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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