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

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?

• Immediately after you reduce the volume the pressure is doubled. But the forward reaction then leads to a reduction in amount of gas, and therefore in pressure (at constant V and T). Feb 10, 2020 at 8:09

1 Answer

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)