There are times when we need to convert between K_c and K_p.

Consider the following general reaction.

aA + bB ⇄ cC + dD

$\displaystyle K_p\;=\;\frac{P_C^cP_D^d}{P_A^aP_B^b}$
 
Each reactant and product can be written in terms of partial pressures. Starting with reactant A, the partial pressure can be written as:

P_AV = nRT and $\displaystyle P_A\;=\;\frac{nRT}{V}$
 
The quantity n/V is equal to the concentration in molarity. We can write it as [A].

$\displaystyle P_A\;=\;[A]RT$
 
The partial pressures of the other reactant and the product is set up below.

$\displaystyle P_B\;=\;[B]RT\;\;\;\;\;P_C\;=\;[C]RT\;\;\;\;\;P_D\;=\;[D]RT$
 
Plug the partial pressures into the K_p expression.

$\displaystyle K_p\;=\;\frac{([C]RT)^c([D]RT)^d}{([A]RT)^a([B]RT)^b}$
 
Rearrange the equation.

$\displaystyle K_p\;=\;\frac{[C]^c[D]^d}{[A]^a[B]^b}\times (RT)^{(c+d)-(a+b)}$
 
We see that $\displaystyle \frac{[C]^c[D]^d}{[A]^a[B]^b}$ is equal to K_c, and (c + d) – (a + b) is equal to the number of moles of product gas minus the number of reactant moles of gas, Δn.

Δn = (moles of product gases – moles of reactant gases)

The equation can now be written as:
 
$\displaystyle K_p\;=\;K_c(RT)^{Δn}$

Temperature, T, is in Kelvin and R = 0.0821 (atm⋅L)/(mol⋅K). From the equation, K_p is equal to K_c if the same number of moles of gas are on both sides of the chemical equation.

For the reaction, 2 SO₂ (g) + O₂ (g) ⇄ 2 SO₃ (g), Kc = 2.45 x 10² at 300 ^oC. To calculate Kp, first calculate Δn. There are 2 moles of gaseous products and 3 moles of gaseous reactants.

Δn = 2 moles – 3 moles = -1

Next, convert 300^oC to Kelvin.

K = ^oC + 273 = 573 K

$\displaystyle K_p\;=\;K_c(RT)^{Δn}\;=\;2.45 \times 10^2 \times\Biggl(0.0821 \frac{Latm}{molK}\times 573\;K\Biggr)^{-1}\;=\;5.21$

Watch the Following Video

Exercises

Exercise 1. The following reaction has K_c = 1.8 x 10^-8 at 45.6 ^oC. What is K_p?

 

Check Answer/Solution to Exercise 1

Exercise 2. Calculate K_c for the following reaction.

N₂ (g) + 2 H₂ (g) ⇄ N₂H₄ (g)   K_p = 4.09 x 10^-29 at 45.5^oC.

Check Answer/Solution to Exercise 2

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