AS/A-Level Chemistry- Enthalpy (ii): Hess's Law

Hess's Law

• The enthalpy change is independent of the path taken

• applying Hess’s Law enables one to calculate enthalpy changes from other data
• used for calculating changes which can’t be measured directly  - Lattice Enthalpy
• used for calculating enthalpy change of reaction from bond enthalpy, enthalpy change of reaction from ∆Η°c, enthalpy change of formation from ∆Ηf                                                                                              

Let's see how it is used to calculate A-Level Chemistry enthalpy of formation and combustion:

Enthalpy of formation: If you formed the products from their elements you should need the same amounts of every substance as if you formed the reactants from their elements.

∆Η_r = Σ∆Η_{f (Products)} - Σ∆Η_{f (Reactants)}

example:

Calculate  the  standard  enthalpy  change  for  the  following  reaction,  given  that  thestandard enthalpies of formation of water, nitrogen dioxide and nitric acid are -286,+33 and -173 kJ mol-1 respectively.  

2H₂O(l)   +   4NO₂(g)   +   O₂(g)    ——>    4HNO₃(l)

applying Hess’s Law ...∆Η°r =    [ 4(-173) ]   -   [ 2(-286) + 4(+33) + 0 ]    =  -252 kJ

 [oxygen’s value is zero as it is an element ]

Enthalpy of combustion: If you burned all the products you should get the same amounts of CO₂ and H₂O etc. as if you burned the reactants.

∆Η_r = Σ∆Η_{c (Reactants)} - Σ∆Η_{c (products)}

example:

Calculate  the  standard  enthalpy  change  of  formation  of  methane,  given  that  thestandard  enthalpies  of  combustion  of  carbon,  hydrogen  and  methane  are  -394,-286 and -890 kJ mol-1 respectively.

C(graphite)   +   2H₂(g)     ——>   CH₄(g)

applying Hess’s law ...∆Η°r =   [ (-394) +  2(-286) ]   -   [ (-890) ]   =   -74 kJ mol-1

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Drafted by Eunice Wong(Chemistry)