Thermodynamic reversibility is only satisfied at thermodynamic equilibrium, and approximately satisfied near equilibrium. Since matter and anti-matter are not in equilibrium the process is not reversible. Annihilation is "reversible" in the weak sense that it satisfy the CPT-symmetry. 

According to Rolf Haase in the case pf small deviations from equilibrium is considered  the terms isentropic and adiabatic  are practically identical  In the closed, homogeneous, isotropic system, in which, according to the assumptions, electrification, magnetization, and dissipative effects (frictions,  current passage, etc) are excluded , an adiabatic-isochoric state change  U (energy) = constant,   V = constant. In the case of an adiabatic -isobaric state change:  H (enthalpy) = constant and  P = constant. 

Actually here implicitly It has assumed that the process or change is reversible since dissipative effects are excluded. There is actually  a connection between the inter entropy production and dissipation of energy such as:    Tdin S /dt  = Zeta > 0  where Zeta is called power dissipation.  din S  > is natural process 'îrreversible'    din  S = 0  equilibrium process which is reversible. 

Since there is no dissipation of energy and the mass is assumed to be a  form of energy than the  process  is   adiabatic, equilibrium process and it is reversible. 

Here we enlarged the concept of energy by including mass as a new form.

Actually thermodynamics claims  that only the natural process takes place spontaneously in nature  in the absence of  any  external manipulations. 

Normally annihilation is not reversible and some entropy is produced. A classic example is electron positron annihilation. Assume that electrons and positrons are sent against each other with energies between 2 and 4 GeV. The annihilation creates myons or hadrons in the proportion 1:2.2 so the output is a mixture altough the input was not. In this case some entropy has been created. 

Only a mixture of high energetic particle may have reversible annihilation. In the very early universe annihilation was reversible. Also the creation and annihilation of virtual particles is reversible, but by defintion these particles cannot be observed.

In the most ideal case the annihilation of matter and antimatter can be isentropic. Consider an electron and positron colliding with negligible kinetic energy. The result will be two 511 KeV gamma rays emitted in opposite directions. Since there are 2 particles in both the initial and final states, there is no increase in entropy, even if the process occurs at room temperature, vastly below its equilibrium temperature. But most annihilations away from thermodynamic equilibrium are not ideal and thus there is an increase in entropy. As Peter Harremoes noted at extremely high temperatures annihilation is in thermodynamic equilibrium with its reverse and hence the system is already at maximum entropy, so the process is then isentropic simply because entropy cannot increase further.

Since the 2 photons are emitted in opposite directions to conserve momentum, in the ideal case there need not be any increase in entropy. With specular reflection mirrors could reflect the photons back at zero thermodynamic cost. By contrast, the expansion of a gas cannot be reversed without thermodynamic cost. The photons are emitted radially outward from the point of annihilation of the electron and positron, and so if specular reflection is possible they can be reflected back for free by a spherical-shell mirror. By contrast, in the expansion of a gas the molecules move in random directions and so cannot be reflected back for free even with specular reflection. Specular reflection may be impossible for 511 KeV gamma rays because the wavelength is smaller than atoms, but it would be possible if the experiment could be modified so that longer-wavelength photons were emitted.

Even if the electron and positron converge from the left they will produce two photons! To see this we notice that annihilation is Lorentz invariant and that the electron and the positron hit each other head on in the center of mass system.

As the previous 2 answers state, even if the electron and positron converge from the left, say +/- 45 degrees from the horizontal in the laboratory frame, they converge head on in the center-of-mass frame. Since there must be 2 photons to conserve momentum in the center-of-mass frame, there must also be 2 photons in any frame: 2 is a pure number (a zero-order tensor) that transforms into itself when changing from one frame to another.

Merely covering a greater distance per unit time does not contribute to entropy if there is no increase in a particle's delocalization. This would be the case in the classical limit, wherein quantum-mechanical uncertainty could be neglected.

Quantum mechanically, the uncertainty principle requires an increase in a particle's delocalization with time, but this depends on how well it is initially localized, not on its speed. Let a particle initially be localized to within (delta Xo). Then the uncertainty in its position at time t is [delta X(t)] = (delta Xo) + t(delta V) = (delta Xo) + t(delta p)/m = (delta Xo) + ht/[m(delta Xo)].

The answers above/ till now are from the point of view of statistical mechanics and where there arise some complications, recourse to quantum mechanics is taken.

Let me attempt an answer from classical thermodynamics point of view. Chemical reactions are processes which transform chemical energy into other forms of energy. If the other forms include, heat, the process is considered irreversible, else reversible. By way of example, one can consider galvanic cell reactions (ideal cases with no transformation of energy into heat) as reversible processes.

If we consider matter-antimatter anihilation process in an isolated system, the process of production of 'r' rays could be considered irreversible, if the rays are considered heat. On the other hand, if the binary collision is considered as an elastic process, then it could be reversed and the initial matter antimatter particles produced. The process then is deemed reversible. 

If covering of greater volume per unit time be a reason of increasing entropy, then it naively appears that in pair production it should decrease !

Because of the speed of the annihilation reaction this is definitely an adiabatic change that means external entropy contribution is zero. Secondly this process takes place without any external manipulations therefore it is  either a natural process or equilibrium or reversible change. In the first place internal entropy production should be positive definite quantity(irreversible change), and in the second case internal entropy change is zero. But according to the postulate advocated by Max Born this can never takes place in nature.

ANSWER:  The change is adiabatic and irreversible change. 

Note: If this would be isothermal process (not isentropic) then there should be definitely energy dissipation in the case of irreversible change since there is linear connection between  the internal entropy production 

and energy dissipation rate.  dSin/dt = -1/T dQ/dt  >0  Power Dissipation (See: Haase)

Entropy: A concept that is not a physical quantity

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

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