Which Methanol State to Use for Heat of Reaction Calculations?

[gfd43tg]

Hello,

I am confused about the following scenario.

Suppose I have the reaction

CO (g) + 2H2 (g) → CH3OH (g)

and I am asked to solve for the heat of reaction at 800 C. My query is the following: when finding the heat of reaction at standard state, which heat of formation I should use for the methanol? The heat of formation for gaseous or for liquid methanol?

The reason I may argue for liquid is because at standard state, methanol exists as a liquid. On the other hand, the reaction gives it as a gas. Which one do I choose and why? What's the rule about this sort of situation?

 

[Adithyan]

Heat of formation of gaseous methanol must be chosen. The temperature is given as 800 degrees. You should only look at the final state of the product molecule and not bother about what its state is at the standard conditions. And your question's answer is as follows:, liquid methanol is converted to gaseous (in the intermediate stage)- so a certain heat is required for a change of state. So the final heat required for this reaction is more than that of a reaction yielding liquid methanol.