# What is Kp? What goes on in a Kp equation?

Jun 18, 2018

${K}_{P}$ is the equilibrium constant conventionally used to describe gas-phase reactions in which both products and reactants are generated simultaneously.

Conventionally, it is reported in implied units of $a t m$.

It is used in a mass action expression to determine equilibrium partial pressures. Here is a nice example...

$1 \text{NH"_4"Cl"(s) rightleftharpoons color(red)(1)"NH"_3(g) + color(red)(1)"HCl} \left(g\right)$

$\text{I"" "-" "" "" "" "" "0" "" "" } 0$
$\text{C"" "-" "" "" "" "+P_i" } + {P}_{i}$
$\text{E"" "-" "" "" "" "color(white)(//.)P_i" "" } {P}_{i}$

In this case, we see a solid goes into the expression as $\boldsymbol{1}$, because its standard concentration is equal to its molar density. (It has nothing to do with coefficients.)

I have purposefully highlighted the coefficients, as they become the exponent for gaseous reactants.

The mass action expression is therefore:

${K}_{P} = \frac{{\left({P}_{N {H}_{3}}\right)}^{\textcolor{red}{1}} {\left({P}_{H C l}\right)}^{\textcolor{red}{1}}}{1}$

But since the coefficients for both products are the same, ${P}_{N {H}_{3}} = {P}_{H C l}$. Thus, both exponents are the same as well.

As a result, if the ${K}_{P}$ at a certain temperature is $6.25$, then

$6.25 = {P}_{N {H}_{3}} {P}_{H C l} = {P}_{i}^{2}$

So, the partial pressures at equilibrium are each:

$\implies {P}_{i} = {P}_{N {H}_{3}} = {P}_{H C l} = \sqrt{6.25} = \underline{\text{2.5 atm}}$

CHALLENGE: What is the total pressure at equilibrium? HINT: What is Dalton's law?