# Relationship between thermodynamic reversibility and reactions

Often we're concerned with changes in thermodynamic properties during reactions, like $\Delta H_p = Q_{reaction}$ (change in enthalpy for an isobaric process). What seems odd to me is that in the derivation of this relationship,

$\Delta H_p = \Delta U + P\Delta V=Q-W+P\Delta V$

we assume the work is given by $W=P\Delta V$, which requires that the process is reversible in the thermodynamic sense. Then the change in enthalpy simplifies to

$\Delta H_p=Q - P\Delta V + P\Delta V=Q$.

Supposedly, measuring the heat transfer during a reaction at constant pressure will give the enthalpy. Even though enthalpy is a state function in general, our case requires that the process is reversible.

To what extent does the restriction of thermodynamic reversibility apply to the reaction? Do we have to assume the reaction happens slowly and can be undone, leaving only an infinitesimal change in the universe? Or does it not matter?

Please correct me if any of my points are wrong. Classical thermodynamics usually doesn't mention kinds of matter, so I'm a bit confused when chemistry comes into it.

• Here one assumes that the changes are quasistatic, so slow that there is at each instant negligible departure from thermodynamic equilibrium within the system. In other words, δW = -PdV where P is pressure and V is volume. - Wikipedia . Read this up for a better understanding of your own question . – Del Pate Mar 7 '15 at 17:26

The process does not have to be considered reversible, and, if a reaction is occurring in the system, it is, of course, not reversible. We are trying to find the change in enthalpy between the following two states:

State 1: Reactants at temperature $T_0$ and pressure $P_0$

State 2: Products at temperature $T_0$ and pressure $P_0$

We would like to measure the $\Delta H$ in such a way that, for the process path we select, all we need to do is measure the amount of heat transferred to the system Q. The process path we select involves keeping the system in contact with a constant temperature bath at $T_0$ and controlling the motion of the portion of the system boundary at which the volume change occurs in such a way that the force per unit area is always constant and equal to $P_0$. Since, in general, the work done by the system on its surroundings is determined by the integral of $dW=P_{ext}dV$ (where $P_{ext}$ is the force per unit area acting on the portion of the boundary at which the volume change is occurring), we have selected $P_{ext}$ such that $P_{ext}=P_0$; and the total cumulative work done by the system when it reaches its final equilibrium state will be just $W=P_0\Delta V$.

From the first law, it then follows that, for this irreversible path: $\Delta U=Q-P_0\Delta V$. So $\Delta U+P_0\Delta V=\Delta H =Q$

Reversible reactions are considered to be ideal, since in our everyday life we don't get to see anything broken come back together by its own. So you have to remember one thing, which I also think it's the answer youre looking for : every reversible reaction we talk about in thermodynamics, happens infinitezimal slowly. For example : Reversible gas compression in constant pressure, happens by removing the heat, lowing the temperature of the ambient in infinitezimal amounts, so that the temperature of the system changes from T1 to T2. So your assumtion was right, infinitezimal is the key word in reversible reactions.

In chemistry dU = dQ + dW ( not dQ = dU + dW) and dW = -Pdv.The general expression for enthalpy is H = U + PV . Enthalpy is the amount of heat content used or released in a system at constant pressure. If there is no non-expansion work on the system and the pressure is still constant, then the change in enthalpy will equal the heat consumed or released by the system (q) i.e. (dH = Q). But,you can calculate change in enthalpy every case throuugh equation (H = U + PV) i.e. dH = dU + d(PV).