# Homework Help: Is this correct Hess' law, thermodynamics

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1. Feb 29, 2016

### Diamond101

1. The problem statement, all variables and given/known data
Calculate ∆rH° for the reaction
c2h50h+3o2-->2co2+3h20
Given that ∆rU° = - 1373 kJ mol-1 at 298K.

2. Relevant equations
c2h50h+3o2-->2co2+3h20

3. 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)

2. Feb 29, 2016

### Staff: Mentor

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

3. Feb 29, 2016

### Diamond101

no clue.

4. Feb 29, 2016

### Staff: Mentor

What is the definition of ΔH in terms of ΔU?

5. Feb 29, 2016

### Diamond101

i do know DH=DU+PDV

6. Feb 29, 2016

### Staff: Mentor

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?

7. Feb 29, 2016

### Diamond101

WAIT ITS A COMBUSTION REACTION

8. Feb 29, 2016

### Diamond101

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

9. Feb 29, 2016

### Staff: Mentor

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. Feb 29, 2016

### Diamond101

Change will be 2. so the relationship between enthalpy and internal energy is DH=DU=DNgasRT

11. Feb 29, 2016

### Staff: Mentor

What if I told you that I get a change in the number of moles of gas as -1, not 2?

12. Feb 29, 2016

### Diamond101

how? you didn't cater for h2o as a gas?

13. Feb 29, 2016

### Diamond101

because water isnt gas in its standard state ?

14. Feb 29, 2016

### Staff: Mentor

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. Feb 29, 2016

### Diamond101

my mistake water is supposed to be a gas.
so dh= -1373* 2 mol*8.314*298 k

16. Feb 29, 2016

### Diamond101

i got -3582.14 KJ thank you for your assistance. (-1373+ (2 MOLES * 8.314* 298))

17. Feb 29, 2016

### Staff: Mentor

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

18. Feb 29, 2016

### Diamond101

OH IT WILL BE JOULES EVERYTHING ELSE CANCELS ?

19. Feb 29, 2016

### Staff: Mentor

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. Feb 29, 2016

### Diamond101

I did compare the enthalpy of formation against the enthalpy of reaction . so the unit is Joule per kelvin?

21. Feb 29, 2016

### Staff: Mentor

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

22. Feb 29, 2016

### Staff: Mentor

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. Mar 2, 2016

### Diamond101

hey i actually adds up the formation value for ethanol is -277.6 according to my chem3 text book by burrows