Chemical Equilibrium - Le Chatelier's Principle, Change in Volume

Question

Question: Consider the following reaction at equilibrium at a total pressure that we will call $P_1$.

$$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}$$

Suppose the volume of the system is compressed to $\frac 12$ its initial volume and then equilibrium is reestablished. The new equilibrium total pressure will be:

a. Twice $P_1$

b. Three times $P_1$

c. 3.5 times $P_1$

d. less than twice $P_1$

e. unchanged

My thought process: Since I knew the volume had decreased, that meant the pressure had increased and so the reaction will head towards the side with less moles, which would be the products. I thought that since the volume halved, that meant the pressure should increase two times.

However, the answer is D. Could someone please explain this to me?

Answer 1

Expanding on what Buck Thorn said in the comments and providing a more rigorous answer.

The partial pressure would change as follows: $$\begin{array}{c|c|c|c} & \ce{SO2} & \ce{O2} & \ce{SO3} \\ \hline \text{Initial state} & P_A & P_B & P_C \\ \text{On halving volume} & 2P_A & 2P_B & 2P_c\\ \text{Final state} & 2P_A-2x & 2P_B-x & 2P_c + 2x\\ \end{array}$$

where $P_A + P_B + P_C = P_1$

Since initial and final states are in equilibrium, $$ K_p = \frac{P_C^2}{P_A^2 \cdot P_B} = \frac{(2P_C+2x)^2}{(2P_A-2x)^2 \cdot (P_B+x)}$$

We do not need to solve this equation: looking at both the sides, we can see that the value of x would be positive (for a negative x, the numerator would decrease and denominator would comparatively increase due to the $(2P_A - 2x)^2$ term increasing more than the $(P_B + x)$ term due to the square). You could go ahead and solve the equation for a more accurate answer but looking at the options, we can make do with an approximation.

Adding the partial pressures at the final state, we have $$P_2 = 2(P_A + P_B + P_C) - x = 2P_1 - x < 2P_1$$

Which corresponds to option (d)