2017 H2 Chemistry JC1 & 2 students post your questions here

last SG · 2016-12-14T03:15:46+00:00

http://www.bedokfunlandjc.com ------------------------------------------------------------------------------This thread is open to all 2017 Singapore JC stud...

UltimaOnline SG

Hello, I have a question, how exactly does a non-polar solvent dissolve a non-polar solute?

Whether the solute and solvent are polar or non-polar, the same consideration below applies. It's just that "like dissolves like" is a shortcut guideline that usually works.

In terms of enthalpy change (delta H), heat released is comparable with heat absorbed (overall enthalpy change preferably exothermic, but can be slightly endothermic), in terms of physical interactions (van der Waals, and if applicable, hydrogen bonds or ionic bonds) formed and broken.

In terms of entropy change (delta S), there will be a favourable positive entropy change upon mixing.

Overall quantitatively, Gibbs free energy change must be negative for the dissolving to be thermodynamically favourable, according to the formula delta G = delta H - T x delta S.

When you (ie. all students using this forum) ask me qns, and if you don't fully understand my replies, I won't spoonfeed you on the forum, so you'll have to research it further online (Google by thy sword, Wikipedia be thy shield), and/or ask your school teacher or private tutor, and/or join my BedokFunland JC tuition for more detailed, personalized guidance.

Jh2424 SG

Whether the solute and solvent are polar or non-polar, the same consideration below applies. It's just that "like dissolves like" is a shortcut guideline that usually works.

In terms of entropy change (delta S), there will be a favourable positive entropy change upon mixing.

Overall quantitatively, Gibbs free energy must be negative for the dissolving to be thermodynamically favourable, according to the formula delta G = delta H - T x delta S.

So the 'energy' explaination for why something dissolves is more relevant than describing the physical process?( like how the polar water molecules pulls away the NaCl crystals)

UltimaOnline SG

So the 'energy' explaination for why something dissolves is more relevant than describing the physical process?( like how the polar water molecules pulls away the NaCl crystals)

Of course, because the physical process is able to occur only because the overall thermodynamics is favourable enough to allow it.

Jh2424 SG

Of course, because the physical process is able to occur only because the overall thermodynamics is favourable enough to allow it.

Oh I see.. Thanks :)

UltimaOnline SG

Jh2424 SG

Hello, does water form 2 or 4 hydrogen bonds? Im not too sure about this, thanks

UltimaOnline SG

Hello, does water form 2 or 4 hydrogen bonds? Im not too sure about this, thanks

Depends on state. As gaseous (assuming ideal gas behavior), 0. As liquid, fluctuates (many times per second) between 2 to 4. As solid, 4 (for the specific crystalline structure of ice relevant to A level syllabus).

Jh2424 SG

Thank you, I have another question, for ammonia, i dont understand why it forms just 2 hydrogen bonds, why can't the two hydrogens(left over after nitrogen and hydrogen) take part in hydrogen bonding?

UltimaOnline SG

They can, but when you're talking about intermolecular H bonding between NH3 molecules, notice the ratio of H bond acceptors (ie. the N atoms) to H bond donors (ie. the H atoms) is 1:3, while the required ratio per H bond is 1:1.

Analogously, in a class of 1 guy and 3 girls, how many (monogamic heterosexual) couples can you have at most? Concordantly, on average only 2 atoms (1 N and 1 H) can participate in H bonding between NH3 molecules.

Jh2424 SG

Sorry, i dont quite understand your explanation , going by your analogy, if 1 'guy' gets 1 'girl', whats going to stop the other 'girls' from getting other 'guys'? (unless you are talking about just 3 ammonia molecules)

UltimaOnline SG

Then you'll run out of guys much sooner than girls, in a school of 10 classes, each class with 1 guy and 3 girls.

Jh2424 SG

Then you'll run out of guys much sooner than girls, in a school of 10 classes, each class with 1 guy and 3 girls.

