This is the boss. No calculator, and the paper prints the mental-math forms: at 300 K take $RT = 25\ \mathrm{L \cdot atm/mol}$ for the pressure crossing, $RT = 2500\ \mathrm{J/mol}$ for the free-energy one, and $\ln 10 = 2.3$. Every concentration is pinned so $K$ lands on a power of ten, which is exactly why $\ln K = 2.3 \times (\text{number of decades})$ is enough to finish the problem in your head. Chemists have used that shortcut for a century; today you get to.

One reaction — the contact process, $2\mathrm{SO_2(g)} + \mathrm{O_2(g)} \rightleftharpoons 2\mathrm{SO_3(g)}$ — and every answer feeds the next: build $K_c$ from the settled converter, judge $Q$ on the fresh one, call the shift, cross to $K_p$, let Gibbs put a number on it, and then answer the question the examiner actually asked. Write each line down. One slip in line one rolls all the way to the verdict, and the finish will show you where the cascade started.

5★ under par here is the chapter crown. Clock in.