Is this correct Hess' law, thermodynamics

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SUMMARY

The forum discussion centers on calculating the standard enthalpy change (∆rH°) for the combustion reaction of ethanol (C2H5OH) using the provided internal energy change (∆rU° = -1373 kJ/mol) at 298K. Participants debated the correct approach to derive ∆H without using heats of formation, ultimately establishing the relationship ∆H = ∆U + ∆(PV) and calculating the change in the number of moles of gases (Δn). The final calculations yielded a discrepancy in values, indicating potential errors in the provided ΔU or the enthalpy of formation for ethanol, which was noted as -277.6 kJ/mol according to a referenced textbook.

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Diamond101
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Homework Statement


Calculate ∆rH° for the reaction
c2h50h+3o2-->2co2+3h20
Given that ∆rU° = - 1373 kJ mol-1 at 298K.

Homework Equations


c2h50h+3o2-->2co2+3h20

The Attempt at a Solution


delta H f CO2(g) = -393.5 kJ/mole
delta H f H2O(l) = -241.8 kJ
delta H f O2(g) = 0 these values are from my textbook delta H reaction = ((2 moles CO2)(-393.5 kJ/mole) + (3 moles H2O)(-241.8kJ/mole)) - ((1 mole C2H5OH)(delta H f C2H5OH(l))

-1373 kJ = (-787.0 kJ - 725.4 kJ) - (delta H f C2H5OH(l)
-139.4 kJ = delta H f C2H5OH(l)
 
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You are supposed to do this problem without using heats of formation. Any thoughts on how?
 
Chestermiller said:
You are supposed to do this problem without using heats of formation. Any thoughts on how?
no clue.
 
What is the definition of ΔH in terms of ΔU?
 
Chestermiller said:
What is the definition of ΔH in terms of ΔU?
i do know DH=DU+PDV
 
Diamond101 said:
i do know DH=DU+PDV
ΔH=ΔU+Δ(PV)
At constant temperature, for ideal gases, Δ(PV)=RTΔn.

What is the change in the number of moles of gases Δn between reactants and products for you chemical reaction? If you neglect the volume of the liquids relative to the gases, what is the change in the PV in going from your reactants to going to your products?
 
Chestermiller said:
What is the definition of ΔH in terms of ΔU?
WAIT ITS A COMBUSTION REACTION
 
Chestermiller said:
ΔH=ΔU+Δ(PV)
At constant temperature, for ideal gases, Δ(PV)=RTΔn.

What is the change in the number of moles of gases Δn between reactants and products for you chemical reaction? If you neglect the volume of the liquids relative to the gases, what is the change in the PV in going from your reactants to going to your products?
Well 1 mole of c2h5OH and 3 molews of 02 , 2 Moles of CO2 and 3moles of h20 so the change is 1 mole 5- 4 moles
 
Diamond101 said:
Well 1 mole of c2h5OH and 3 molews of 02 , 2 Moles of CO2 and 3moles of h20 so the change is 1 mole 5- 4 moles
I asked for only the change in the number of moles of gases. The changes in PV resulting from the changes in the liquids is negligible.
 
  • #10
Chestermiller said:
I asked for only the change in the number of moles of gases. The changes in PV resulting from the changes in the liquids is negligible.
Change will be 2. so the relationship between enthalpy and internal energy is DH=DU=DNgasRT
 
  • #11
Diamond101 said:
Change will be 2. so the relationship between enthalpy and internal energy is DH=DU=DNgasRT
What if I told you that I get a change in the number of moles of gas as -1, not 2?
 
  • #12
Chestermiller said:
What if I told you that I get a change in the number of moles of gas as -1, not 2?
how? you didn't cater for h2o as a gas?
 
  • #13
Chestermiller said:
What if I told you that I get a change in the number of moles of gas as -1, not 2?
because water isn't gas in its standard state ?
 
  • #14
Diamond101 said:
because water isn't gas in its standard state ?
I see (l)'s next to your H2O's for the reaction. Apparently, they are saying that the product water is in the liquid state.
 
  • #15
Chestermiller said:
I see (l)'s next to your H2O's for the reaction. Apparently, they are saying that the product water is in the liquid state.
my mistake water is supposed to be a gas.
so dh= -1373* 2 mol*8.314*298 k
 
  • #16
Chestermiller said:
I see (l)'s next to your H2O's for the reaction. Apparently, they are saying that the product water is in the liquid state.
i got -3582.14 KJ thank you for your assistance. (-1373+ (2 MOLES * 8.314* 298))
 
  • #17
Diamond101 said:
i got -3582.14 KJ thank you for your assistance. (-1373+ (2 MOLES * 8.314* 298))
Not correct. Watch out for the units on that PV term.
 
  • #18
Chestermiller said:
Not correct. Watch out for the units on that PV term.
OH IT WILL BE JOULES EVERYTHING ELSE CANCELS ?
 
  • #19
Diamond101 said:
OH IT WILL BE JOULES EVERYTHING ELSE CANCELS ?
But the Delta U is in kJ. The units of the two terms have to be consistent.

Also, as a check, how does your answer compare with what you get using heats of formation? (I know you must have thought of doing this check)
 
  • #20
Chestermiller said:
But the Delta U is in kJ. The units of the two terms have to be consistent.

Also, as a check, how does your answer compare with what you get using heats of formation? (I know you must have thought of doing this check)
I did compare the enthalpy of formation against the enthalpy of reaction . so the unit is Joule per kelvin?
 
  • #21
The units of heats of formation and heats or reaction are J/mol or kJ/mol
 
  • #22
ΔH=2(-241.8)+3(-393.5)-(-277.7)=-1386.4 kJ/mol (Heat of formation calculation)

ΔH=-1373+2(0.008314)(298)=-1368.5 kJ/mol (From ΔU provided)

They don't quite match. Something is not quite right. Maybe the ΔU they provided was wrong.
 
  • #23
Chestermiller said:
ΔH=2(-241.8)+3(-393.5)-(-277.7)=-1386.4 kJ/mol (Heat of formation calculation)

ΔH=-1373+2(0.008314)(298)=-1368.5 kJ/mol (From ΔU provided)

They don't quite match. Something is not quite right. Maybe the ΔU they provided was wrong.
hey i actually adds up the formation value for ethanol is -277.6 according to my chem3 textbook by burrows
 

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