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Topic: How do you actually work out these thermochemical equations?  (Read 4956 times)

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Munch

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How do you actually work out these thermochemical equations?
« on: January 16, 2006, 06:58:32 AM »
Hi - i'm a chemistry beginner and finding enthalpy and gas laws just so hard

Can someone please help with write the following thermochemical equations that relate to the following enthalpy changes

CH3OH(l) = -715kJmol-1 - which i think is C(s) + 2H2(g) + 1/2O2

CH3COCH3(l) = -216kJmol-1 - which i think is 3C + 3H2 + 1/2O2

but the problem one is

(NH4)3*PO4(s) = -1681kJmol-1

How do you actually work out these thermochemical equations. Do you have to break them down into their componants?

Thanks so much for any help
« Last Edit: January 16, 2006, 03:28:29 PM by Mitch »

Offline plu

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Re:Thermochemical equations
« Reply #1 on: January 16, 2006, 07:23:31 AM »
I believe you are talking about enthalpies of formation.  In that case, your first two equations are correct.  For the third, simply break the compound down into its basic elements like you did for the first two and add them

Munch

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Re:Thermochemical equations
« Reply #2 on: January 16, 2006, 07:33:41 AM »
Add what though?  thats the problem i dont understand.

(NH4)3*PO4(s)

2N2 + 3H2 + P + 2O2

When i look at the table for the standard entahlpys of formation they are all 0!!


Offline plu

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Re:Thermochemical equations
« Reply #3 on: January 16, 2006, 01:08:34 PM »
You just wrote the equation!  I do not see what the problem is...

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