Question

C (graphite)+O2(g) --> CO2(g) delta H rxn = -393.5 kJ/mol
H2(g)+1/2O2(g) --> H2O(l) delta H rxn = -285.8 kJ/mol
2C2H6 (g) +7O2(g) --> 4CO2(g) +6H2O (l) delta H rxn =-3119.6 kJ/mol

calculate the enthalpy change for the reaction below:
2C (graphite) +3H2(g) --> C2H6(g)

Answer 1

This question is asking for the formation energy

First off you know the enthalpy of formation for CO2 and H2O it is given by the first two reactions.

Secondly you know the reaction enthalpy for the break down of 2C2H6 to CO2 and H2O

1. flip the reaction for C2H6 (you want formation energy) remember to change the enthalpies sign whenever you flip a reaction)

2. Notice it is in terms of 2 moles of C2H6 so simply divide all the constants in the equation by 2 as well as the enthalpy by 2

3. Since you know formation enthalpies for H2O, and CO2 you can simply apply hesses law to solve for formation of C2H6

This goes over hesses law
http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Standard_Enthalpy_Of_Formation

Answer 2

for step number 2 I am talking about the 3rd chemical reaction equation

Answer 3

also note that O2 has an enthalpy of formation equal to 0 because it is in standard state

Answer 4

So 393.5 + -285.8 +-1559.8?

Answer 5

You have delta E_reaction from the 3rd equation
you have formation enthalpies for H2O and CO2
You know formation enthalpy of O2 is 0

So you need to set up Hess equation to solve for C2H6

Answer 6

Look at this example and I think you will see how to solve it

Answer 7

H subscript f is just enthalpies of formation

you can check your answer from a table of enthalpy formations I assume

Answer 8

So I take 2(-393.5)-3(-285.8)+1(-3119.6)

Answer 9

Actually just go over what I wrote

Answer 10

you skipped steps

Answer 11

I think I'm getting myself confused

Answer 12

2C2H6 (g) +7O2(g) --> 4CO2(g) +6H2O (l) delta H rxn =-3119.6 kJ/mol
You want formation enthalpy so you need to write it in terms of CO2 and H2O as reactants
-RULE WHEN FLIPPING REACTIONS ENTHALPY CHANGES SIGNS
So Flip it
4CO2(g) +6H2O -> 7O2(g) + 2C2H6 (g) delta H rxn =3119.6 kJ/mol

Now you need 1 moles of C2H6 you have 2 moles of C2H6 in your reaction divide everything by 2!!!

2CO2(g) +3H2O -> 7/2 O2(g) + C2H6 (g) delta H rxn = (3119.6/2) kJ/mol

Answer 13

Now you need to set up the equation to solve specifically for C2H6 formation, you should be able to do this step without my help just look at the example question and look how they set their problem up

Answer 14

How do you know which numbers go with which compound or element? For example the NO was 90.25 in the example you gave

Answer 15

Okay when I'm setting up the equation do I just write what doesn't cancel or am I thinking of something completely different?

Answer 16

So the equation basically says


Reaction enthalpy = SUM OF products reaction enthalpy - (SUM OF reactants reaction enthalpy)

So for example if you had the reaction

Br2 + C3H6 -> C3H6Br2
you knew formation of Br2 is 0 (standard state)
C3H6 we will just say has a enthalpy of formation of -500kJ (Not really)
C3H6Br2 has formation enthalpy of -700kJ (Not really)

You would set up the equation as

reaction enthalpy

Reaction Enthalpy = (-700kJ)*1 - [(-500)*1 + (-700kJ)*1
you could then solve for reaction enthalpy

IN your case you know the reaction enthalpy you dont know the formation of C2H6

Answer 17

So would it be solved differently then?

Answer 18

yes

Answer 19

not exactly the same just plug in everthing you know into the formula and solve for what you dont know which is the formation enthalpy of C2H6

Answer 20

That is what is being asked in this question because, this reaction:

2C (graphite) +3H2(g) --> C2H6(g)

uses two elements in standard states and whenever you see something like this you are getting enthalpy of formation of whatever isnt in its standard state

Answer 21

By doing the 2(-393.5)-3(-285.8)+1(-3119.6) is that what I'm plugging the numbers into?

Answer 22

Because you use the coefficients of the equation and plug in the given kJ/mol right?

Answer 23

the -3119.6 is your reaction enthalpy (which is incorrect read what I wrote) not your formation enthalpy for C2H6 which is what you need to solve for.

Answer 24

So just 2(-393.5)-3(-285.8)

Answer 25

If so I got 71.4

Answer 26

no

Answer 27

YOu need to include all reactants and all products in the equation

Answer 28

The only difference between the example I posted and your question is that you know the reaction enthalpy and you dont know one of the enthalpies of formation

Answer 29

dont get frustrated you can get this and once you do you will be like damn so easy

Answer 30

I'm more exhausted than frustrated.

Answer 31

Okay so delta H are the reaction enthalpy?

Answer 32

And I don't know C2H6

Answer 33

the third reaction gives its reaction enthalpy, you know the H2O and CO2 formation enthalpies from reaction 1 and 2 (It is just their reaction enthalpies)

Answer 34

So that isn't the numbers by the delta H?

Answer 35

huh?

Answer 36

What is a reaction enthalpy?

Answer 37

It is the amount of energy or heat absorbed or produced by a reaction

since the reactants aren't in standard state the reaction enthalpy is not equal to the the formation enthalpy for C2H6

Answer 38

So is that what the -285.8 and -393.5 kJ/mol is?

Answer 39

Oh! I see now

Answer 40

I would take the -393.5 by 2 =-787 kJ then -285.8 by 3=-857.4 kJ

Answer 41

and the third one would be divided by 2 since it got flipped

Answer 42

After plugging those into the formula I got -84.6 kJ

Answer 43

you change sign when flipping a reaction
you divide by 2 to make sure its just 1 mole of C2H6

Answer 44

Forgot to mention that. Sorry

Answer 45

Thank you for your help and patience!

Answer 46

I'm going to go pass out now

Answer 47

(3119.6/2) =[x*1] - [-285.8*6 + -393.5*4]

solve for x and you have your answer I hope you understand how I got this