5) Chemical energetics (2016)

You may like to take a look at these A2 component Chemistry Revision Videos by Allery Tutors (and if you wish, send them a donation. Note: I have no connection whatsoever to Allery Tutors other than be an admirer of their videos)

Introduction to Entropy (9:30)

Entropy Change 1 (7:50)

Entropy change 2 (4:21)

Gibbs Free energy change 1 (11:19)

Gibbs Free energy change 2 (7:40)


Below = suggested answers to the entropy review questions given out 23-Sept-15 (sorry about the typo)
Suggested ansters to Entropy Review Q’s Hodder A2



6 thoughts on “5) Chemical energetics (2016)

  1. Hi Sir,
    There is this question from the handout you gave us on the topic thermochemistry, photostatted from the Advanced Chemistry, Spread 10.10, page 161:

    Butane and 2-methylpropanone both have the molecular formula C4H10. Would you expect their standard enthalpy changes of combustion to be identical? Explain your answer.

    By using the average bond enthalpies to calculate the ^Hc( standard enthalpy change of combustion), the answer will be identical because the numbers of C-C bond and C-H bond are the same, however if we were to compare the intermolecular forces between these two isomers, the Van der Waal’s forces is stronger in butane than in 2-methylpropanone due to the greater number of contact points. This is the main reason why butane has a higher boiling point than 2-methylpropanone, so can we conclude that the ^Hc of the two will not be identical since more energy is required to overcome the stronger intermolecular force in butane? And from here, can we say that the bond dissociation energy of C-C bond or C-H bond in butane is higher than that in 2-methylpropanone? Do we even consider the intermolecular force of a compound when calculating the enthalpy changes?


  2. Technically, this is posted in the wrong place, given you’ll be doing AS 2015 syllabus. But I guess it will still be relevant to the 2016 syllabus also.
    Well my internet access is at times awfully slow – worse then KY’s (!) so apologies for not getting back sooner. I made a reply in this document…
    Let me know if you still have any questions about it.


  3. Apologise for me mistake, should be 2-methylpropane xD
    So the ^Hc would be different because both chemicals have different chemical environment? I kind of googled the ^Hc value for both isomers, butane has a ^Hc value of -2876.5 kJ/mol, 2-methylpropane -2868.5 kJ/mol, the difference is almost negligible to me, but do we still say that they have different ^Hc values in the exam?


  4. If one to actually measure them experimentally then yes, the Hc for each molecule would be different. – As I see you’ve found out already. But the difference wasn’t likely to be so great between them for the two reasons in the document. Indeed as you’ve discovered, experimentally there’s only 8 kJ mol-1 difference between them.

    So in an exam if you were using BEV’s – a theoretical, paper based exercise, you would assume the isomers had the same Hc because in the reactants and products, they had the same initial bonds, same number of bonds.

    Experimentally, you would not assume them to have the same because the experimental results reflect the reality of what’s going on and the molecules are indeed different.

    What often happens in some exams, is that you are told to work out a theoretical value (and possible comment on it, saying something like “it gives a reasonable/workable approximation as to the real values” {that is afterall why we bother to use them in the first place –> errr don’t write that in the exam! :p}). Then you are often given the experimental value and asked to compare the two values, saying which is more accurate and to account for the difference. Here you would point out the problems with the fact that BEV’s are just average values, not reflecting the exact bond enthalpies in any particular compound.

    I said the BEV’s were useful – as demonstrated that they almost reflected the reality whereby butane and methylpropane had almost the same values, but notice however the calculation I did gave -2146 kJ mol-1 but the experimental values you quoted are almost 700 kJ mol-1 different! (more exothermic) Wow! that IS a significant difference. BEV’s don’t look so good now huh? Well let me blab on a bit about probably reasons why this is.

    The C=O and O-H bonds in the products formed are in “extreme molecules”, extremely small H2O and CO2 molecules, with high levels of attraction and high levels of repulsion. The environment in which they exist is quite different from most other C=O (aldehydes/ketones) other OH’s (like alcohols) containing molecules, so the deviation of each C=O and O-H in carbon dioxide and water from the average C=O and O-H, is at a maximum ‘outlying’ range amongst the typical values making up the average, and, we have a LOT of these bonds, 10 O-H and 8 C=O’s which multiplies the error up by about 18 times. The cumulative error here is therefore quite large. But we can ‘fool’ ourselves (as scientists often do for the sake of simplicity) and say “well… it’s in the -2000’s”, lol, and maybe we can be content with that 😉


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