The change in pressure depends on the design of the reactor, specifically its boundary conditions like pressure and volumetric flow rate at the inlet and outlet. Without any kind of compressor at the inlet, a reaction like the one you described will generally result in the reactor operating at the same pressure as the feed, but with the exit velocity and volumetric flow rate of the outlet stream being larger/faster than the feed. If the reactants and products are all in the gas phase and are ideal gasses, and the temperature is held constant, you can calculate the change in inlet flow rate to outlet flow rate (volumetric) with this equation:
$$
\Delta \dot V = \xi \left( \nu_p - \nu_r \right)
$$
Here, $\xi$ is the extent of reaction, and $\nu_p$, $\nu_r$ are the total stoichiometric numbers of the products and reactants. Note that if $\nu_p=\nu_r$, the reaction is equimolar and the volumetric flow rate entering the reactor is the same as the flow rate exiting the reactor (which is intuitive).
To better explain why the velocity increases while the pressure stays more or less the same, consider that you always need a negative pressure gradient through the reactor (assuming steady-state operation) so that the fluid will always flow from inlet to outlet. That is, the outlet pressure, $P_{out}$, must always be less than the inlet pressure $P_{in}$.
Doing a mass balance on the reactor shows that the mass flow rate coming out must be exactly equal to the inlet flow rate (no gains/losses):
$$
\dot m = \rho_{in} u_{in} A = \rho_{out} u_{out} A
$$
where $\rho$ is density, $u$ is velocity, and $A$ is the cross-sectional area of the reactor (constant). We also have the ideal gas equation of state (assuming the reactants and products are in-fact ideal gasses):
$$P=\rho RT = \rho \frac{\bar R}{MW} T$$
where $P$ is pressure, $R$ is specific gas constant, $\bar R$ is universal gas constant, $MW$ is the average molecular weight, and $T$ is temperature. We can solve the equation of state for $\rho$ and plug it in to the mass balance:
$$P_{in} \, MW_{in} \, u_{in} = P_{out} \, MW_{out} \, u_{out} $$
where $T$, $A$, and $R$ have canceled out. If we assume that the pressure gradient is very small, $P_{in} \approx P_{out}$ (just enough to get a little bit of flow), then we can also cancel the pressure terms and rearrange:
$$\frac{MW_{in}}{MW_{out}} = \frac{u_{out}}{u_{in}}$$
Now it should be more apparant that, if the outlet average molecular weight is less than the inlet (i.e. products have a higher stoichiometric number than the reactants), then the outlet velocity must be higher to satisfy our mass balance and equation of state.