Question #cd6d2

2 Answers
Apr 15, 2016

To calculate #Delta H# for a reaction, you must substract enthalpy of formation of products to enthalpy of formation of reagents.

Explanation:

Let us see it with an example:

#"CH"_4 ("g") + 2 "O"_2 ("g") rightarrow "CO"_2 ("g") + 2 "H"_2 "O" ("g")#

We must know enthalpy of formation of all compounds and elements:

  • #Delta H_"form" ("CH"_4) = -17.9 " kJ/mol"#, from the reaction:
    #"C" + 2 "H"_2 rightarrow "CH"_4#

  • #Delta H_"form" ("H"_2"O") = -241.82 " kJ/mol"#, from the reaction:
    #"H"_2 + 1/2 "O"_2 rightarrow "H"_2 "O"#

  • #Delta H_"form" ("CO"_2) = -393.5 " kJ/mol"#, from the reaction:
    #"C" + "O"_2 rightarrow "CO"_2#

  • #Delta H_"form" ("O"_2) = 0 " kJ/mol"#, like every element at its natural state.

So, now:

#Delta H_"reaction" = #
#= sum n cdot Delta H_"form" ("products") - sum m cdot Delta H_"form" ("reagents") =#
#= [(-393.5) + 2 cdot (-241.82)] - [(-17.9) + 2 cdot 0] = -859.24 " kJ/mol"#

where #m, n# represent the coefficients in the reaction.

Note: albeit there are other methods to obtain enthalpy (such as Hess's law, for example), I think this is the one which better adjusts to what you are looking for.

Apr 24, 2016

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