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

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:

$\text{Hydrocarbon"(l"/"g) + "O"_2(g) -> "CO"_2(g) + "H"_2"O} \left(g\right) + \Delta$

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. ${\text{CO}}_{2}$ at room temperature is a gas, but water at room temperature is a liquid.

So although ${\text{CO}}_{2}$ 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 ${\text{CO}}_{2}$ 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.