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Posted: Fri Jan 12, 2018 7:02 pm
Professor Lavelle mentioned that Δ H=q at constant pressure but how could that be true if Δ H is a state property while q (heat) is not a state property?
Re: Δ H=q
Posted: Fri Jan 12, 2018 8:00 pm
You can use Dr. Lavelle's analogy of a person climbing a mountain. The change in altitude is a state function but the actual distance traveled is not a state function because it depends on what path the person takes (the climber could have taken a shortcut or a longer, more scenic route). The change in altitude would be delta H while the distance traveled on foot is q. The fact that q is not a state function makes sense if you think of q as the "path" that heat travels to and from substances.
Re: Δ H=q
Posted: Fri Jan 12, 2018 9:10 pm
Another way to think about state properties is to think about them as properties that belong to a system. For example, internal energy "belongs" to a system and so it is a state property. In contrast, heat is not a state property because it is simply energy that is transferred in and out of a system, and after that transfer, it does not define the system in any way (we don't look at the heat of a system, we look at its internal energy or enthalpy).
Now, for your question: regardless of the path the transfer of heat takes, enthalpy is a state function, so it does not depend on the path the heat takes to transfer to the system. Rather, the change in enthalpy only depends on initial and values. Therefore, at constant pressure, enthalpy will be equal to heat.
Re: Δ H=q
Posted: Sat Jan 13, 2018 6:31 pm
You are right to determine that enthalpy is a state function while heat is not. The only reason that they are equivalent in the equation (Δ H=q) is because it is at a constant pressure, mathematically leading allowing for the equation to be derived. Just because they are different types of functions does not necessarily mean that they are unequal.