Step 1: Identify the gaseous species in the reaction. The given reaction is: $$\text{A}_{(s)} + 2\text{B}_{(g)} \rightleftharpoons 2\text{C}_{(aq)} + \text{D}_{(g)}$$ 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. Step 2: Count the total moles of gas on each side of the reaction. 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. Step 3: Apply Le Chatelier's Principle. 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. Step 4: Determine the direction of the shift. 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. Step 5: Evaluate the given options. 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 $\boxed{\text{C}}$