Finding the pressure equilibrium constant of the reverse reaction

Question

At $1000~\mathrm{K}$, $K_\mathrm p = 1.85$ for the reaction. $$\ce{SO2 + \frac12O2 <=> SO3}$$

What is the value of $K_p$ for the reaction
$$\ce{SO3 <=> SO2 + \frac12O2}$$

Now, since the task gives us $K_\mathrm p$, it seems strange to ask for it again with the equilibrium put the other way around. Does this make sense? Is it a typo? Either that or my understanding is flawed. It would make more sense to ask for $K_\mathrm c$.

The chapter has this equation: $K_\mathrm p = K_\mathrm c(RT)^{\Delta n}$, I think this is relevant given the temperature is available to solve it.

Source: task 15.21 in the 11th edition of Chemistry: The Central Science, and the answer is suppose to be $0.541$.

Answer 1

They are actually just looking for the equilibrium constant for the reverse reaction.

$$K_p(-\xi) = K_p^{-1}(\xi) = \frac{1}{1.85} = 0.541$$