4. Hess Cycle using Combustion Enthalpies
Uploader Comments (jamesmungall)
Top Comments
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Explained better than how my teacher explained it
All Comments (20)
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All I can say is... thank you sir! Simple and straight to the point, no hay-wire nonsense! Much better than my teacher!
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Thanks very much for the explanation. Currently taking a break from revising for AS exam by.. watching revision videos. Hahaha. Hess's law is kind of like using a combination of vectors and enthalpy change calculations, which I find helpful to think of, as vectors are something we use regularly in mathematics, and it helps one get one's head around the diagram.
Thanks again, and hope you have a very pleasant new years. :)
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Its kinda confusing. I prefer my teacher's method. Maybe cuz she uses white board and white is step by step
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im an a level student and my chemistry teacher sux... thnx 4 uploading these videos on hess law and calculations based on it...they r really helpfull...
arent all combustions should get minus value ?
SinhaleseForever 1 year ago
@SinhaleseForever The data which you are using in the question are combustion enthalpies, so all of these values are negative. However, the enthalpy which you are being asked to work out is *not* a combustion enthalpy and therefore isn't necessarily -ve. Hope that makes sense, James.
jamesmungall 1 year ago 2
Thanks for this I understand this concept much better but there is still one thing that I can't get my head around...how do you know that -1300 should change to +1300?
sasukerulez04 2 years ago
Hi there, I guess I'm saying that we can go along a theoretical route in which the reactants are burnt (exothermic) and then these combusted products are 'unburnt' (endothermic) to get the products. Hope that makes some sort of sense, James
jamesmungall 2 years ago
I'm a little confused... Combustion is an exothermic reaction (heat is given out) therefore the resulting figure should be a negative number! Or maybe this video should be called enthalpy of formation?
richard93goodman 2 years ago
Hi Richard, So in this example the enthalpy which you are calculating *is* an enthalpy of formation! (enthalpies of formation can be either positive or negative). However you are using enthalpies of combustion to do the calculation. Hope that makes sense, James.
jamesmungall 2 years ago