When #"32.4 mL"# of liquid benzene (#C_6H_6#, #rho = "0.879 g/mL"#) reacts with #"81.6 L"# of oxygen gas, measured at #"1.00 atm"# and #"25°C"#, #1.19xx10^3 "kJ"# of heat is released at const. pressure. What is #DeltaH^@# for the following reaction?

(#R = "0.0821 L"cdot"atm/K"cdot"mol"#)
#2"C"_6"H"_6(l) + 15"O"_2(g) -> 12"CO"_2(g) + 6"H"_2"O"(l)#

1 Answer
Apr 11, 2018

#DeltaH_(rxn)^@ = -3.264 xx 10^3 "kJ/mol"#

Why is it important (i.e. useful!) to use #"kJ/mol"# instead of #"kJ"#?


At constant pressure, the heat released IS the #DeltaH# of the reaction. The combusted quantity is

#32.4 cancel"mL" xx "0.879 g"/cancel"mL" = "28.48 g benzene"#

#28.48 cancel"g benzene" xx ("1 mol benzene")/(78.1134 cancel"g benzene") = "0.3646 mols benzene"#
vs. #"2.735 mols O"_2# needed

and we can probably tell that the #"O"_2# is in excess... assuming ideal gases,

#n = (PV)/(RT) = ("1.00 atm" cdot "81.6 L")/("0.08206 L"cdot"atm/mol"cdot"K" cdot "298.15 K")#

#= "3.335 mols O"_2#

Indeed, it is in excess by #"0.6006 mols"#. We base the mols off of the limiting reactant benzene.

The reaction apparently releases #"1190 kJ"# of heat at constant pressure, so

#color(blue)(DeltaH_(rxn)^@) = -"1190 kJ"/"0.3646 mols benzene"#

#= color(blue)(-3.264 xx 10^3 "kJ/mol")#