Edexcel - Chemistry Unit 2 - 4 June 2013

OP

Is this correct....

2HI + 2H2SO4 ------> H2S + I2 + S + 4H+

??? Thanks in advance :smile:

Reply 102

Gnome :)

2

Original post by posthumus

Is this correct....

2HI + 2H2SO4 ------> H2S + I2 + S + 4H+

??? Thanks in advance :smile:

Nope- you made the oxygen disappear :tongue:

8HI + H2SO4 ----> H2S + 4I2 + 4H2O ?

Reply 103

OP

Original post by Gnome :)

Nope- you made the oxygen disappear :tongue:

8HI + H2SO4 ----> H2S + 4I2 + 4H2O ?

Woop

Thanks for pointing that out !

It stats: "Hydrogen iodide reduces concentrated sulfuric acid to sulfur and hydrogen sulfide"... so just the sulfur is missing :frown:

2HI + 2H2SO4 ----> H2S + I2 + S + 8H2O

So could this be right ? :smile:

Okay just realized that is no way right ! :tongue:

too many Hydrogen on the right :tongue:

Reply 104

Gnome :)

2

Original post by posthumus

Woop

Thanks for pointing that out !

It stats: "Hydrogen iodide reduces concentrated sulfuric acid to sulfur and hydrogen sulfide"... so just the sulfur is missing :frown:

2HI + 2H2SO4 ----> H2S + I2 + S + 8H2O

So could this be right ? :smile:

Okay just realized that is no way right ! :tongue:

too many Hydrogen on the right :tongue:

Just had a quick look at the textbook and chemguide (gonna marry whoever made that, seriously!)

Iodide ions are stronger reducing agents than bromide ions are. They are
oxidised to iodine by the concentrated sulphuric acid.

The reduction of the sulphuric acid is more complicated than before. The
iodide ions are powerful enough reducing agents to reduce it

•

first to sulphur dioxide (sulphur oxidation state = +4)

•

then to sulphur itself (oxidation state = 0)

•

and all the way to hydrogen sulphide (sulphur oxidation state = -2).

The most important of this mixture of reduction products is probably the
hydrogen sulphide.

The reaction isn't mentioned in the textbook (the Hodder one), so I think we just need to know that I is a stronger reducing agent than Br and Cl and so results in the formation of S/H2S :smile:

Reply 105

OP

Original post by Gnome :)

Just had a quick look at the textbook and chemguide (gonna marry whoever made that, seriously!)

The reaction isn't mentioned in the textbook (the Hodder one), so I think we just need to know that I is a stronger reducing agent than Br and Cl and so results in the formation of S/H2S :smile:

Ah okay thanks! :biggrin:

I guess I should just stick to what there is :tongue:

it's not in the George Facer book either :smile:

Reply 106

phaseshift

12

Original post by posthumus

Ah okay thanks! :biggrin:

I guess I should just stick to what there is :tongue:

it's not in the George Facer book either :smile:

You would need to know the reaction specifically if you were on AQA (if I remember correctly) but it isn't in our spec. :smile:

Reply 107

Katy1704

Hello

has anyone made any revision materials on the different reactions we have to remember (Edexcel unit 2) ? e.g. reactions of halides , reactions of group 2 metals & things like what colour solutions the halogens make in different solvents. Its just I want them altogether so they are easier to memorise. If not , no worries !

Reply 108

Zoeyyy

Guys, if anyone's got something to share like the possibility of a certain question showing up in the paper or something likely to be asked in this or that way bla bla , do share thanks :biggrin:

Reply 109

Hey, I have just come across this question and wonder if any of you could shed some light, it's in Unit 2. 'Why does platinum speed up the reaction between ammonia and oxygen'

Thanks in advance :biggrin:

Reply 110

OP

Original post by Cattey watty

Hey, I have just come across this question and wonder if any of you could shed some light, it's in Unit 2. 'Why does platinum speed up the reaction between ammonia and oxygen'

Thanks in advance :biggrin:

Because it finds an alternative/ quicker route which lowers the activation energy required by the molecules :smile:

I think that would be fine (for 2 marks? I assume), but in further detail;

It has active sites on its surface that become saturated by reactants which then slowly convert it into products. This means that the rate of reaction is not altered by an increase of gaseous reactants.

Reply 111

Original post by posthumus

Because it finds an alternative/ quicker route which lowers the activation energy required by the molecules :smile:

I think that would be fine (for 2 marks? I assume), but in further detail;

It has active sites on its surface that become saturated by reactants which then slowly convert it into products. This means that the rate of reaction is not altered by an increase of gaseous reactants.

That's fab! Thanks very much! And yeah it's a 2 mark question so that should really be fine :biggrin:

Reply 112

Katy1704

Hello

for edexcel do we need to know how to distinguish between primary, secondary and tertiary halogenoalkanes by experiment? Eg mix halogenoalkane with water to make an alcohol. Add silver nitrate solution, which gives a silver halide precipitate. Set up 3 test tubes each containing a different halogenoalkane, ethanol (as solvent) and silver nitrate solution in a water bath. Tertiary - precipitate forms instantly. Secondary - precipitate takes several seconds to form. Primary - precipitate takes several minutes to form.

Just don't want to memorise this if I won't need it - thank you for any help you can provide.

Reply 113

Original post by Katy1704

Hello

for edexcel do we need to know how to distinguish between primary, secondary and tertiary halogenoalkanes by experiment? Eg mix halogenoalkane with water to make an alcohol. Add silver nitrate solution, which gives a silver halide precipitate. Set up 3 test tubes each containing a different halogenoalkane, ethanol (as solvent) and silver nitrate solution in a water bath. Tertiary - precipitate forms instantly. Secondary - precipitate takes several seconds to form. Primary - precipitate takes several minutes to form.

Just don't want to memorise this if I won't need it - thank you for any help you can provide.

Hi!

Yeah, we need to know what they are and how to distinguish between them, my teachers have been going on about them for ages. I'm not sure if we need to know how to by experiment. I don't remember doing much on it with experiments. But our spec does say to be able to distinguish the difference of tertiary, primary and secondary by observations of a reaction 'Interpret give data and observations comparing the reactions and reactivity of primary, secondary and tertiary compounds' . Look at the spec, it's under organic chemistry in unit 2 section.

http://www.edexcel.com/migrationdocuments/GCE%20New%20GCE/Content1245-GCE-Chemistry.doc

Hope this helps and good luck! :biggrin:

(edited 10 years ago)

Reply 114

Katy1704

That helps a lot thank you very much :smile:

good luck in the exam!

Reply 115

Original post by geor

How are you guys revising the green stuff? I hate it!

I have made some mind maps, I know this wont work for everyone but it has helped me. For example, i have made a mind map of how you you get an Aldehyde from a ketone etc. Mind maps really helped me with this, also Flash cards :biggrin: