Identify the gaseous species in the reaction.

The given reaction is:

When considering the effect of pressure on equilibrium, we only look at the gaseous reactants and products. Solids ($\text{A}_{(s)}$) and aqueous solutions ($\text{C}_{(aq)}$) are not significantly affected by changes in pressure.

• Reactant side: We have $2\text{B}_{(g)}$, which means there are 2 moles of gas.
• Product side: We have $\text{D}_{(g)}$, which means there is 1 mole of gas.

Le Chatelier's Principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

• A decrease in pressure favors the side of the reaction with a greater number of moles of gas.
• An increase in pressure favors the side of the reaction with a fewer number of moles of gas.

On the reactant side, there are 2 moles of gas.
On the product side, there is 1 mole of gas.
Since the reactant side has more moles of gas (2 moles > 1 mole), a decrease in pressure will cause the equilibrium to shift towards the reactant side. This means the backward reaction will be favored.

• A. rate of forward reaction will increase: Incorrect. The equilibrium shifts to the left, favoring the backward reaction.
• B. rates of forward and backward reaction are not affected: Incorrect. Pressure changes affect equilibria involving gases.
• C. rate of backward reaction will increase: Correct. When the equilibrium shifts to the left, the rate of the backward reaction increases relative to the forward reaction to re-establish equilibrium.
• D. the equilibrium will shift: While true, option C provides a more specific and accurate description of how the equilibrium shifts in terms of reaction rates.

The final answer is $\boxed{\text{C}}$