Rates of Reaction: Question 7

Worked solution

Step 1: Recall the factors that affect rate

Five factors change the rate of a reaction: concentration of solutions, pressure of gases, surface area of solids, temperature, and the presence of a catalyst. For a reaction between gases, pressure has the same kind of effect that concentration has for reactions in solution.

Step 2: Check each option

• Option A (reducing the volume, increasing pressure): squeezing the same number of gas particles into a smaller volume increases the number of particles per unit volume, so $\text{SO}_2$ and $\text{O}_2$ particles collide more frequently. This option would increase the rate.
• Option B (increasing the volume, reducing pressure): spreading the same number of particles over a larger volume reduces the number of particles per unit volume, so collisions happen less often. This option would decrease the rate, not increase it.
• Option C (lowering the temperature): a lower temperature always gives particles less kinetic energy on average, so they collide less often and fewer collisions have enough energy to succeed. This option would decrease the rate.
• Option D (larger catalyst pieces, same total mass): larger pieces have a smaller total surface area than the same mass split into small pellets, so less of the catalyst’s surface is available to the reacting gases at any moment. This option would decrease (not increase) how effectively the catalyst speeds up the reaction.

Step 3: Select the answer

Only reducing the volume of the vessel, which raises the pressure of the gas mixture, increases the rate of this reaction.

Final answers

• $\boxed{\text{option A}}$. Increasing the pressure by reducing the volume of the reaction vessel increases the rate, since the gas particles are packed more closely together and collide more often.