Changes in entropy of the universe in a reversible reaction
First, a defintion of terms:  The meaning of reversible in "reversible reaction" is different from the thermodynamic meaning of reversible.  A reversible reaction is one that is kinetically reversible.  I.e., one where you can adjust the relative concentration of products and reactants back and forth as you change conditions.   By contrast, an example of an irreversible reaction would be cooking an egg, which causes the proteins to denature.  The proteins can't be readily undenatured, even if you change the conditions to thermodynamically favor denaturation because, kinetically, the proteins can't find their way back to their undenatured state.

Anyways, let's suppose we have a "reversible reaction", $$\ce{A &lt;=&gt; B},$$  in which the equilbrium constant is, under certain conditions (e.g., at a certain temperature and pressure):

$$K_{eq} = \frac{[B]}{[A]} = 10,$$ i.e., at equilibrium, we will have 10x as much B as A.

This represents the state at which the entropy of the universe is a maximum under those conditions.  Thus if, under those conditions, $\frac{[B]}{[A]} &lt;10$, the reaction will proceed in the forward direction until it reaches equilibrium, increasing the entropy of the universe in the process.  Likewise if, under those condtions, $\frac{[B]}{[A]} &gt; 10$, the reaction will proceed in the reverse direction until it reaches equilibrium, also increasing the entropy of the universe in the process.

So it is not the case that if, say, $\frac{[B]}{[A]} = 9$, then the entropy of the universe would increase in both directions. Rather, it would increase in the forward direction (until it reached equilibrium), and (theoretically) decrease in the reverse direction.  If we instead had $\frac{[B]}{[A]} = 11$, the opposite would happen: decreased entropy of the universe in the forward direction, increased entropy of the universe in the reverse direction (again, until it reached equilibrium).