How much heat is produced in burning 1 mol of ethylene (c2h4) under standard conditions?

Question

If reactants and products are brought to 298k and h20(l) is formed? And what is the maximum amount of useful work that can be accomplished under standard conditions by this system?

I have no idea how to solve this question. I'm assuming it will probably use some enthalpies of formation, so the enthalpies of formation are:
c2h4(g) 52.30
h20(l) -285.83
c02(g) -393.5
02 0
Thanks alot for the help

The enthalpies of formation are in kJ/mol at 298 K

Answer 1

You don't give units, so I assume kJ/mol. The heat of formation of C, H, and O in their standardstates is 0 kJ/mol.

2H2 + 2C(graphite) ===> C2H4(g) 52.30kJ/mol

H2 +1/2 O2 ===> H2O -285.83kJ/mol

C(graphite) + O2 ===> CO2 -393.5kJ/mol

Also, overall:

C2H4 + 3O2 ===> 2CO2 + 2H2O

Next, arrange the equations so that they add up to the desired reaction, reversing the sign of the enthalpy for the reverse reaction. Add up all the reactions, and cancel on either side of the equation.

C2H4(g) ===> 2C(graphite) + 2H2 -52.30kJ

2O2(g) + 2C(graphite) ===> 2CO2(g) 2x(-393.5kJ/mol) = -787kJ

O2 + 2H2 ===> 2H2O 2x(-285.3) = -575.6kJ
----------------------------------------------------
Add the equations, cancel on both sides

-52.30 kJ
-787 kJ
-575.6kJ
-------------
-1415 kJ

Next, you must convert all gases from 273K to 298K, and raise water from 273K to 298K. The corrected result will be the maximum amount of useful work, except of course for the entropy.

Answer 2

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Answer 3

C2H4 + 6O2 --> 2CO2 + 4H2O

Total heat = (total enthalpy of products) - (total heat of reactants)
= (2(-393.5) + (4(-285.83)) - (52.30 + 6(0))
=-1878.02
Meaning that the system loses 1878.02J of heat, meaning that 1878.02J of heat is prodcued.

By the way you have to multiply that number by the number of moles of C2H4. You don't have to worry about that since there is one mole only, in this case.