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Reply 2520
why can't enthalpy change be calculate sometimes? eg in this question:

The enthalpy change of formation, ΔHf , of glucose, C6H12O6, cannot be determined directly. The equation for this enthalpy change is shown below.
6C(s) + 6H2(g) + 3O2(g) C6H12O6(s)
Suggest why the enthalpy change of formation of C6H12O6 cannot be determined
directly
Which page is the 2013 paper?

Dw. got it :biggrin:
(edited 10 years ago)
Original post by SAS18
why can't enthalpy change be calculate sometimes? eg in this question:

The enthalpy change of formation, ΔHf , of glucose, C6H12O6, cannot be determined directly. The equation for this enthalpy change is shown below.
6C(s) + 6H2(g) + 3O2(g) C6H12O6(s)
Suggest why the enthalpy change of formation of C6H12O6 cannot be determined
directly

You can say that the activation energy is too high.
Reply 2523
Original post by purplemind
You can say that the activation energy is too high.


how would you know that though or is it the same answer for all those type of questions?
Reply 2524
What are the condition for harbour process?

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Original post by SAS18
why can't enthalpy change be calculate sometimes? eg in this question:

The enthalpy change of formation, ΔHf , of glucose, C6H12O6, cannot be determined directly. The equation for this enthalpy change is shown below.
6C(s) + 6H2(g) + 3O2(g) C6H12O6(s)
Suggest why the enthalpy change of formation of C6H12O6 cannot be determined
directly


The reaction was too slow
Reply 2526
If you were asked to draw something like 2-methylpentan-1-oll(just made that up) but it just said structure would you be expected to draw the bonds in the methyl and oh group?
Reply 2527
Original post by SAS18
why can't enthalpy change be calculate sometimes? eg in this question:

The enthalpy change of formation, ΔHf , of glucose, C6H12O6, cannot be determined directly. The equation for this enthalpy change is shown below.
6C(s) + 6H2(g) + 3O2(g) C6H12O6(s)
Suggest why the enthalpy change of formation of C6H12O6 cannot be determined
directly

You can say the reactants, react to form different products.
If we get a question on how temperature affects rate of reaction do we put more successive collisions or more frequent collisions per second?


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Original post by SAS18
how would you know that though or is it the same answer for all those type of questions?

It is just one answer that the mark schemes give to this type of question. :smile:
Reply 2530
Original post by Jimmy20002012
If we get a question on how temperature affects rate of reaction do we put more successive collisions or more frequent collisions per second?


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temperature would always be more successful collisions but pressure increase one is more chances or probability of successful collisions :smile:
Reply 2531
Original post by purplemind
It is just one answer that the mark schemes give to this type of question. :smile:


Thanks :smile:
Reply 2532
Original post by wndms
What are the condition for harbour process?

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450 degrees and 200 atmospheres, an iron catalyst is also required
Reply 2533
What are the conditions for hydrolysis?

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Reply 2534
Original post by Mikoto
450 degrees and 200 atmospheres, an iron catalyst is also required


please don't tell me we need the Haber process!:eek:
Reply 2535
Original post by wndms
What are the conditions for hydrolysis?

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bit late but here you go

EDIT: ignore the amine
Original post by SAS18
please don't tell me we need the Haber process!:eek:


I just discovered it in the depths of my notes and thought the exact same thing D:


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Do we need to know haber process?


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Reply 2538
Original post by Jimmy20002012
If we get a question on how temperature affects rate of reaction do we put more successive collisions or more frequent collisions per second?


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For temperature it's more successful collisions, for pressure its more collisions per second, for a catalyst it's more successful collisions.
Maybe I'm just being stupid but I don't understand the answer to 4 e) ii) on the may 2012 paper. Could anyone explain it to me please?