment is not required', which i assume we have to explain without numbers/ formulas... and this is where i am having problems with... this is my explanation: when temperature increased for an exothermic reaction, TdeltaS also increases. Hence deltaG becomes more negative since deltaG = deltaH - TdeltaS and deltaH is negative. Hence the reaction occurs more readily and more products are formed. Since K = [Products]/[Reactants], and [Products] increases while [Reactants] decreases, K increases correspondingly, thus the equilibrium shifts to the right... do you think my explanation is correct?
Just because 'a quantitative treatment is not required', does NOT mean you "*have* to explain without numbers/ formulas", and it also does NOT mean that Cambridge won't accept your explanation using mathematical formulae, as long as the formulae is valid, correct and relevant, even if the formulae isn't taught within the syllabus.
Lastly, to 'show understanding that the position of equilibrium is dependent on the standard Gibbs free energy change of reaction' as specified in the syllabus, all you need to do is simply state in words, that when the position of equilibrium lies on the LHS, Gibbs free energy change (ΔG) is positive and cell potential (E°cell) is negative ; conversely and concordantly when position of equilibrium lies on the RHS, Gibbs free energy change (ΔG) is negative and cell potential (E°cell) is positive; no further explanation in either words or formulae is required.
Originally posted by Flying grenade:what flows through the salt bridge? is it electrons and spectator ions ?
Thank you UltimaOnline
cs toh writes Amides are slowly hydrolysed by heating under reflux with Aqueous alkali or acid
my cher say reagents and conditions is 6moldm-3 , prolonged heating
so in Cambridge exam write conc acid/alkali or aq?
Originally posted by Flying grenade:cs toh writes Amides are slowly hydrolysed by heating under reflux with Aqueous alkali or acid
my cher say reagents and conditions is 6moldm-3 , prolonged heating
so in Cambridge exam write conc acid/alkali or aq?
Can you figure out what is your misconception before I reveal it?
Originally posted by UltimaOnline:
Aha! You've shown you've a serious misconception right there! One that almost all Singapore JC students (and maybe even some JC teachers) have!Can you figure out what is your misconception before I reveal it?
as in, either acid or alkaline hydrolysis , for acid hydrolysis, use H2SO4 or HCl(aq) , for alkaline hydrolysis use NaOH(aq) (from cs toh books)
is it cannot concentrated acid or alkaline cos the R groups may/will get protonated/deprotonated?
Originally posted by Flying grenade:as in, either acid or alkaline hydrolysis , for acid hydrolysis, use H2SO4 or HCl(aq) , for alkaline hydrolysis use NaOH(aq) (from cs toh books)
is it cannot concentrated acid or alkaline cos the R groups may/will get protonated/deprotonated?
!?!?
so mind boggling
reagents and conditions for the hydrolysis of amides also have a deeper aspect to consider??
will clarify with god soon enough, in the meantime,
what is the real/proper reagents and conditions for the hydrolysis of amides?
2CrO42- + 2H+ 《》 Cr2O72- + H2O
i know H2O is produced.
how is this an acid base reaction??
which species is the bronsted acid or base??
clarified
CrO42- : Cr2O72-
2 : 1
O in CrO42- functions as a proton acceptor bronsted base, H2SO4 function as the bronsted acid.
hence acid base reaction.
Originally posted by Flying grenade:!?!?
so mind boggling
reagents and conditions for the hydrolysis of amides also have a deeper aspect to consider??
will clarify with god soon enough, in the meantime,
what is the real/proper reagents and conditions for the hydrolysis of amides?
is Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
considered an acid base neutralisation reaction?
O in Mg(OH)2 act as bronsted base, HCl act as bronsted acid ah??
Originally posted by Flying grenade:is Mg(OH)2 + 2HCl -> MgCl2 + 2H2O
considered an acid base neutralisation reaction?
O in Mg(OH)2 act as bronsted base, HCl act as bronsted acid ah??