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OCR Salters Chemistry F335 12th June 2013 Exam revision thread

Reply 300

martynsteel

11

Original post by hannah1994

What is the mechanism for the AlCl3 and CH3Cl?

The mechanism's shown here :-)

Reply 301

super121

5

Original post by hannah1994

What is the mechanism for the AlCl3 and CH3Cl?

It forms CH3+ and AlCl4-. The electrophile then reacts with benzene by electrophilic substitution to form CH3 bonded to benzene and HCl. The catalyst, AlCl3, is reformed.

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Reply 302

RainieXD

Hi, I'm quite new to talking on TSR ...
Emm could anyone help me with question 5d June 2011... I really don't understand it :frown:

And for 5bi why is it assumed that H+ concentration is the same as the HPO4 ^2- concentration?
And also... I really don't understand dilutions.. could someone talk me through it? I'd be really gratefull ^^ ..

Reply 303

strawberri

8

Original post by RainieXD

Hi, I'm quite new to talking on TSR ...
Emm could anyone help me with question 5d June 2011... I really don't understand it :frown:

And for 5bi why is it assumed that H+ concentration is the same as the HPO4 ^2- concentration?
And also... I really don't understand dilutions.. could someone talk me through it? I'd be really gratefull ^^ ..

Answer to Question 5d! :smile:

You know that, because it is in equilibrium:

Ka = [H⁺][HPO₄^2-]/[H₂PO₄^-]

So....

Ka/[H+] = [HPO₄^2-]/[H₂PO₄^-]

In a solution with pH 7:
pH = -log[H⁺]
[H+] = 10^-pH = 10^-7

It tells you what Ka is so you can write this equation:

[HPO₄^2-]/[H₂PO₄^-] = 6.2 x 10^-8 / 1 x 10^-7
= 0.62

This is how many times the concentration of [H₂PO₄^-] goes into the concentration of [HPO₄^2-] .

The question tells us that the concentration of (Na₂)[HPO₄^2-] is 0.1moldm^-3.

0.1/0.62 = [H₂PO₄^-] = 0.161
Seeing as this is in a solution of 1.0dm³, we can take this as the number of moles of NaH₂PO₄^-.

number of moles = mass / Mr
mass = number of moles x Mr = 0.161 x 120 = 19.32 = 19.3g

Hope this helped! :smile:

Also for 5bi...in a weak acid (indicated by the equilibrium), we assume that [H+] = [A-] because we assume that the amount of H₂PO₄^- has dissociated equally into H+ and HPO₄^2-, due to the 1:1 molar ratio between the products.

(edited 10 years ago)

Reply 304

hannah1994

7

Original post by martynsteel

The mechanism's shown here :-)

thankyou!!!

Just one quick question:
When the R-Cl reacts with AlCl3, is the pair of electrons that form a dative covalent bond with aluminium from the lone pair on Cl or the bonding pair R-Cl. Another mechanism I saw had the electrons coming from the bonding pair ...

Reply 305

hannah1994

7

What is the half equation for ammonium ions to nitrate (III) ions

I have wrote down:

NH4+ + O2 ---> NO2- + 4H+ +2e-

But surely as the nitrogen is going from an oxidation state of -3 to an oxidation state of +3, it needs to lose 6 electrons, not 2?

Reply 306

stealth_writer

4

are delta s system and delta s surr always different signs?

Also what happens to a buffer soln when small extra amounts of OH is added.

Reply 307

Jlane5000

4

Could someone please confirm for me the conditions and reactants for making a diazonium salt? Is it okay to say ice cold temp, nitrous acid, phenylamine, and HCl? Or do you have to state the starting materials you need in order to make the nitrous acid? :confused:

Reply 308

Salmonidae

1

Original post by Jlane5000

Could someone please confirm for me the conditions and reactants for making a diazonium salt? Is it okay to say ice cold temp, nitrous acid, phenylamine, and HCl? Or do you have to state the starting materials you need in order to make the nitrous acid? :confused:

I don't think you have to write the starting materials, as the actual reactant is the nitrous acid - it is just made in situ.

Reply 309

abzy1234

14

Original post by hannah1994

What is the half equation for ammonium ions to nitrate (III) ions

I have wrote down:

NH4+ + O2 ---> NO2- + 4H+ +2e-

But surely as the nitrogen is going from an oxidation state of -3 to an oxidation state of +3, it needs to lose 6 electrons, not 2?

