Question #71f02

You can calculate the enthalpies of combustion for these two reactions by using the standard enthalpies of formation for the species involved in the reactions.

You can find these values here:

https://en.wikipedia.org/wiki/Standard_enthalpy_change_of_formation_(data_table)

Now, the idea is that you can determine the change in enthalpy of a given reaction by taking into account the standard enthalpies of formation and the number of moles of each species that takes part in the reaction.

Standard enthalpies of formation are always calculated for the formation of one mole of a compound.

So, let's take the first reaction. The standard enthalpy of formation for magnesium oxide is listed as #"-601.6 kJ/mol"#.

This tells you that when one mole of magnesium oxide is being formed from its constiuent elements in their most stable form, which happen to be magnesium and oxygen gas, #"601.6 kJ"# of energy as heat are being given off.

Now take a look at your reaction

#2Mg_((s)) + O_(2(g)) -> color(red)(2)MgO_((s))#

Since the reaction produces #color(red)(2)# moles of magnesium oxide, it will of course give off twice as much energy as the standard enthalpy change of formation.

#color(red)(2)color(blue)(cancel(color(black)("moles MgO"))) * "-601.6 kJ"/(1color(blue)(cancel(color(black)("mole MgO")))) = color(green)(-"1203.2 kJ")#

You have to do a little more calculations to get the heat of combustion for the second reaction.

This time, you need to use the standard enthalpy change of formation for two species, carbon monoxide and carbon dioxde, because the reaction does not describe the formation of #"CO"_2# from its constituent elements in their most stable form.

The formula you'll use is

#color(blue)(DeltaH_"rxn" = sum(n * DeltaH_"prod"^@) - sum(m * DeltaH_"react"^@)#, where

#n#, #m# - the number of moles of the product and of the reactants, respectively;

The standard enthalpy change of formation for carbon monoxide is listed as #"-110.5 kJ/mol"#, and for carbon dioxide as #"-393.5 kJ/mol"#. The standard enthalpy change of formation for elements in their most stable form is equal to zero.

So, this means that you have

#2CO_((g)) + O_(2(g)) -> 2CO_(2(g))#

#DeltaH_"rxn" = 2color(blue)(cancel(color(black)("moles CO"_2))) * (-393.5"kJ"/color(blue)(cancel(color(black)("mole CO"_2)))) - [2color(blue)(cancel(color(black)("moles CO"))) * (-110.5"kJ"/color(blue)(cancel(color(black)("mole CO"))))]#

#DeltaH_"rxn" = -"787 kJ" - (2"221 kJ") = color(green)(-"566 kJ")#