Enthelpy and Heat State Functions

[chem1234]

I'm sure this topic has been asked before so I'd be happy if you'd just direct me to the thread.
I know U+PV=H and if the only work done is PV work then H=q-P∆V+P∆V=q. And I know that pressure must be constant. I also know that a state function is a function where the value of the function only depends on the change between the last and first states. I have 2 questions.
1. If pressure were not constant wouldn't the PV work=0 and enthalpy would still equal q?
2. I understand why enthalpy is a state function. The heat lost or gained only depends on the change between the first and last states. If enthalpy always equals q then why isn't q a state function? For example if 2 reactions happened one released 20 kJ of heat and one absorbed 10 kJ of heat the change in q or the heat lost or gained in the system is -20 kj + 10 kj =10 kJ lost.

I feel like my reasoning is probably off somewhere. Can someone help me please?

 

[juanrga]

I'm sure this topic has been asked before so I'd be happy if you'd just direct me to the thread.
I know U+PV=H and if the only work done is PV work then H=q-P∆V+P∆V=q. And I know that pressure must be constant. I also know that a state function is a function where the value of the function only depends on the change between the last and first states. I have 2 questions.
1. If pressure were not constant wouldn't the PV work=0 and enthalpy would still equal q?
2. I understand why enthalpy is a state function. The heat lost or gained only depends on the change between the first and last states. If enthalpy always equals q then why isn't q a state function? For example if 2 reactions happened one released 20 kJ of heat and one absorbed 10 kJ of heat the change in q or the heat lost or gained in the system is -20 kj + 10 kj =10 kJ lost.

I feel like my reasoning is probably off somewhere. Can someone help me please?

dH = dU + pdV + Vdp

Only mechanical work (dU = δQ - pdV)

dH = δQ + Vdp

If p=constant

dH|_p = δQ|_p