Question

Question #8349f

Answer 1

Yes, of course.

Explanation:

As you know, `K_p` depends on the equilibrium partial pressures exerted by the gaseous chemical species that take part in an equilibrium reaction.

So, for example, the following equilibrium reaction

`2"SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO"_ (3(g))`

will have

`K_p = ( ("SO"_3)^2)/( ("SO"_2)^2 * ("O"_2))`

If you use atmospheres to express the equilibrium partial pressure of the three gases, you will have, using only units

`K_p = color(red)(cancel(color(black)("atm"^2)))/(color(red)(cancel(color(black)("atm"^2))) * "atm") = "atm"^(-1)`

As you can see, `"atm"^(-1)` can indeed be a unit for `K_p`.