Lurkers, register with SgForums and ask ur H2 Chem qns here!
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thank you god for your patience
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https://www.dropbox.com/s/6ebkwa8la584e8e/20160926_185432-1.jpg?dl=0
small +ve E°value means redox reaction does occur, but is it occurs very slowly, or extent of forward reaction is very small, or both?
Both, but more directly & reliably about thermodynamics + equilibrium, less directly & reliably about kinetics. If Cambridge asks, focus on the thermodynamics + equilibrium, and bring in kinetics only if the question suggests for it.eh for this p3 qn, is write in words, so do we in fact write s2o32- is oxidised to so2 by cl2 or just s2o32- oxidise to s4o62- ?
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https://www.dropbox.com/s/or00wwm2n25dtdc/20160927_093326-1.jpg?dl=0
for IV, should be I and Br(instead of CH3) Z or E to each other right
since Ar of C less than Br
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Originally posted by Flying grenade:
Both, but more directly & reliably about thermodynamics + equilibrium, less directly & reliably about kinetics. If Cambridge asks, focus on the thermodynamics + equilibrium, and bring in kinetics only if the question suggests for it.eh for this p3 qn, is write in words, so do we in fact write s2o32- is oxidised to so2 by cl2 or just s2o32- oxidise to s4o62- ?
Look up the TYS answer booklet urself lah! -
Originally posted by Flying grenade:
https://www.dropbox.com/s/or00wwm2n25dtdc/20160927_093326-1.jpg?dl=0
for IV, should be I and Br(instead of CH3) Z or E to each other right
since Ar of C less than Br
You're wrong. Shows you don't understand how to apply E/Z. Either read up on it properly, or skip it entirely (since it's not compulsory for H2 syllabus). -
tys qn say predict the products formed. cs toh say so2 wont be formed , even when the E°cell is +0.03V. so what do i write?
Cl2 can indeed oxidise s2o32- to so2, but so2 formed is in very small amounts?
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ok
https://en.m.wikipedia.org/wiki/E-Z_notation
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Originally posted by UltimaOnline:
You're wrong. Shows you don't understand how to apply E/Z. Either read up on it properly, or skip it entirely (since it's not compulsory for H2 syllabus).i got it alr.
because the qn require us to construct a pair of geometric isomers using the molecular formula of C3H4BrI. what cs toh drew, is I and Br connected to one C atom of one side of the double bond.
so in my pic, what i written is also acceptable, I and CH3 Z and for the other, E to each other
what i meant was for C3H4BrI , i can construct the isomer such that Br and I lies on different side of the double bond. and there can also be EZ isomers for that
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Originally posted by Flying grenade:
tys qn say predict the products formed. cs toh say so2 wont be formed , even when the E°cell is +0.03V. so what do i write?
Cl2 can indeed oxidise s2o32- to so2, but so2 formed is in very small amounts?
Obviously. -
Originally posted by Flying grenade:
i got it alr.
because the qn require us to construct a pair of geometric isomers using the molecular formula of C3H4BrI. what cs toh drew, is I and Br connected to one C atom of one side of the double bond.
so in my pic, what i written is also acceptable, I and CH3 Z and for the other, E to each other
what i meant was for C3H4BrI , i can construct the isomer such that Br and I lies on different side of the double bond. and there can also be EZ isomers for that
As long as you realize that your molecules (constructed the way you described) would then be constitutional structural isomers vis-Ã -vis the original molecules in the MCQ options, and no longer geometric stereoisomers. Just so you're clear. -
okay thanks god !!
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is this correct? CaO + H2O -> Ca(OH)2
cstoh writes BaO with water
https://www.dropbox.com/s/o4a7wj1qa53gx8a/20160927_181840-1.jpg?dl=0
https://www.dropbox.com/s/okb8b3lr05nxh81/20160927_182145-1.jpg?dl=0
2005 p3 qn 5c
BaO2 + 2H2O -> Ba(OH)2 + H2O2
this is unusual, that heating Ba in air generates Barium peroxide. but yes ok question leads us to get BaO2 formula.
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Originally posted by Flying grenade:
is this correct? CaO + H2O -> Ca(OH)2
cstoh writes BaO with water
https://www.dropbox.com/s/o4a7wj1qa53gx8a/20160927_181840-1.jpg?dl=0
https://www.dropbox.com/s/okb8b3lr05nxh81/20160927_182145-1.jpg?dl=0
2005 p3 qn 5c
BaO2 + 2H2O -> Ba(OH)2 + H2O2
this is unusual, that heating Ba in air generates Barium peroxide. but yes ok question leads us to get BaO2 formula.
