7.75b reaction states

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104341943
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7.75b reaction states

Postby 104341943 » Thu Jan 23, 2014 3:32 pm

In 7.75 it gives us CH3OH, and we are to find the enthalpy. When writing the reaction equation, how do we know that the reactants are in the gas phase?

Chem_Mod
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Re: 7.75b reaction states

Postby Chem_Mod » Mon Jan 27, 2014 10:48 am

Enthalpy of formation relates to the formation of the compound from its elements in their most stable forms. Most elemental compounds are most stable as diatomic gases, such as H2 and N2.

All bond enthalpies relate to breaking the bonds of a compound in its gas phase.

Denise 3L
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Re: 7.75b reaction states

Postby Denise 3L » Sun Jan 11, 2015 9:27 pm

For this same question, the solutions manual also inserts a number to atomize C(gr) at 717kj/mol, where did they find this number, I don't see it in the 3 tables, and what does it mean?

Niharika Reddy 1D
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Re: 7.75b reaction states

Postby Niharika Reddy 1D » Mon Jan 12, 2015 12:20 am

That value is the enthalpy of sublimation of carbon, and it is found in the problem itself. Enthalpy of sublimation is the molar enthalpy change when a solid converts directly into a gas.

Denise 3L
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Re: 7.75b reaction states

Postby Denise 3L » Mon Jan 12, 2015 1:44 pm

Oh my eyes totally flew past that part, great thank you!

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Re: 7.75b reaction states

Postby Chem_Mod » Mon Jan 12, 2015 2:31 pm

In this problem you are asked to calculate standard enthalpies of formation from information given in several tables. For CH3OH (methanol) the elements that form it are carbon ("C", solid, given the enthalpy of formation in the problem), oxygen (O2, gas), and hydrogen (H2, gas). The standard states for several molecules are given in the periodic table such as nitrogen (gas), fluorine, (gas), bromine (liquid), and iodine (liquid). However, I would suggest looking at the table for bond energies to successfully complete this problem and calculate the enthalpies of the reactions (in this case formation of the compounds from its elements).


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