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Topic: Enthalpy of fuels  (Read 16590 times)

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Offline UG

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Re: Enthalpy of fuels
« Reply #15 on: November 07, 2009, 03:02:02 AM »
So you multiply 0.9375 by 1045^2 = 11140J or 11.14 Kj/mol?????
What? Where did all these numbers come from?
The answer is in joules. Kilojoules only if you convert the mass into kilograms.

Offline hoppingma-d

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Re: Enthalpy of fuels
« Reply #16 on: November 07, 2009, 03:20:26 AM »
So you multiply 0.9375 by 1045^2 = 11140J or 11.14 Kj/mol?????
What? Where did all these numbers come from?
The answer is in joules. Kilojoules only if you convert the mass into kilograms.

why 32/3 I thought it was m/MM which is 3/32 which was what it was then you've swapped it (sorry that's where the crazy number came from....)

Offline UG

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Re: Enthalpy of fuels
« Reply #17 on: November 07, 2009, 03:25:25 AM »
Yes, the amount of substance (n) = m/M so you have 3 grams of methanol, the n = 3/32 = 0.09375 moles. So 0.09375 moles was consumed. The energy released was 1045 Joules. The enthalpy of combustion is the enthalpy change per mole of substance, so in order to find this, multiply 1045 by 32/3 or 1/0.09375. So you get enthalpy of combustion as -11140 J mol-1 or -11.14 kJ mol-1

Offline hoppingma-d

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Re: Enthalpy of fuels
« Reply #18 on: November 07, 2009, 04:00:11 AM »
Yes, the amount of substance (n) = m/M so you have 3 grams of methanol, the n = 3/32 = 0.09375 moles. So 0.09375 moles was consumed. The energy released was 1045 Joules. The enthalpy of combustion is the enthalpy change per mole of substance, so in order to find this, multiply 1045 by 32/3 or 1/0.09375. So you get enthalpy of combustion as -11140 J mol-1 or -11.14 kJ mol-1

1045 : 0.09375
x : 1

got it...

Now is this seriously inaccurate considering:
http://en.wikipedia.org/wiki/Heat_of_combustion#Heat_of_combustion_tables

says it's supposed to be -726kj/mol

Offline UG

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Re: Enthalpy of fuels
« Reply #19 on: November 07, 2009, 04:02:21 AM »
Yes, that's because it was an example and I made all of them numbers up  ;D
most of them anyway...

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