Enthalpy - Hess' Law for Combustion data (video)

This is a VERY short clip that is meant to go with my flashcards on enthalpy (https://getrevising.co.uk/revision-cards/energetics-23) and energetics, just showing an example of caluctaing Hess' Law Cycles as I know it's difficult to describe in writing!

For As Chemistry (edexcel)

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  • Created by: LouiseG
  • Created on: 01-06-16 17:03

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Important points:

- Consider the direction the arrows face; either the species in your equation under investigation are the products, (arrows point UP) or the reactants.

- Add enthalpy values given in the question, considering the stoichiometric equation: in this example, the enthalpy of combustion of H2 was multiplied by 2 as two moles of H2 were present in the equation.

-To find your missing enthalpy value, you need to move around the cycle via your intermediate (the species written below the line) in the same direction as the equation as written (i.e. from left to right). Therefore it is important to consider whether you are moving "with" an arrow (from tail to tip/head) or against it, in which case you reverse the sign.

- The enthalpy data given is not always of a type you know, such as combustion and neutralisation (i.e. it will be an enthalpy of "reaction"). The key point is that the enthalpy change equations referred to will be given and there will be some common species among them, such as a salt product which is formed in all reactions of the different products/reactants in your equation, and this will be the "intermediate" species. For example, to work out the enthalpy change of the thermal decomp. of CaCO3;

CaCO3 (s) ----> CaO(s) + CO2 (g)
The enthalpies of reaction of calcium carbonate and calcium oxide can be used:
CaCO3 (s) + 2HCl (aq)---> CaCl2 (aq) ---> CaCl2 (aq) + H2O (l) + CO2 (g)  -17 kJmol-1

CaO(s) + 2HCl (aq) ---> CaCl2 (aq) + H2O (l) -195 kJmol-1

Your intermediates are CaCl2 (aq), H2O(l) and CO2 (g). Notice that although CO2 isn't formed in the reaction of CaO, it is a product of the thermal decomp. reaction therefore can be considered part of the reaction but as a species which does not change, i.e.

CaO(s) + 2HCl + CO2 ---> CaCl2 (aq) + H2O (l) + CO2 (g). 

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