Hmm, I think I understand what you mean, thanks!

UltimaOnline SG

"READ THIS BEFORE YOU SHOP FOR CNY MANDARIN ORANGES, FULL OF SULPHUR & CHEMICALS"

https://www.allsingaporestuff.com/article/read-you-shop-cny-mandarin-oranges-full-sulphur-chemicals

Note : it's now spelled sulfur, not sulphur. The Brits (including Cambridge) have agreed to the American spelling. So JC students reading this, don't get penalized, yo.

UltimaOnline SG

In the 1st experiment, what's cool is that the slug fragmented into 2 equally lethal parts *in mid air* before reaching the target, *both* fragments subsequently penetrating the solid metal armor. In the 2nd experiment, what's cool is that upon hitting the solid metal armor, the slug fragmented into many dozens of molten metal shrapnel fragments, radiating spectacularly sideways (imagine if you were standing just beside where the slug hit), as illustrated by the explosions of the 2 filled soft drink bottles to the sides of the metal armor. Chemistry note : a "carbide" is a compound composed of carbon and a more electropositive element (eg. a metal), and isn't a metal per se, as erroneously stated by the YouTube video host. But let's not fault him, and appreciate his video making and sharing efforts instead. He can leave the Chemistry checks to us... makes you appreciate taking H2 Chemistry more and more everyday, don't it? ;Þ

https://www.youtube.com/watch?v=voM13PNFYZI

ArJoe SG

Hi ultimaonline, sorry for disturbing you again.

in the recent 2016 h2 chem paper 2, there was one qn on explain how heterogeneous catalyst works. Would it be sufficient if one just writes partially filled 3d orbitals, reactant molecules adsorbed onto xatalyst surface, weakening the bonds, lower ea, increase frequency of effective collisions, without saying desorption? Qn was 3 marks for explanations

cu is less reactive than k. Because cu have higher nuclear charge amd shielding effect than k, but 3d electrons in cu do not provide as effeftive shielding effect as compared to s or p electrons, since 3d orbitals are more diffused. So valence electrons in cu are more strongly attracted to the nucleus, hence cu less reactive. Is this correct?

Thanks!

UltimaOnline SG

Hi ultimaonline, sorry for disturbing you again.

Thanks!

Without desorption, you'll be penalized 1 mark, regardless of how many other points you write, because desorption is an essential point.

Yes, correct for K vs Cu.

cienhan SG

hello ultimaonline,

when heated with carbon, azurite produces copper, carbon dioxide and steam as the only products. write a balanced eqn for the reaction of azurite with hot carbon. azurite is Cu3(OH)2(CO3)2

thank you in advance.

UltimaOnline SG

hello ultimaonline,

when heated with carbon, azurite produces copper, carbon dioxide and steam as the only products. write a balanced eqn for the reaction of azurite with hot carbon. azurite is Cu3(OH)2(CO3)2

thank you in advance.

Decomposition :

OH-, OH- [proton transfer] O 2-, H2O.

CO3 2-, CO3 2- [elimination] O 2-, CO2, O 2-, CO2.

Cu2+, Cu2+, Cu2+, O 2-, O 2-, O 2- as ionic solid + CO2, CO2, H2O as molecular gases.

Redox :

Cu2+, Cu2+, Cu2+, O 2-, O 2-, O 2- + 1.5 C [electron transfer + covalent bond formation] Cu, Cu, Cu as metallic solid + 1.5 CO2 as molecular gas.

Hence overall balanced equation is :

1 Cu3(OH)2(CO3)2 (s) + 1.5 C (s) ----> 3 Cu (s) + 3.5 CO2 (g) + 1 H2O (g)

This is the BedokFunland JC way, seeing clearly the mechanisms involved (proton transfer, elimination, electron transfer, covalent bond formation) in the type of reactions occurring (decomposition, redox) and recognizing the underlying thermodynamic motivation (positive entropy change) for it.

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