The value of DH0 for the reaction below is #-6535# kJ. __________ kJ of heat are released in the combustion of #16.0# g of #C_6H_6#(l)? 2C6H6(l) + 15O2 (g) --> 12CO2 (g) + 6H2O(l)

#2C_6H_6(l) + 15O_2 (g) -> 12CO_2 (g) + 6H_2O(l)#

#1.34xx10^"3"#
#669#
#-6535#
#5.25xx10^4#
#2.68xx10^3#

1 Answer
anor277
Aug 1, 2017

#DeltaH_"rxn"^@# are always quoted per mole of #"REACTION"# as written. And thus #669*kJ# of energy are #"RELEASED"# from the reaction.

Explanation:

We gots.........

#2C_6H_6(l) +15O_2(g) rarr 12CO_2(g) + 6H_2O(l) +Delta#

#DeltaH_"rxn"^@=-6535*kJ*mol^-1#, and I reiterate, that this energy value is PER MOLE of reaction as written, i.e. the enthalpy change associated with the combustion of #2*mol# benzene, i.e. a #154*g# mass.

We gots, #(16.0*g)/(78.11*g*mol^-1)=0.205*mol# with respect to benzene......and thus the heat associated with the combustion of this molar quantity is.........

#DeltaH_"rxn"^@=0.205*molxx1/2xx-6535*kJ*mol^-1=-669*kJ#

Note that when we say that this amount of energy is #"released from the reaction"#, in effect we are saying that the reaction is exothermic, and #DeltaH^@# is a #"NEGATIVE QUANTITY"#.

Happy?

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