Question #8349f

1 Answer
Mar 17, 2017

Answer:

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#.