C6H14(l)->C6H14(g) ->6CO2(g) + 7H2O(g) ->6CO2(g) + 7H2O(l)

Hexane is a liquid at room temp. It can be converted into carbon dioxide and water through the above processes. There are 3 steps. Steps 2 & 3 are exothermic, right? I think step 2
C6H14(g)-„³6CO2(g) + 7H2O(g) is something like combustion so it is exothermic.Please help.

correct.

But step 1 is endothermic.

To determine if a reaction is exothermic or endothermic, we need to consider the change in enthalpy (heat) of the system. In step 2, the reaction is indeed a combustion reaction, where hexane (C6H14) is reacting with oxygen (O2) to produce carbon dioxide (CO2) and water (H2O).

The general equation for the combustion of a hydrocarbon is:

CnHm + (n + m/4)O2 -> nCO2 + m/2H2O

In this case, we have hexane (C6H14), so n = 6 and m = 14.

C6H14 + (6 + 14/4)O2 -> 6CO2 + 7H2O

Since combustion reactions typically release heat, we can conclude that step 2 is indeed exothermic.

In step 3, the reaction is the formation of liquid water (H2O) from gaseous water (H2O) along with the remaining carbon dioxide (CO2). This is typically a condensation process, where water vapor converts into a liquid state.

The enthalpy change for the condensation of water is negative, indicating that heat is released when the gas phase changes to the liquid phase. So, step 3 is also exothermic.

Therefore, both step 2 and step 3 in the given process are exothermic reactions.

It is important to note that while step 1 is not explicitly mentioned in your question, the given equation implies that the liquid hexane (C6H14) is being vaporized to form hexane gas (C6H14). This is an endothermic process as it requires energy to break the intermolecular forces holding the hexane molecules together.