Dear Colleagues,

The following question asks me: Calculate the Enthalpy of Reaction for ( using Hess's Law ):

C

_{6}H

_{12}O

_{6} ----->2C

_{2}H

_{5}OH+2CO

_{2}Using the following data ( Enthalpy of Combustion ): Glucose=-2820 Ethanol -1367 The answer is:

The answer then states that there is a linear method available for obtaining the same answer:

Δ

_{r}H=ΣΔ

_{c}H

_{reactant}-ΣΔ

_{c}H

_{reactant}Which is : -2820 - ( 2 x -1367 ) = - 86 KJ mol

My confusion is: if the question is asking for the Enthalpy of Reaction ( Delta H ) - assuming its for

the whole reaction, then why is the other product CO

_{2} not included in the calculation ? ( why is only 2C

_{2}H

_{5}OH included in the calculation ? ).