Chem question - Combustion enthalpies

Factual Questions
1

I’ve got a new chemistry assignment, which I’m a bit stuck with. I think my teacher makes some of the questions a little too hard, but they’re challenging, and that’s why I like them. However, if I’m getting stuck, I dread to think how some of the other students are getting on.

Anyhow, it’s supposedly organic chemistry, but the thermochemistry that I’m stuck with.

The data necessary for the question is here:

Alkane ¦No. Of C atoms ¦Relative Molecular Mass ¦Enthalpy Change of combusions/kJ per mol

Methane ¦ 1 ¦16 ¦-890
Ethane ¦ 2 ¦ 30 ¦-1560
Propane ¦ 3 ¦ 44 ¦-2220
Butane ¦ 4 ¦ 58 ¦-2877
Pentane ¦ 5 ¦ 72 ¦-3509
Hexane ¦ 6 ¦ 86 ¦-4195

I’ve been told to draw a graph of the enthalpy change against no. of carbon atoms, and that I’ve done. I’ve also been told to work out the intercept on the deltaH axis which is about -238kj per mol.

What is the physical significance of this value? This is the question that I’ve been stuck on. The only thing I’ve worked out so far is that it’s roughly one deltaHc of the CH2 structural group less than methane, as you would expect, since it’s a straight line graph, well near enough (y = mx + c, and C is obviously this value). I can’t work out what significance this value has, but it’s probably kickmyself obvious, and I’m going to look really silly when someone explains it to me.

If anyone can please help, that’d be great, it’s a shame there are so many people who can help with physics on SDMB, but far fewer chemists!

2

dammit that ‘table’ of data looked prettier in preview

3

Use the “code” tag. Like “[” “code” “]” table of stuff “[/” “code” “]”.

4

aha, thankyou

5

Just out of curiosity, what’s the enthalpy of combustion pf H[sub]2[/sub]?

6

ooh, let me look…finds book

deltaHc [H2(g)] = -286kJ per mol

is there a way I can put the delta symbol into a message rather than writing it?

7

Perhaps -238 should be the deltaHc of Hydrogen?
Given that you are graphing DHc against No. C,
it seems reasonable to conclude that where n©=0 and therefore n(H)=2, DeltaHc will be the DeltaHc of H2.
Just a WAG.
-Oli

8

You’ve probably gotten all that you need, Pretend My Name Is Witty. Work out the final step yourself.

9

Theres nothing very pretty about the significance of the intercept.
Nothing elegant at all.

Think about this: at each step you’re adding one carbon; and at each step the enthalpy change is roughly the same (in fact, except for rounding off, each step should be exactly the same because these are only calculated figures).

For each step write out how many of each type of bond are being formed and how many broken. You should spot a pattern.

Now your line is y=mx+c and m=(delta y)/(delta x) but you can set (delta x) = 1 for each step and then m = (delta y) = the enthalpy change at each step. And c = y-m where x=1.

Look for the pattern of bonds, that’s the key along with one other thing…think hard about what you do with the C-C bonding.

I hope this helps.

10

OK, basically, when I started this thread it was the start of the half term and I had a week to do this, with no hope of going to a chemistry workshop for a while.

I did think about the significance of it being the combustion of H[sub]2[/sub], but since I had that value given in my textbook, and found them to be roughly 50kJ apart, I thought it must be something else. However, I went back to college today after the holiday and talked to my teacher. He told me that my initial idea of it being the combustion of H[sub]2[/sub] was correct (as were the helpful dopers who attempted to re-steer me back toward this, but whose advice I didn’t follow!). A difference of 50kJ is not really that significant, and anything between 200 and 300 would be acceptable as the intercept.

Thankyou all for your help, if only I had trusted my instincts in the first place, I would have been able to answer it anyway!

11

I probably shouldn’t do this, but I think your teacher is wrong. Work it through:

Use the following rounded-off values of bond energies:

(H-H) = 436 kJ/mol…………………………………(O=O) = 498
(C-C) = 347…………………………………………(C=O) = 805
(C-H) = 413…………………………………………(H-O) = 464

Methane: CH(4) + 2O(2) ---- CO(2) + 2H(2)O

Bonds broken = 0(C-C) + 4(C-H) +2(O=O)
Bonds formed = 2(C=O) + 4(H-O)
Delta H = -818

Ethane: C(2)H(6) + 3.5 O(2) — 2CO(2) + 3H(2)O

Bonds broken = 1(C-C) + 6(C-H) + 3.5(O=O)
Bonds formed = 4(C=O) + 6(H-O)
Delta H = -1436

Propane: C(3)H(8) + 5 O(2) — 3CO(2) + 4H(2)O

Bonds broken = 2(C-C) + 8(C-H) + 5(O=O)
Bonds formed = 6(C=O) + 8(H-O)
Delta H = -2054

Butane: C(4)H(10) + 6.5 O(2) — 4CO(2) + 5H(2)O

Bonds broken = 3(C-C) + 10(C-H) + 6.5(O=O)
Bonds formed = 8(C=O) + 10(H-O)
Delta H = -2672

Notice that at each step Delta H changes by –618.

