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Hess cycles

so here's my working what have i done wrong and why?
(edited 7 years ago)
Reply 1
Can you try and re upload the picture because it's too blurry to see anything.
Reply 2
Original post by B_9710
Can you try and re upload the picture because it's too blurry to see anything.


better?
Reply 3
Original post by thefatone
better?

Much better.
At the bottom of the cycle should be the gaseous atoms with the bonds all broken. So at the bottom of the cycle should be N and 3H and the arrows should be going towards the bottom of the cycle - if you get what I mean - so to work out the enthalpy of formation you should be doing sum of enthalpy of bonds broken - sum of enthalpy of bonds made.
(edited 7 years ago)
Reply 4
Original post by B_9710
Much better.
At the bottom of the cycle should be the gases atoms with the bonds all broken. So at the bottom of the cycle should be 2N and 6H and the arrows should be going towards the bottom of the cycle - if you get what I mean - so to work out the enthalpy of formation you should be doing sum of enthalpy of bonds broken - sum of enthalpy of bonds made.


why do the arrows go down? surely it's a formation enthalpy so the arrows go up from 2C and 6H not down?
my understanding if down is that, that would be an enthalpy of combustion which includes oxygen as one of the reactants in the main equation and what would be produced is water and CO2
but the equation doesn't show that so surely it's a formation enthalpy change and the arrows go up from 2C and 6H?
Reply 5
Original post by thefatone
why do the arrows go down? surely it's a formation enthalpy so the arrows go up from 2C and 6H not down?
my understanding if down is that, that would be an enthalpy of combustion which includes oxygen as one of the reactants in the main equation and what would be produced is water and CO2
but the equation doesn't show that so surely it's a formation enthalpy change and the arrows go up from 2C and 6H?


Yes but this is slightly different as we are using the mean bond enthalpies rather than enthalpies of combustion. So we break bonds of the N2 and 1.5H2 to form 2N and 3H (so the arrow go down). And then we breaking the bonds to go from NH3 to N and 1.5H2 (so the arrow goes down). Then you just reverse the arrow on the right - which is where bonds broken - bonds made comes from.
Upload a combustion cycle question if you want me to show you the slight differences. :wink:
Reply 6
Original post by B_9710
Yes but this is slightly different as we are using the mean bond enthalpies rather than enthalpies of combustion. So we break bonds of the N2 and 1.5H2 to form 2N and 3H (so the arrow go down). And then we breaking the bonds to go from NH3 to N and 1.5H2 (so the arrow goes down). Then you just reverse the arrow on the right - which is where bonds broken - bonds made comes from.
Upload a combustion cycle question if you want me to show you the slight differences. :wink:


i'll have to come back to this a bit later since i'm going out shopping for some waterproofs and stuff xD
btw thanks a lot for the help
Reply 7
Original post by B_9710
Yes but this is slightly different as we are using the mean bond enthalpies rather than enthalpies of combustion. So we break bonds of the N2 and 1.5H2 to form 2N and 3H (so the arrow go down). And then we breaking the bonds to go from NH3 to N and 1.5H2 (so the arrow goes down). Then you just reverse the arrow on the right - which is where bonds broken - bonds made comes from.
Upload a combustion cycle question if you want me to show you the slight differences. :wink:


ok what i don't get is where the oxygen comes from in the first cycle

Reply 8
Original post by thefatone
ok what i don't get is where the oxygen comes from in the first cycle



Well you are given combustion data - so you just burn each of the reactants in oxygen and you get an enthalpy change and you get given the value in the table. The oxygen though is not important because the enthalpy of formation and combustion of oxygen is 0 (well there is no value for oxygen - you can't burn oxygen in oxygen). So you combust C2H4 and H2 to CO2 and H2O and so the arrow goes from the reactants to CO2 and H2O.
Same for C2H6 - you burn it in oxygen to CO2 and H2O so the arrow goes from C2H6 to CO2 and H2O. You may want to look at this on chemguide.
Reply 9
Original post by B_9710
Well you are given combustion data - so you just burn each of the reactants in oxygen and you get an enthalpy change and you get given the value in the table. The oxygen though is not important because the enthalpy of formation and combustion of oxygen is 0 (well there is no value for oxygen - you can't burn oxygen in oxygen). So you combust C2H4 and H2 to CO2 and H2O and so the arrow goes from the reactants to CO2 and H2O.
Same for C2H6 - you burn it in oxygen to CO2 and H2O so the arrow goes from C2H6 to CO2 and H2O. You may want to look at this on chemguide.


ok understood thanks
Original post by thefatone
so here's my working what have i done wrong and why?


i had the same problem myself

for b)ii) you took it away wrong way -- this is how it is
=[(3/2 x436) + (1/2 x944)]- (388x3)
=-38KJmol-1

i'd love to show you my working but my phone's not working :redface: hope this helps

and for c just follow the working in the mark scheme carefully and you'll get it soon :smile:
Reply 11
Original post by hideNfreak
i had the same problem myself

for b)ii) you took it away wrong way -- this is how it is
=[(3/2 x436) + (1/2 x944)]- (388x3)
=-38KJmol-1

i'd love to show you my working but my phone's not working :redface: hope this helps

and for c just follow the working in the mark scheme carefully and you'll get it soon :smile:


yup thanks :biggrin:

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