Question #4416b

1 Answer
Dec 13, 2015

Answer:

#DeltaH_"comb" = -"17,517 kJ"#

Explanation:

As it is written, the problem does not provide enough information to allow you to calculate the enthalpy change of combustion, #DeltaH_"comb"#, for that reaction.

More specifically, you need to know the standard enthalpy change of formation of methyl myristate, #"C"_13"H"_27"CO"_2"CH"_3#. I was able to find this problem here - page #801#, Alternative fuels section

https://books.google.ro/books?id=98UTCgAAQBAJ

The standard enthalpy change of formation for methyl myristate is said to be equal to #-"771.0 kJ/mol"#. The standard enthalpy change of formation for water and carbon dioxide can be found here:

http://nshs-science.net/chemistry/common/pdf/R-standard_enthalpy_of_formation.pdf

So, you can calculate the enthalpy change of reaction by using the standard enthalpy changes of formation of the products and of the reactants

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

#n#, #m# - the stoichiometric coefficients of the products and of the reactants, respectively
#DeltaH_"f prod"^@#, #DeltaH_"react"^@# - the standard enthalpy changes of formation for the products and for the reactants, respectively

In your case, the balanced chemical equation for this combustion reaction looks like this

#2"C"_13"H"_27"CO"_2"CH"_text(3(l]) + 43"O"_text(2(g]) -> 30"H"_2"O"_text((g]) + 30"CO"_text(2(g])#

So, plug in your values and calculate #DeltaH_"comb"# - keep in mind that the standard enthalpy change of formation for oxygen is zero!

#DeltaH_"comb" = [30 color(red)(cancel(color(black)("moles"))) * (-393.5 "kJ"/color(red)(cancel(color(black)("mol")))) + 30color(red)(cancel(color(black)("moles"))) * (-241.8 "kJ"/color(red)(cancel(color(black)("mol"))))] - [2 color(red)(cancel(color(black)("moles"))) * (-771.0 "kJ"/color(red)(cancel(color(black)("mol"))))]#

#DeltaH_"comb" = ( -"11,805 kJ" - "7,254 kJ") -"1542 kJ"#

#DeltaH_"comb" = color(green)(-"17,517 kJ")#