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AQA CHEM4 and CHEM5 revision thread 2016

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Using Ka for a diprotic acid e.g. H2A --> 2H+ + A2-
Would the expression be
Ka = [H+]^2 * [A2-] / [H2A]? very confused about this
I know a similar thing is done with Kc
Original post by Thisshouldbeapun
Using Ka for a diprotic acid e.g. H2A --> 2H+ + A2-
Would the expression be
Ka = [H+]^2 * [A2-] / [H2A]? very confused about this
I know a similar thing is done with Kc


No

Kc I'd different
With kc, the formula is products over reactants to the power of the number of moles

With ka the formula is always - [H+] x [A-] / [HA]



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Help please :smile:
2bi :smile:

Thank you

I sort of understand the answer
But I would like to know why

ImageUploadedByStudent Room1464960162.278414.jpg


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Original post by High Stakes
Aye I lav organic synthesis.


Mo u r an odd one :wink:
How would you name that?

ImageUploadedByStudent Room1465043523.218942.jpg


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Original post by Glavien
How would you name that?

ImageUploadedByStudent Room1465043523.218942.jpg


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would it be N,N-ethylpropylbutylamine?
Original post by theboss1998
would it be N,N-ethylpropylbutylamine?


I got N-ethyl-N-propylbutan-1-amine but I'm not sure if it's correct.


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Original post by Glavien
I got N-ethyl-N-propylbutan-1-amine but I'm not sure if it's correct.


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That seems right! How did you do it?
Original post by theboss1998
That seems right! How did you do it?


Hi,
Since it's a tertiary amine it's missing a + charge on the N - how I know it's a tertiary amine is the 3 alkyl chains attached to the nitrogen -

So you name it according to the rules of nomenclature - butylethylpropylamine or N-butylethylpropylamine





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(edited 7 years ago)
Original post by Glavien
How would you name that?

ImageUploadedByStudent Room1465043523.218942.jpg


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Also personally I wouldn't have the N but as its not a N subsitututed amide


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does anyone know whether a catalyst effects the rate constant, also is the rate expression and the rate equation the same thing?
Original post by theboss1998
Only temperature affects the value of K. Also, yes they are.


thank you!
Reply 72
Avatar for B_9710
B_9710
18
Original post by Superbubbles
does anyone know whether a catalyst effects the rate constant, also is the rate expression and the rate equation the same thing?


k=Ae−Ea/(RT) \displaystyle k=Ae^{-E_a/(RT)} .
The activation energy would be lower in presence of a catalyst, so the rate constant would increase.
Original post by Superbubbles
thank you!


Sorry my mistake I thought that was Kc, yes catalyst does affect it as B_9710 stated above. Sorry once more.
Original post by theboss1998
Sorry my mistake I thought that was Kc, yes catalyst does affect it as B_9710 stated above. Sorry once more.


no worries!
Original post by B_9710
k=Ae−Ea/(RT) \displaystyle k=Ae^{-E_a/(RT)} .
The activation energy would be lower in presence of a catalyst, so the rate constant would increase.


thank you for clearing up the confusion! very much appreciated
does anyone know what pka is actually measuring, there doesnt seem to be a proper definition for it in the AQA text book :/
Reply 77
Avatar for B_9710
B_9710
18
Original post by Superbubbles
does anyone know what pka is actually measuring, there doesnt seem to be a proper definition for it in the AQA text book :/


Acidity. Lower the pKa the higher the acidity.
Original post by B_9710
Acidity. Lower the pKa the higher the acidity.


I know at the half- neutralisation point A- = HA so pH= pKa but why is this only for weak acids or does it also apply for strong acids??, in the indicators section of the text book it says "the colour change of most indicators takes place over a pH range of around 2 units, centred around the value of pka for the indicator" but it doesnt talk about weak acids in this context
Chemistry unit 5
January 2010
Question 3h
Is it okay yo say:
Electrolysis of water to obtain hydrogen uses fossil fuels so not carbon neutral


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