# Question 8349f

Mar 17, 2017

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

$2 {\text{SO"_ (2(g)) + "O"_ (2(g)) rightleftharpoons 2"SO}}_{3 \left(g\right)}$

will have

${K}_{p} = \left(\left({\text{SO"_3)^2)/( ("SO"_2)^2 * ("O}}_{2}\right)\right)$

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, ${\text{atm}}^{- 1}$ can indeed be a unit for ${K}_{p}$.