Help understanding the pressure's effect on equilibrium and Le Chatelier's principle
What the textbook means when it is saying one occupies more volume than the other is that the reactants (in this case) will occupy less volume than the product under standard conditions (~ atmospheric pressure). Therefore you have to use more force (pressure) to confine the products to the given volume than you need to do so for the reactants.
Therefore there is less enforcement on the reactants than there is on the products. Therefore it is energetically more favorable to be in the reactants-form.  

In order to understand this a little better you can have a look at the ideal gas law:
$$
p = \frac{nRT}{V}
$$
As you can clearly see the pressure is linearly dependent on the amount of particles in the system. Given that during the reaction (in the way you stated it in your question) this amount increases (doubles if the reaction did run through completely). Thus the pressure increases as well. However a higher pressure is connected to a higher energy state of the particles in the "box" as they repulse one another.
Why don't the reactants stay as they are then as this is the state that produce the least pressure? Because by forming the products, energy is released ($\Delta_rG &lt; 0$). Therefore the reaction can take place until the increase in pressure results in a "raise of energy" that is equal to the energy released during the reaction. Any further reaction would be energetically unfavorable.
And this point is the one that can be influenced by the volume. Let's say that the reaction can take place until the pressure reaches $1.5p_0$ and $p_0 = \frac{n_0RT}{V_0}$ (and for simplicity let's assume a isothermal reaction). This means that 50% of the the reactants can react until that limit is reached. If we increase the volume, this point is reached only after having a higher particle count than with a lower volume which is why the equilibrium shifts to the side with more gaseous particles if the volume is increased and vice versa.