# Question #82dc7

##### 1 Answer

Here's what I got.

#### Explanation:

The equilibrium reaction given to you looks like this

#color(blue)(2)"NH"_ (3(g)) rightleftharpoons "N"_ (2(g)) + color(red)(3)"H"_ (2(g))#

Now, the equilibrium constant for this reaction can be expressed using **equilibrium concentrations**, in which case you have **equilibrium partial pressures**, in which case you have

For

In your case, you have

#K_c = (["N"_ 2] * ["H"_2]^color(red)(3))/(["NH"_3]^color(blue)(2))#

For

In your case, you have

#K_p = (("N"_2) * ("H"_2)^color(red)(3))/(("NH"_3)^color(blue)(2))#

Notice that *concentration* is expressed using **brackets** and *partial pressure* is expressed using **parentheses**.