Question

Can someone check my answer?
http://oi64.tinypic.com/2iw0yus.jpg

Answer 1

@LeibyStrauss

Answer 2

Ok, this one is easy. The standard enthalpy of formation (at 25° -298 K, 1atm)
\[\Delta H ^{°} _{f} \] is the energy required to form a mol of the compound from its elements. Methane has \[\Delta H ^{°} _{f} \] different than 0 because the O2 and C need be reordered to form CH4,
which is not the case of O2(g) or H2(g) for example, since they are in their most stable state, there isn't energy involved in their formation.
I hope this is clear enough. If not let me know.

Answer 3

Still confused.

Answer 4

Think about it this way, Oxigen is a gas at room temperature, Sulfur is a solid at room temperature, mercury is a liquid.
Would you require any energy at all to keep them this way?
Not really, because they are in their most stable form. Therefore the heat of formation at 25° C is O, for every element in their most stable form.

I have another example:
Carbon can be present in solid form as graphite or diamond
however the most stable form is graphite, so the enthalpy (or heat) of formation for graphite at 25° C and 1 atm is O
since diamond is stable but still not the most stable its enthalpy of formation is 2,1 kJ/ mol

Answer 5

Is that an O or 0?

Answer 6

sorry, it is 0 (zero), whenever you write Oxygen, you have to use the name or the \[O _{2}\] followed by the state \[O _{2} ({g})\]
if not is just the number 0 zero.

Answer 7

So what about something like H (g)?

Answer 8

I have to quote this book Chemistry Chang, 10th edition. section 6.6 page 252

"There is no way to measure the absolute value of the enthalpy of a substance. Only values relative to an arbitrary reference can be determined.
Chemists have agreed on an arbitrary reference point for enthalpy.
The reference point for all enthalpy expressions is called the standard
enthalpy of formation (DH8f). Substances are said to be in the standard state at 1 atm, †
hence the term “standard enthalpy.” The superscript “°” represents standard-state.
conditions (1 atm), and the subscript “f” stands for formation. By convention, the standard enthalpy of formation of any element in its most stable form is zero. Take the elementoxygen as an example. Molecular oxygen (O 2 ) is more stable than the other allotropicform of oxygen, ozone (O 3 ), at 1 atm and 25°C. Thus, we can write DH°f (O2) 5 0, butDH°f (O3) 5 142.2 kJ/mol.
Similarly, graphite is a more stable allotropic form of
carbon than diamond at 1 atm and 25°C, so we have DH°f (C, graphite) 5 0 and
DH°f(C, diamond) 5 1.90 kJ/mol. Based on this reference for elements, we can now
define the standard enthalpy of formation of a compound as the heat change that
results when 1 mole of the compound is formed from its elements at a pressure of 1 atm."
I have the pdf book if you want I can send you a screenshot of exactly the same page.

Answer 9

I think I'm getting confused how to take that information and apply it to the problem. How do I know it's heat and such from it's state or formation?

Answer 10

as for your question, sorry I didn't answer it before, the hydrogen atom isn't too stable
so it's enthalpy of formation is 218 kJ/ mol.

Answer 11

I'm currently studying chemistry, and we always carry tables with this information, each course has it's own data set, we are encouraged to always use them

Answer 12

Mine doesn't have it. : (

Answer 13

So out of the options would it be CH4 (g), H (g), S8 (monoclinic)

Answer 14

I see, check this out, ignore the spanish.
YES!

Answer 15

So I had the right answers selected?

Answer 16

yes, I guess what I was trying to do was justify the selection, I'm not used to multiple choice questions, sorry again.

Answer 17

Okay. For some reason it's been counting those three wrong.

Answer 18

It shouldn't, anyway chemistry is hard. I wonder if there is something else in the question that we're not seeing.

Answer 19

I agree with it being hard.

Answer 20

@CShrix

Answer 21

I honestly don't remember that much from my college level chemistry days. Sorry bud

Answer 22

@Photon336

Answer 23

@Brrandyn

Answer 24

@Photon336

Answer 25

@ganeshie8

Answer 26

@LeibyStrauss

Answer 27

I don't think S8 (as an elemental solid at that temperature) has a non-zero enthalpy of formation.

Answer 28

So it should just be the CH4 and H?

Answer 29

That's my understanding...H(g) is split from H2(g), so that takes some energy, and the CH4(g) appears in my table with a non-zero EoF

Answer 30

What table are you given?

Answer 31

I'm looking at Table 8.3 Standard Enthalpies of Formation at 25 C (kJ/mol) of Compounds at 1 atm, Aqueous Ions at 1 M from Masterton & Hurley: Chemistry, 5th edition

Answer 32

Oh

Answer 33

the chemistry book I happened to have handy. You can also look up many values on wolframalpha.com website

Answer 34

I'll have to do that. I've been getting different answers.

Answer 35

but not all of the items in this problem appear in either place

Answer 36

so which ones are you thinking would qualify?

Answer 37

I think after you uncheck S8 that your answer will be marked correct.

Answer 38

Thank you

Answer 39

in my table, monoclinic sulfur has en enthalpy of formation of +0.33 kJ/mol
and rhombic sulfur is the most stable with an enthalpy of formation of 0 (zero)

Answer 40

Thanks guys!

Answer 41

You're welcome

Answer 42

Ah, yes, looking more closely, the value I reported for sulfur was not tagged as the monoclinic form, even though I requested it...sorry about that! Makes sense that the monoclinic form has a positive enthalpy of formation vs. the rhombic form.