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unusual H2O2 is produced. but yeah becos Barium peroxide is generated,which is relatively rare
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https://www.dropbox.com/s/wvvsxce2hkjff2j/20160927_200958.jpg?dl=0
why first step higher Ea? both species are not same charged, no repulsion. step 1 involves breaking of a pi bond and forming a bond.
step 2 involves breaking of a sigma bond
https://www.dropbox.com/s/9x3h7qzb0k6upvd/20160927_201057-1.jpg?dl=0
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how do u know that hydrolysis of ester is an exo reaction, the products of hydrolysis have a lower energy content of ester(reactant) ?
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Originally posted by Flying grenade:
https://www.dropbox.com/s/wvvsxce2hkjff2j/20160927_200958.jpg?dl=0
why first step higher Ea? both species are not same charged, no repulsion. step 1 involves breaking of a pi bond and forming a bond.
step 2 involves breaking of a sigma bond
https://www.dropbox.com/s/9x3h7qzb0k6upvd/20160927_201057-1.jpg?dl=0
Your points are valid, but at the same time, bear in mind that the Ea for the 1st step is somewhat higher than you would expect, due to electrostatic repulsions between the partial negatively charged O atoms between both reactants. The Ea for the 2nd step to cleave the C-O bond is somewhat lower than you would expect, because of electrostatic repulsions between the (partial and formal) negatively charged O atoms in the unstable intermediate.Finally, For MCQs, always choose the best answer. Options A & D can be ruled out for obvious reasons. Between C and B (both of which are not an accurate portrayal of the energy magnitudes involved), B is more likely, because in option C, the products are so thermodynamically unstable, that you would unlikely get much alkaline hydrolysis of the ester (despite the non-reversible step 3), and yet you should already be familiar with the fact that alkaline hydrolysis of esters occurs readily (ie. thermodynamically feasible) upon heating.
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Originally posted by Flying grenade:
how do u know that hydrolysis of ester is an exo reaction, the products of hydrolysis have a lower energy content of ester(reactant) ?
The options given are inaccurate portrayals. Also, it's actually the final products (after proton transfer in step 3) that have less energy than the initial reactants.Fyi, thermodynamic energy / enthalpy and entropy wise, because esterification and hydrolysis are not redox reactions (the OS of the C atoms are exactly the same pre and post esterification, pre and post hydrolysis), and because there is no dramatic difference in steric strain or resonance stabilization energy in the reactants vs products, as well as number of reactants vs products, any difference in thermodynamic energy / enthalpy and entropy, won't be too dramatic.
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ok thanks. u are uber(zai,outstanding, not private car for hire)
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The options given are inaccurate portrayals. Also, it's actually the final products (after proton transfer in step 3) that have less energy than the initial reactants.I've been looking at the energy profile diagram for some time. maybe idk how read it. isnt the diagram drawn such that the final products at the horizontal line i.e. nothing wrong with the diagram?
is my interpretation correct,
first intermediate(species after first step) at first hump,
the two species at the 2nd hump,
the species after step 3 is at the products horizontal line?
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Originally posted by Flying grenade:
The options given are inaccurate portrayals. Also, it's actually the final products (after proton transfer in step 3) that have less energy than the initial reactants.I've been looking at the energy profile diagram for some time. maybe idk how read it. isnt the diagram drawn such that the final products at the horizontal line i.e. nothing wrong with the diagram?
is my interpretation correct,
first intermediate(species after first step) at first hump,
the two species at the 2nd hump,
the species after step 3 is at the products horizontal line?
There are 3 steps, hence 4 species. Hence, the full energy profile diagram will have reactant --> intermediate --> intermediate --> product. But the MCQ is only asking for the 1st 2 steps only, hence the correct option only has reactant --> intermediate --> intermediate. -
ok i see alr. thanks
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2005 p1 qn 21
As long as you realize that your molecules (constructed the way you described) would then be constitutional structural isomers vis-à-vis the original molecules in the MCQ options, and no longer geometric stereoisomers. Just so you're clear.
Yes, thank you so much.
yes, it's a MCQ qn but the options are open ended, it gave us molecular formula , e.g. C3H4BrI, and we can constuct the molecules ourself
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NaBD4 or NaBH4 need to be in anhydrous conditions right???
https://www.dropbox.com/s/vjp7sikfsz4zytr/20160928_114100.jpg?dl=0