A 2.200 g sample of quinone, C6H4O2, is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/°C. The temperature of the calorimeter increased from 22.54°C to 29.67°C. What is the heat of combustion per gram of quinone?
The way to do this is multiply the change in temperature during combustion by the heatcapacity of the calorimeter. This is the heat evolved for your sample weight of 2.200g, so you need to divide by 2.200 to get the heat of combustion for 1g.
[(29.67-22.54 C) x 7.854kJ/C]/2.200g = 25.45kJ/g
For the second part, you just need to take the kJ/g value calculated above and multiply by the number of grams in a mole of quinone:
(12.01 x 6) + (1.01 x 4) + (16.00 x 2) = 108.1g/mole quinone
25.45 kJ/g x 108.1g/mole = 2751kJ/mol
Both answers are wrong you have to think of the reaction, calculations are both right but they are also negative.
-25.5 Kg/J Heat combustion per gram of quinone ( combustion is always negative because is exothermic reaction.
heat combustion per mole -2750 kJ/mol
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