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ENTHALPY HELP! I am confused about energy cycles and Q=mcdeltaT

Hello everyone please help me!

I have been asked to calculate enthalpy change of formation for N2O using data given.

table says:
C + N2O -> CO +N2 -193 KJ mol-1 (not specified whether C or F)
C +1/2O2 -> CO -111 KJ mol-1 (again not specified)


how to do I know whether to draw a Hess' cycle where the arrows point upwards from the elements (formation) or whether to draw one where they point down to combustion products (combustion) IF DONT KNOW WHETHER THE TABLE VALUES ARE FOR COMBUSTION OR FORMATION ENTHALPY CHANGE??? PLEASE HELP ME EVEN MY TACHER COULD NOT EXPLAIN THIS TO ME!!!!!!!!!!! a clear description would help :smile: Can u tell me how to figure out if its a combustion cycle AND ALSO how to figure out if its a formation cycle! thanks!

and secondly,

in Q=MCdeltaT you determine the sign (-/+) by yourself by looking at whether the temperature has increased or decreased. I did a question where two solutions were mixed (NaOH and H2SO4) the temperature went from 20 degrees and rose to 29 degrees. The answer in the book said use a + sign as it's endothermic? can someone explain this?

another question mixed solid iron powder to copper sulphate in a polystyrene cup. The sign here needed was - when the temperature went from 21 to 25 degrees? here the temperature s increased but the sign is - , why is it EXOTHERMIC NOW??? I am so confused please explain clearly thanks.
Reply 1
The cycle will look like this (fingers crossed I can get the formatting anywhere near right)...


C + N2O -> CO + N2
....arrow.........arrow
...+1/2O2......+1/2O2
.....to............to
......CO + N2O

i.e. the two arrows pointing down represent:
C + 1/2O2 + N2O -> CO + N2O
and
CO + N2 +1/2O2 -> CO + N2O

You have two of the values, work out the third.

I teach delta T = final temp. - initial temp., i.e. temp. rises -> +ve delta T and temp. drops -> -ve delta T
q = m c delta T, i.e. temp rises -> +ve q and temp. drops -> -ve q (now I admit that you can't have a -ve q, but OCR (and I'd bet other exam boards) will overlook this).
Then delta H = - q / n, i.e. temp. rises -> -ve delta H and temp. drops -> +ve delta H (the minus sign for exothermic reactions no longer appears from nowhere).

For your two reactions, both involve temp. rises and therefore both should have been exothermic, i.e. -ve delta H.

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