Is this correct Hess' law, thermodynamics

[Diamond101]

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)

 

[Chestermiller]

You are supposed to do this problem without using heats of formation. Any thoughts on how?

 

[Diamond101]

You are supposed to do this problem without using heats of formation. Any thoughts on how?

no clue.

 

[Chestermiller]

What is the definition of ΔH in terms of ΔU?

 

[Diamond101]

What is the definition of ΔH in terms of ΔU?

i do know DH=DU+PDV

 

[Chestermiller]

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?

 

[Diamond101]

What is the definition of ΔH in terms of ΔU?

WAIT ITS A COMBUSTION REACTION

 

[Diamond101]

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

 

[Chestermiller]

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.

 

[Diamond101]

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

 

[Chestermiller]

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?

 

[Diamond101]

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?

 

[Diamond101]

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 ?

 

[Chestermiller]

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.

 

[Diamond101]

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

 

[Diamond101]

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))

 

[Chestermiller]

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.

 

[Diamond101]

Not correct. Watch out for the units on that PV term.

OH IT WILL BE JOULES EVERYTHING ELSE CANCELS ?

 

[Chestermiller]

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)

 

[Diamond101]

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?

 

[Chestermiller]

The units of heats of formation and heats or reaction are J/mol or kJ/mol

 

[Chestermiller]

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.

 

[Diamond101]

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