Question

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

Answer 1

`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 `atm`.

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

`1"NH"_4"Cl"(s) rightleftharpoons color(red)(1)"NH"_3(g) + color(red)(1)"HCl"(g)`

`"I"" "-" "" "" "" "" "0" "" "" "0`
`"C"" "-" "" "" "" "+P_i" "+P_i`
`"E"" "-" "" "" "" "color(white)(//.)P_i" "" "P_i`

In this case, we see a solid goes into the expression as `bb1`, 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 = ((P_(NH_3))^color(red)(1)(P_(HCl))^color(red)(1))/1`

But since the coefficients for both products are the same, `P_(NH_3) = P_(HCl)`. 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_(NH_3)P_(HCl) = P_i^2`

So, the partial pressures at equilibrium are each:

`=> P_i = P_(NH_3) = P_(HCl) = sqrt(6.25) = ul"2.5 atm"`

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