# Confusion regarding enthalpy and heat

I'm having difficulty resolving three statements from my thermodynamics class.

1) The enthalpy difference, ΔH, between two states of a system can only take one value, since H is a state function.

2) The heat exchanged when taking the system between two states depends on the path, e.g. a reversible transformation will lead to a different quantity of heat being absorbed than an irreversible one, even if the initial and final states are the same.

3) The enthalpy change is equal to the heat exchanged in a transformation at constant pressure.

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?

• I think may be because its just a state function so it only depends on the initial and final state and not the path through which reaction proceeds – Archis Welankar Feb 6 '17 at 6:00

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.

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.

• The equation you gave for the adiabatic work applies only to an adiabatic reversible process. – Chet Miller Feb 6 '17 at 20:26
• Yes, it's natural, together with the second part of the answer and conclusion provides a more general answer. I do not want to elaborate more on the first part because it's not actually the OP's question. – UKH Feb 7 '17 at 0:16
• My point was that, to avoid confusion, you should edit the answer to include the word reversible. – Chet Miller Feb 7 '17 at 1:19
• Its given in the first paragraph itself. Please see the section where I stated First law of thermodynamics, from which the expression for adiabatic process is followed. – UKH Feb 7 '17 at 11:40
• @UKH Thank you. Yes, that makes sense to me. I think the overall takeaway is that if we identify the "heat delivered reversibly at constant pressure", we are actually specifying the path being taken (a constant-pressure path). So while the amount of heat delivered between two states in general can be anything, the amount of heat delivered "reversibly along a constant-pressure path between two states is a single unique value. – Rogworth Feb 8 '17 at 2:26

There are many paths that can take the system from the initial to the final state in case 3. One of these is a constant pressure path. Others will involve more or less heat than the constant pressure path, but all of them will have the same change in H.

• +1 Suggested improvement: "Others will involve more or less heat than the constant pressure path, but all of them will have the same change in H" but then $\Delta H$ is not equal to heat exchanged in those cases. – Deep Feb 7 '17 at 3:56
• But here's the thing: when we do calorimetry experiments, we basically allow a chemical system to go between two states irreversibly at constant pressure and then equate the resulting heat exchange to the ΔH for the system. So we are saying that Q(irrev, constant P) = ΔH. But in theoretical calculations and textbooks, we calculate the heat exchanged reversibly to obtain the ΔH, and say Q(rev, constant P) = ΔH. But Q(rev) does NOT equal Q(irrev) in general, so the two ΔH values obtained are inconsistent when they shouldn't be. What am I missing? – Rogworth Feb 8 '17 at 2:36
• If the system is in contact with a constant temperature reservoir (to a first approximation the calorimeter, except for the small sensible heat change of the reaction mixture) and the calorimeter is operating at constant pressure (i.e., $P_{ext}$ equal to the initial and final pressures of the reaction mixture), then it doesn't matter whether the reaction was allowed to proceed spontaneously or whether it was carried out reversibly (say using an Van't hoff equilibrium box) between the initial and final states. $Delta H$ and Q will be the same in both cases. – Chet Miller Feb 8 '17 at 14:44

The enthalpy considering the mechanical energy one can extract from a system. It is define regarding the work done in a constant pressure transformation. If you define $Q$ the heat and $W$ the work, and $E$ the internal energy of the system you have: $$dE=\partial Q+\partial W$$

As such, at constant pressure between two states $A$ and $B$ the work is equal to:$$\Delta W = -P\Delta V=-P(V_B-V_A)$$ You end up with the relation:$$\Delta U = \partial Q -P\Delta V$$ Rewritten as: $$(U_B+PV_B)-(U_A+PV_A)=\partial Q$$ You can then define you state function $H=U+PV$ which is the enthalpy of your system. You can now easily see that $\Delta H=\partial Q$ at constant pressure. The function in itself is path independant as it is defined for all states at given entropy and pressure (you have at least one extensive variable so don't worry).

Its differential form is such that: $$dH=TdS+VdP$$ So you can clearly see that for a constant pressure transformation $dP=0$. In this case $dH = \partial Q$.