Why do we often write the water product as a gas in complete combustion reactions?

1 Answer
Nov 13, 2016

Well, let's put it this way.

Combustion reactions are generally quite exothermic. That means it releases heat. We can represent that like this:

#"Hydrocarbon"(l"/"g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O"(g) + Delta#

where the hydrocarbon is generally a liquid if it is massive (larger than butane, usually), and is generally a gas if it is small. #Delta# on the products side would represent the heat released due to the reaction itself.

Since water is produced, the water can immediately absorb that heat. #"CO"_2# at room temperature is a gas, but water at room temperature is a liquid.

So although #"CO"_2# could absorb some of the heat, water also absorbs the heat that is produced in such a short amount of time.

The heat has a greater effect on the water, since the liquid water molecules had less energy to begin with than the gaseous #"CO"_2# molecules, so water is a gas when the combustion reaction first finishes.

It is true that in a closed system, water vapor could condense onto the insides of a bomb calorimeter, but we usually just talk about the moments immediately after the reaction occurs.