Edexcel A2 Chemistry 6ch04/05 JUNE 2015

Original post by MeeraP07

Hey guys...
For Kc, if you have eqm moles of the reactants and products you divide them by the volume and then put those values in the Kc expression, right?
But if you have eqm concentrations, do you still need to divide by the volume?

Always divide by V (volume), even if the units for concentration cancel out because you're calculating Kc, where the c is concentration.

Reply 1042

carboxylic_acid_preparation.gif8.6KB

Original post by Nautic4l

No H2O is sometimes not used when it's a 'solven't but I don't get when it acts as a solvent and when as a reactant.. ANYONE?

Also does anyone PLEASE have diagrams(labelled) of distillation and reflux apparatus?

If your reactants and products have an "aq" state symbol (apart from water, obv) then you exclude water from the Kc calculation. If water is being formed, say for instance in esterification (assuming you're using dry reagents), then you include water in the Kc calculation. Hope that's helped!

Reflux: http://what-when-how.com/wp-content/uploads/2011/06/tmp1703_thumb1.png

and Distilling: http://upload.wikimedia.org/wikipedia/commons/thumb/7/70/Fractional_distillation_lab_apparatus.svg/2000px-Fractional_distillation_lab_apparatus.svg.png

Reply 1043

pilebunker

8

Original post by Nautic4l

No H2O is sometimes not used when it's a 'solven't but I don't get when it acts as a solvent and when as a reactant.. ANYONE?

Also does anyone PLEASE have diagrams(labelled) of distillation and reflux apparatus?

I'd draw an electric heater rather than a bunsen burner for each of them, don't think it matters though. Also, determining when H2O is the solvent, I'm not exactly sure :s-smilie:

1.png45.0KB

Reply 1044

Original post by Wahid1

Jan 2015 q24aiii anyone?

Posted from TSR Mobile

moles of NaOH- 1x45/1000=0.045
moles of HCl- 5x1/1000=0.005
moles of ethanoic acid at eqm- 0.045-0.005=0.04

change in moles- 0.120-0.04=0.08
eqm moles of alcohol- 0.220-0.08=0.14
eqm moles of ester- 0+0.08=0.08
eqm moles of water- 0.278+0.08=0.358

plug these values in Kc expression to get 5.11

Reply 1045

Original post by MoHoosen9167

Always divide by V (volume), even if the units for concentration cancel out because you're calculating Kc, where the c is concentration.

so just to clarify... whether its moles or concentration just always divide the values by volume

Best of luck all!

Original post by MeeraP07

so just to clarify... whether its moles or concentration just always divide the values by volume

Sorry, I misread the question! No, if you've got concentration then you don't need to divide because you're already inputting concentrations into the equation for Kc. So sorry!

Reply 1048

Wahid1

ImageUploadedByStudent Room1433928782.683383.jpg

Thanks in advance

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Reply 1049

Original post by MoHoosen9167

Sorry, I misread the question! No, if you've got concentration then you don't need to divide because you're already inputting concentrations into the equation for Kc. So sorry!

No problem! thanks so much

Reply 1050

i posted it above

Original post by Wahid1

Thanks in advance

Posted from TSR Mobile

Reply 1051

frozo123

12

Original post by MeeraP07

No problem! thanks so much

Wait do we have to divide by the volume or not?

Reply 1052

ImageUploadedByStudent Room1433929320.047521.jpg

For anyone who needs Log laws! :smile:

hope it helps!

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Reply 1053

Original post by frozo123

Wait do we have to divide by the volume or not?

only if youve got the eqm moles you divide by vol. if you have the eqm conc you dont need to

Reply 1054

veniceswan

2

Can somebody please explain this question
4.70=4.80+log(CH3COO-)
-0.1=log(CH3COO-)
e^-0.1=(CH3COO-)
0.9048={CH3COO-}
then I found the moles and mass, but the mark scheme says {CH3COO-}=0.794
Please can somebody correct me

Reply 1055

frozo123

12

Original post by MeeraP07

only if youve got the eqm moles you divide by vol. if you have the eqm conc you dont need to

yeah ofc, but was wondering because surely you'll get the same answer doing it with moles but probably lose marks for not showing you know what's going on?

Reply 1056

ImageUploadedByStudent Room1433929648.981287.jpg

Here are some calculations! :smile:

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Reply 1057

pilebunker

8

Original post by veniceswan

Can somebody please explain this question
4.70=4.80+log(CH3COO-)
-0.1=log(CH3COO-)
e^-0.1=(CH3COO-)
0.9048={CH3COO-}
then I found the moles and mass, but the mark scheme says {CH3COO-}=0.794
Please can somebody correct me

It should be 10^-0.1, not e^-0.1, since you're using logs

Reply 1058

Original post by veniceswan

Can somebody please explain this question
4.70=4.80+log(CH3COO-)
-0.1=log(CH3COO-)
e^-0.1=(CH3COO-)
0.9048={CH3COO-}
then I found the moles and mass, but the mark scheme says {CH3COO-}=0.794
Please can somebody correct me

ImageUploadedByStudent Room1433929830.491966.jpg

Posted from TSR Mobile

Reply 1059