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Yes, provided what you call the standard state refers its elements at the standard state of 298 and 1 atm.Red_CCF said:With regards to ho(T), per the first quote, isn't that the same as the enthalpy defined relative to standard state?
I was unable to find the equation I was referring to in the quote. But, in any event, your statement is incorrect. ho(T) is defined at temperature T relative to the the stoichiometric proportions of its elements at 298 and 1 atm (the standard reference state). On the other hand, hf° is defined at temperature T relative to the stoichiometric proportions of its elements at its actual T.For the second quote, my interpretation is that ho(T) is defined relative to the the stoichiometric proportions of its elements at its actual T (so I need to subtract the sensible heat of the elements from 298K to T of the compound) as opposed to just at 298K (per standard state definition), is this correct?
No. This is the definition of hf°(T).This implies h°(T)=H°(T)-∑niH°i(T) which for elements is 0 for all T which didn't make much sense to me.
This is incorrect.Is h°(T)=H°(T)-∑niH°i(T)+∑nih°i(T) (based on the equality in hf°(T))?
This is not done correctly. And, if done correctly, the sensible heats do not cancel.For H°(T)-∑n_iH_i°(T)=h°(T)-∑n_ih_i°(T), per definition of H°(T) and h°(T), the sensible components cancel as
H°(T)-h°(T)=∑n_iH_i°(T)-∑n_ih_i°(T)
and the equation reduces to H°(298)-∑n_iH_i°(298)=h°(298)-∑n_ih_i°(298) = h°(298). What I'm left with is that h_f°(T) is constant for all T?
Let's start with the equation I gave for hf°(T):
hf°(T)=H°(T)-∑niH°i(T)
Now, [itex]H°(T)=H°(298)+\int_{298}^T{C_pdT}[/itex]
and, for its elements, [itex]H_i°(T)=H_i°(298)+\int_{298}^T{C_{pi}dT}[/itex]
If we substitute this into the equation for hf°(T), we obtain:
[tex]h_f°(T)=(H°(298)-∑n_iH_i°(298))+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}[/tex]
But, [itex](H°(298)-∑n_iH_i°(298))=h°(298)[/itex]
So,
[tex]h_f°(T)=h°(298)+\int_{298}^T{C_pdT}-∑n_i\int_{298}^T{C_{pi}dT}[/tex]
But, [itex]h°(298)+\int_{298}^T{C_pdT}=h°(T)[/itex]
and [itex]0+\int_{298}^T{C_{pi}dT}=h_i°(T)[/itex] for the elements.
So,
[tex]h_f°(T)=h°(T)-∑n_ih_i°(T)[/tex]
Chet