Standard enthalpy change

$\Delta U=q+w$

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Angel Gomez 1K
Posts: 36
Joined: Fri Sep 29, 2017 7:04 am

Standard enthalpy change

Can someone explain why the deltaH was multiplied by 2 mols?
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Michele 2C
Posts: 23
Joined: Fri Sep 29, 2017 7:06 am

Re: Standard enthalpy change

The reaction you need to calculate ∆H for has 4 moles NO, but in the reaction given, the stoichiometric coefficient of NO is 2, so in order to calculate the ∆H that is asked of you, you need to multiply the q (which is for the stoichiometric coefficients in the given reaction) by 2.

Gwyneth Huynh 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:05 am

Re: Standard enthalpy change

So when you find q and divide that by 4 moles, that's the delta H per one mole of NO. You multiply that by 2 moles because in the overall reaction, 2 moles of NO is used.

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