Can You Calculate The Energy That Is Released When 126 G Of Glucose Reacts With Excess Oxygen. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) + 673 Kcal?


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Yes, if you are doing the experiment at constant pressure (this is what people usually think about),

heat evolved = energy released = -DeltaH0

So just go ahead and calculate

DeltaH0(reaction) = sum[DeltaH0f(products)] - sum[DeltaH0f(reactants)]

If you do all this inside a combustion engine, some of the energy released by the system will be in the form of extra work rather than heat, so that DeltaH0 will no longer be equal to q, but that doesn't matter. DeltaH0 is bound to have the same value and -DeltaH0 (minus because "energy released") is what you're after

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