Answers for the Heat of Combustion of Cellulose
The balanced reaction is:
Glucose (to represent cellulose):
The calculated heats of reaction (heats of combustion):
Glucose:
heat of reaction = [6 x heat of formation of CO2 (g) + 6 x heat of formation of H2O (g)] -
[heat of formation of glucose]
Heat of reaction = [6 x (-393.5 kJ/mole) + 6 x (-241.8 kJ/mole)] - [-1250 kJ/mole]
heat of reaction = -2560 kJ/mole
WOW! That was fun!
Let's do one more calculation.
Notice that the value above is in kJ/mole. However, in combustion reactions we generally want to know how much heat is released per gram
of material that we burn, so we need to convert the values into kJ/g. How do we do that? Well ... we can determine from the chemical structure
the mass of one mole and the calculations we've done tell us how much heat is released when a mole burns. That's all we need!
For the glucose example:
From the chemical structure of glucose one mole is 180 grams (the molar mass or molecular weight is 180 g/mole). From the calculation we know
that when one mole, or 180 grams, of glucose burns it releases 2560 kJ of energy (the heat of reaction is -2560 kJ/mole). So on a per gram
basis, the heat released is 2560 kJ divided by 180 g or -14.2 kJ/gram.
Cellulose and how it burns
Activity 4: Calculating Heat of Combustion
Now do cellulose acetate and cellulose nitrate on your own.
In order to balance the reactions, assume that the products of the cellulose acetate combustion are carbon dioxide and water.
Assume that the combustion of cellulose nitrate produces only carbon dioxide and ammonia (NH3). Yes you'll see that it is possible to
balance the reaction without producing water (or adding oxygen!).
Calculate the heat of combustion for each reaction in kJ/mole and in kJ/gram.
(We know you can do it !!)
The data needed to do the calculations can be found here.
When you're done, go here.
