Pressure before equilibrium = Pressure during equilibrium?

Question

For example, let's say we have the equation $$\ce{PCl_3(g) + Cl_2(g) -> PCl_5(g)},$$ and the temperature is held constant.

Would the pressure in the container when equilibrium is reached be greater than, less than, or equal to the pressure when the $\ce{PCl3}$ and the $\ce{Cl2}$ are initially mixed?

I have a few ideas, but they all come up with different conclusions.

1. The pressure increases. This is because the volume is decreasing because of less molecules when equilibrium is established and by Boyle's law, the pressure increases. However, container has a constant volume, so I don't think this is the correct answer.
2. The pressure decreases. This is because less molecules present implies less molecules hitting off the walls of the container.
3. The pressure remains the same. This is because even though less molecules implies less pressure, they will hit the walls of the container with more force. I'm speculating that more force will cancel out the fewer molecules hitting the walls of the container.

Answer 1

Consider $\pu{1 mol}$ of $\ce{PCl3}$ and $\pu{1 mol}$ of $\ce{Cl2}$ at the beginning for the above reaction. Let at equilibrium $x \text{ mol} $ of each reactants get converted to products. Then $\ce{PCl5}$ is $x \text{ mol}$ and $\ce{PCl3}$ is $1-x\text{ mol}$ and $\ce{Cl2}$ is $1-x\text{ mol}$.

$$\begin{array}{ccccc} \ce{PCl3} & + & \ce{Cl2} & \ce{->} & \ce{PCl5} \\ 1 & & 1 & & 0 \\ 1-x & & 1-x & & x \\ \end{array}$$

This means at equilibrium, total number of moles $= 1-x+1-x+x=2-x\text{ mol}$. Initially 2 moles were there and now 2-x. Practically pressure must decrease as all the reactants and products are gaseous.