How can ΔH be a state function and be path-independent, if according to 3. it is numerically equal to a quantity that is, in general, not path-independent?

Before getting into the case of change in enthalpy, I would like to introduce a similar confusion, which is more basic. Consider the first law of thermodynamics for a reversible process:

$$\bar{d}Q_{rev}=dU-PdV$$

where the "bar" sign indicates that it is not exactly differentiable, $dU$ is the change in internal energy and $PdV$ is the work done on the system for an infinitesimal change in volume at a constant pressure.

Internal energy is a state function. If there is no change in heat happened during the process as in the case of an adiabatic one, then we get

$$dU=PdV$$

There comes the confusion. The work done on the system depends on the path traversed by the system (as it is interpreted as the area enclosed by the P-V curve). However, $dU$ does not depend on the path the system followed, but only on the initial and final states. Then how the equality holds? The answer to this question gives the answer to your question.

The answer becomes clear if one solves for the expression of work in an adiabatic process:

$$W_{ad}=\frac{K\left(V_f^{1-\gamma}-V_i^{1-\gamma}\right)}{1-\gamma}$$

where, $\gamma$ and $K$ are constants. As you can see, the work done depends on the initial and final states of the volume. Hence in this case, the work done appears as a state function, due to it's equality with change in internal energy.

Now, to your question.

Enthalpy ($H$) is the measure of total heat content of the system. It is defined as

$$H=U+PV$$

the change in enthalpy is hence given by

$$\Delta H=\Delta U+\Delta (PV)$$

At constant pressure,

$$\Delta H=\Delta U+P\Delta V=\bar{d}Q_{rev}$$

Or, in general

$$\Delta H=\bar{d}Q_{rev}+V\Delta P$$

A close look at the above equations reveals that the enthalpy is a state function and as you can see, the enthalpy change depends on the change in internal energy (which is a state function) and on the change in volume between the initial and final states (again another state function as it is a thermodynamic coordinate). Hence the enthalpy change is a state function. However, heat is not. But it doesn't mean that enthalpy change is not a state function. If that's the case, why should we write them as separate quantities even though the equality holds?

The above equation means that any change in the heat supplied to the system (which is not a state function) will be utilized by the system, under suitable conditions, for the change of state of a system from one to another, both being well-defined states. That is, if you reverse the process, by changing enthalpy into heat, the system will give out the heat by dropping back to the initial state.

If you supply heat, the system stores it in such a way that it is moved to another well defined equilibrium state and leaves no trace on which path the process happened. The enthalpy change is a property of the system and it doesn't care what type of energy you supplied to create that change. The right hand side of the equation, on the other hand, is what you give as an external parameter. They are not the property of the system and hence not, by definition, is a state function.

Conclusion: The equality just means only the numerical equivalence between the two quantities on it's either sides. A quantity is a state function if that quantity represents the energy stored in the system. The external supplies are not properties of the system like work, heat etc.

"[…]Since the temperature change accompanying a chemical reaction does not result from heat transfer via conduction, convection or radiation […]"

I think there's some confusion here. It is confusing to ascribe the temperature change to a flow of heat. The point is, surely, that in exothermic reactions heat escapes from the system of reactants, usually by conduction through the walls of the conducting vessel. The heat flow takes place because, due to the reaction, the temperature of the reactants/products becomes greater than that of the vessel and its surroundings.

In thermodynamics, heat is not a property of a system, but is energy in transit to or from the system. However enthalpy *is* a property of the system. We can equate the enthalpy change to the heat flow only when the pressure is constant.

## Best Answer

It's not entirely wrong to think of enthalpy as the heat content of a system

^{1}, but due to the definition of heat as a certain type of energy transferred between systems, there are subtle differences.Enthalpy is a state function, a priori only well-defined after equilibrium has been established, whereas heat is a characteristic of the specific process $p$ that takes you from initial state $i$ to final state $f$.

Let $$ \Delta H_{i\to f} = H(f) - H(i) $$

In general, we have $$ \Delta H_{i\to f} = Q(p) + \Delta H^\text{uncomp}(p) $$ where we can have an additional change in enthalpy due to entropy production within the system (eg when friction is involved) that is not compensated by entropy loss of the environment due to heat transfer.

For quasistatic processes where state variables are well-defined at all times, we also have $$ \Delta H_{i\to f} = \int_p T \mathrm dS $$

In that case, we may go to an infinitesial description $$ \mathrm dH = T\mathrm dS = \delta Q + T \mathrm \delta S^\text{uncomp} $$ where the $\delta$ forms need only be defined along the specific curve through phase space.

In case of reversible processes, we additionally have $$ \delta S^\text{uncomp} = 0 $$ and thus $$ \mathrm dH = \delta Q \qquad\qquad \Delta H_{i\to f} = Q(p) $$ which was your starting point.

^{1}or rather, as Wikipedia puts it, "the capacity to do non-mechanical work plus the capacity to release heat"