AQA Chemistry Unit 2 (resit) CHEM2 10th June

Questions I can remember:

Magnesium observations were bright white flame and white solid.
simple magnesium and steam equation

MBD graph was displaced to the right with a lower peak.
Decreased pressure increases yield as right side has fewer moles so the equilibrium will shift to the right to oppose the increase in pressure blah blah.
Using 300degrees will keep the rate high whilst keeping the yield high. If temps above 300 were used the equilibrium will shift to the left (endothermic direction) to oppose the increase in temp since the forward reaction is exothermic.

Enthalpy of combustion is the enthalpy change for one mole of a compound to burn completely in oxygen with all reactants and products in standard states

A was butan-2-ol and B was butan-1-ol due to silver mirror etc
T was carboxylic acid and S was aldehyde i think

(Z)-pent-2-ene and (E)-Pent-2-ene I think they were. Geometrical isomers formed due to restricted rotation about the C=C bond.

For mechanism of 2 different reactions the reagent was KOH/NaOH and the conditions were ethanolic and heat under reflux for one and aqueous for the other, i think that what i put. The reagent acted as a base in one and electrophile in the other.

The free-radical substitution question was easy plus the CClF3 was split into two species with a free radical on each.

for the question about why it is dangerous to acidify the solution of nacl and naclo or whatever, i put that when increasing the conc of H+ ions the equilibrium will shift to the left and the conc/moles of Chlorine gas will increase and chlorine gas is toxic in high concs

For the Mg and Ba question where the two solutions mix i said BaSO4 is insoluble and a white ppt would form and MgSO4 is soluble so no ppt would form.
Ba2+ + SO42- -> BaSO4

H2SO4 + 8H+ + 8e- -> H2S + 4H2O (i think H2s was the desired product... i dont even rememebr lol)
2I- -> I2 + 2e-

I wont list the advantages and disadvantages of fermentation, they'll have a range of options

when it asked about wanting to form one product to form over another or something it was KOH/NaOH and alcoholic

an answer for one of the questions about different intermediates forming was that it formed via the secondary carbocation which is more stable than a primary carbocation

The two environmental reasons were no SO2 produced which causes acid rain and the other is no CO2 produced which is a greenhouse gas. The economic reason was that scrap iron is cheap.

Cu2+ (aq) + Fe (s) -> Cu(s) + Fe2+ (aq) i think i put, cant remember

TiO2 + 2Cl2 + 2C -> TiCl4 + 2CO
TiCl4 + 2Mg -> 2MgCl2 + Ti

TiO cannot be extracted using carbon as titanium carbide is formed which is brittle.

ZnS + 1 1/2O2 -> ZnO + SO2

I think i got enthalpy of formation for the first one as low -300 or sumthin, i cant remember so feel free to correct me, It wasnt -1000 like others said as it was calculated from enthalpy of combustion data so u do reactants-products and not the usual products-reactants. The C-C bond i think was 300 and sumthin, i dont remember.



That's all i can remember for now, sorry for the vague questions and answers but hopefully it should help some ppl

It clearly said in the Q both of the product formed also have peak at functional group of C=O


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What did everybody draw for the hydration of prop-1-ene mechanism? I used sulfuric acid instead of water, everybody's answer in my class conflicted with mine but using sulfuric acid is actually the technically correct mechanism.

ugh was it butanone and butanal or butan-1-ol and butan-2-ol?

It was H2SO4 mate, you're right

P and Q were butanone and butanal. All my friends said the question asked for A and B and put the correct alcohols, as did i. May be wrong tho..

I definitely got that propagation step, but I can't remember for the life of me if I put that termination....I hope I did

Yeah, and those were the primary and secondary alcohols right?

Yeah, and those were the primary and secondary alcohols right?

P and Q were butanone and butanal. All my friends said the question asked for A and B and put the correct alcohols, as did i. May be wrong tho..

Agreed. A and B were the alcohols, not the carbonyl compounds.

Yeah

THIS IS WHAT I DONE!!


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Are you 100% sure?

A and B are alcohols. I've got the paper right in front of me and they can literally only be a primary alcohol (B) and a secondary alcohol (A)

100% sure it asked for P Q - they wouldn't want the alcohol - as they told you what reacted with A and B - and also mentioned the C=O functional group in both products


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I never said their not alcohols - but I'm
Saying the Q asked for products of A and B which were p and Q which are butanal and Butanone


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Here's what it says on the paper:

"Use the information about compounds P and Q to identify compounds A and B. Explain your answer with reference to the functional groups in P and Q."