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In this problem we are going to use the idea of heats of formation to find heats of reaction
and move it around a little so that we can find given a heat of reaction one of the heats
of formation of a reactant or a product. In this case what we are going to be looking
for is the heat of formation for C6H6 given the heat of reaction as well as the heats
of formation of CO2 and H2O. Recall that the heats of formation of elements in there most
stable form such as O2 will equal 0. Our delta heat of reaction is going to equal the sum
of the heats of formation of the products. That is 12 times the heat of formation of
CO2 since there are 12 moles of CO2. Plus 6 times the heat of formation of H2O. Make
sure that this is the heat of formation in its liquid form. Minus 2 times the heat of
formation of our C6H6 which we are looking for. This has to equal -6,534 kJ. What we
are looking for is this heat of formation. All we have to do is put in the number that
we are given up above and solve for the heat of formation of C6H6. So -6,534 kJ equals
-6,436.8 kJ, we got that from substituting in the numbers above, minus 2 times the delta
H of formation of C6H6. 97.2 kJ equals 2 times the heat of formation of C6H6. The heat of
formation of C6H6 is positive 48.6 kJ per mol.