*****Notice that at each step the bonds being broken and formed change by the factor:

Broken: 1(C-C) + 2(C-H) + 1.5(O=O)………Formed: 2(C=O) + 2(H-O) and the Delta H for this is –618.**

Now, your line is y = mx + c where y is the heat of combustion and x is the number of carbons.

The slope m = (Delta y) / (Delta x) = -618.

If you set x = 1 then y = -818. You can now calculate c = y – m = (-818) – (-618) = -200

What does this represent? It represents the bond breakage/formation of y (methane) minus the bond breakage/formation of your slope m (see ***).

……………0 (C-C) + 4(C-H) + 2 (O=O)………2(C=O) + 4(H-O)
minus……1 (C-C) + 2(C-H) + 1.5 (O=O)………2(C=O) + 2(H-O)

equals……-1 (C-C) + 2(C-H) + 0.5(O=O)…………………2(H-O)

and the Delta H for this is exactly –200, which is your y - intercept.

The heat of combustion of hydrogen = -243 kJ/mol.

But more importantly, why would the intercept be related to the heat of combustion of hydrogen when throughout this entire exercise we have never broken any hydrogen-hydrogen bonds? It doesn’t make sense.

12

Right, I caught my teacher at a possibly poor moment, and I dont quite know if he understood my disposition.

I understand that each time you add a carbon atom, you increase the enthalpy of combustion by the [delta]H of the functional CH[sub]2[/sub] group. This is logical, since that’s the difference between each member of the homologous series.

I had to work out this difference for an earlier part of the question. To work out the intercept, I found the equation of the line in the form y = mx + c, and used the C value for my intercept, which also is the deltaH[sub]c[/sub] of methane minus the difference calculated earlier. Is this the only physical significance of the intercept? (dH[sub]c[/sub] [CH[sub]4[/sub]] - dH[sub]c[/sub] [CH[sub]2[/sub]])

I took this as given really in answering the question, and tried to find a nicer, prettier significance. The combustion of Hydrogen seemed reasonable, yet I hadn’t thought about there being no H-H bonds in the data before. They obviously can’t make themselves without giving off more energy, so I’m stuck once again.

Do I really need to go any further? Had I been looking for a nicer answer all along, and simply not realised that what I took as a part of the question is really the answer?

Hmm, I’m not sure, I’ve got a nice day at college tomorrow with plenty of free periods so I’ll take it to the workshop. The rest of the assignment is ok, it’s just this one damn question that I seem to have got hung up on.

Maybe I could ask my auntie for help on this, she has a phd in thermochemistry, and has helped me before, I’m just hesitant to ask her incase she thinks I’m losing my touch with chemistry! (Which I dont seem to be, I got my last assignment back today - 100%, sorry, now this is just silly bragging but it’s just that chemistry is the one subject that’s always seemed to come easily to me and I’ve always enjoyed it, hence why this question is bothering me so much!)

13

It sounds to me like you know what’s going on here. I think the problem is that the word “significant” seems to indicate that there is something “profound” about the y - intercept, whereas it appears to me that all the question is really asking is what does the intercept represent.

My answer would be something along the lines of "the heat of combustion of any alkane is going to be a factor of the number of carbons plus “a little something extra” which represents the difference between the initial alkane (methane) and the functional group which is being added each time (the CH(2) group). Note also that you would get a similar trend if, instead of the alkanes you did the alcohols or the carboxylic acids. In each case your line would be the same except you would have different y - intercepts because the “little something extra” would be different.

If that rather poorly worded explanation makes sense to you then I think you’ve probably gotten all you’re going to get out of this exercise. (But talking to a Phd never hurts!)

14

Excellent, thankyou for helping me in this, I’ll talk it over with my teacher before I hand it in, and I’m not going to mention the combustion of hydrogen!

Thanks again, I’ll tell you what he says

15

I can guess what he’s going to say.

It’s approximately equal to the combustion enthalpy of H[sub]2[/sub], but it isn’t exactly equal because because there’s a difference in the net bond enthalpies between the H-H case and the 2 x C-H + C-C case.

16

Right, I went to the workshop today and talked to my favourite chemistry teacher and he admitted that he didn’t like that particular assignment.

The drawing of the graph is a reasonable idea to show the trend of the alkanes but that is all. It shows you the difference between each being the dH[sub]c[/sub] of the CH[sub]2[/sub] group.

However, being discontinuous data, the intercept has no real significance other than the one that I settled on. It’s like using the graph to work out the dH[sub]c[/sub] of an alkane with 2[sup]1[/sup]/[sub]2[/sub] carbon atoms. It just doesn’t have any real value. A bar chart with seperated bars would have neem more suitable for something like this.

He also said that the answer of it being the dH[sub]c[/sub] of H[sub]2[/sub] would have been acceptable from a Lower 6[sup]th[/sup] As-Level student. Seeing that there are no carbon atoms and just two hydrogen atoms, the average student would be forgiven for accepting the value as being dH[sub]c[/sub] of H[sub]2[/sub].

After a debate with the teacher as to what answer would be most suitable to give me the mark, we decided that I should put the dH[sub]c[/sub] [H[sub]2[/sub]] answer to get the mark and not look too big headed. I’m going to put a footnote at the bottom to explain that although my answer was the one written, I know that it doens’t really make much sense, and that the graph’s not suitable for this.

Thankyou all again for your help. I’ll post my result when I get it!