8. When 2.200 g of quinone (CeH4O2) is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 23.44 °C to 30.57 °C. In a separate experiment, the heat capacity of the calorimeter is measured to be 7.854 kJ/2C. Calculate the heat of combustion for this reaction 274 kJ
Molar mass of C6H4O2 (quinone)= 108.092 g/mol
2.200 g quinone / 108.092 g/mol = 0.020353033 mol
Change in temp = 30.57 C - 23.44 C = 7.13 C
qcalorimeter = (Heat Capacity)(Change in Temp) = (7.854 kJ/C)(7.13 C) = 55.99902 kJ
then you could do a couple different things, you could divide by moles to get kJ/mol (but it doesn't specifically say in kJ/mol, however the answer given is very large so I figure that it must be kJ/mol even though the answer key says just kJ.
so 55.99902kJ/0.020353033mol =2751 kJ/mol
or you could divide by the grams of material combusted and then mutiply by the molecular mass which also gives 2751 kJ/mol
(55.99902*108.092)/2.200 =2751 kJ/mol
but the answer key says that it should be 2749 kJ (not kJ/mol) and I do not see how you could possibly get that large of a number unless you are talking about kJ/mol.
Please help!
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