Main thing is that the charges are balanced on both sides. In some cases, the change in oxidation state doesn't correspond to the amount of electrons lost/gained. It's pretty obvious in the exam how much electrons are needed, so I wouldn't worry :smile:

Original post by stealth_writer

are delta s system and delta s surr always different signs?

Also what happens to a buffer soln when small extra amounts of OH is added.

No they can both be the same sign. If entropy increases during the reaction, Ssystem will be positive and if the reaction is endothermic, Ssur will be positive.

The H+ ions react with it, pushing the position of equilibrium to the left hand side, producing more HA ions, whereby more H+ ions are produced. This prevents an increase in PH :smile:

Original post by Jlane5000

Could someone please confirm for me the conditions and reactants for making a diazonium salt? Is it okay to say ice cold temp, nitrous acid, phenylamine, and HCl? Or do you have to state the starting materials you need in order to make the nitrous acid? :confused:

That's all you need. Phenylamine, sodium nitrate and HCL, cold temperatures :smile:

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Reply 310

twizzzzzzter

2

Can anyone explain June 2012 1) d) i) ?

Reply 311

tsr1

Why can't arenes be hydrogenated?

Reply 312

gstarwill

Original post by twizzzzzzter

Can anyone explain June 2012 1) d) i) ?

You have been given the ratio of nitrogen to oxygen (4:1)

so if you use this to get 5 moles multiplied by 24 dm3 = 120dm3.

1mole oxygen/120dm3 = 8.3x10-3

to get concentration of N2 you multiply [O2] by 4 to get 0.0333...

Reply 313

Tikara

Original post by tsr1

Why can't arenes be hydrogenated?

they can can't they?

H2 + finely divided Ni
300degrees and 30 atm

arene to cycloalkane

Reply 314

billybobs

5

Original post by gstarwill

because arenes have a delocalised system of electrons making them very stable, if they are hydrogenated (by addition), they lose stability. Arenes undergo substitution to retain stability

Correct me if I'm wrong but I think thats right

They can be hydrogenated, but just require a lot of energy to do so (300 degrees C) and platinum /finely divided nickel catalyst as mentioned above.

Reply 315

Tikara

Original post by gstarwill

You have been given the ratio of nitrogen to oxygen (4:1)

so if you use this to get 5 moles multiplied by 24 dm3 = 120dm3.

1mole oxygen/120dm3 = 8.3x10-3

to get concentration of N2 you multiply [O2] by 4 to get 0.0333...

ahhh thanks very much! I was also getting stuck on this :P

Reply 316

tsr1

Original post by billybobs

They can be hydrogenated, but just require a lot of energy to do so (300 degrees C) and platinum /finely divided nickel catalyst as mentioned above.

Original post by Tikara

they can can't they?

H2 + finely divided Ni
300degrees and 30 atm

arene to cycloalkane

Thanks

Reply 317

RainieXD

Original post by strawberri

Answer to Question 5d! :smile:

You know that, because it is in equilibrium:

Ka = [H⁺][HPO₄^2-]/[H₂PO₄^-]

So....

Ka/[H+] = [HPO₄^2-]/[H₂PO₄^-]

In a solution with pH 7:
pH = -log[H⁺]
[H+] = 10^-pH = 10^-7

It tells you what Ka is so you can write this equation:

[HPO₄^2-]/[H₂PO₄^-] = 6.2 x 10^-8 / 1 x 10^-7
= 0.62

This is how many times the concentration of [H₂PO₄^-] goes into the concentration of [HPO₄^2-] .

The question tells us that the concentration of (Na₂)[HPO₄^2-] is 0.1moldm^-3.

0.1/0.62 = [H₂PO₄^-] = 0.161
Seeing as this is in a solution of 1.0dm³, we can take this as the number of moles of NaH₂PO₄^-.

number of moles = mass / Mr
mass = number of moles x Mr = 0.161 x 120 = 19.32 = 19.3g

Hope this helped! :smile:

Also for 5bi...in a weak acid (indicated by the equilibrium), we assume that [H+] = [A-] because we assume that the amount of H₂PO₄^- has dissociated equally into H+ and HPO₄^2-, due to the 1:1 molar ratio between the products.

Ohhhh! Thank you so muchhh !!!!
Though I'm quite unsure about the weak acid... how would you know if it's a weak acid or not?? :confused:

Reply 318

twizzzzzzter

2

Original post by gstarwill

You have been given the ratio of nitrogen to oxygen (4:1)

so if you use this to get 5 moles multiplied by 24 dm3 = 120dm3.

1mole oxygen/120dm3 = 8.3x10-3

to get concentration of N2 you multiply [O2] by 4 to get 0.0333...