Calculating Enthalpy  [ENDORSED]

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Priscilla_Covarrubias_HL
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Calculating Enthalpy

Postby Priscilla_Covarrubias_HL » Tue Jan 17, 2017 9:29 pm

For the example problem in class dealing with Standard Enthalpy of formation... why is it that you divide by 2? Is it because there 2 moles of the product?

ssoroush 2B
Posts: 14
Joined: Wed Sep 21, 2016 2:57 pm

Re: Calculating Enthalpy

Postby ssoroush 2B » Tue Jan 17, 2017 9:45 pm

Yes, you have to divide by the number of moles to get the standard enthalpy of formation per mole.

maria_tamondong_1L
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Joined: Wed Sep 21, 2016 2:55 pm

Re: Calculating Enthalpy

Postby maria_tamondong_1L » Wed Jan 18, 2017 12:09 am

For the same question. How did you get -555.38kJ? Was that given?

Jocelyn Sandoval 3B
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Joined: Wed Sep 21, 2016 2:59 pm

Re: Calculating Enthalpy

Postby Jocelyn Sandoval 3B » Wed Jan 18, 2017 1:13 pm

Yes, for that sample problem, the reaction enthalpy is given as -555.38 kJ, but in order to get the standard enthalpy of formation (the same as trying to get the enthalpy of one mole)of graphite, you divide by 2 as stated above.

Henry_Shin_3B
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Joined: Wed Sep 21, 2016 2:55 pm

Re: Calculating Enthalpy  [ENDORSED]

Postby Henry_Shin_3B » Wed Jan 18, 2017 4:52 pm

Essentially the standard enthalpy of formation is how much energy it takes for a bunch of atoms/molecules to form 1 mol of a substance made up of those atoms (w/ the assumption that everything is in standard state)

555.38 kJ is given in the problem, and you can see that there are 2 moles of the product.
If it takes 555.38 kJ to form 2 moles of the product, then it'll take 555.38/2 kJ to form 1 mole of the product.

That's why you divide everything by 2.


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