Relationship between equilibrium constants Kc and Kp

Many exercises on Chemical Equilibrium content include calculations that involve the relationship between the equilibrium constants K_ç (in terms of concentration) and K_P (in terms of gas pressures). If there is any doubt about what these constants represent and how their expressions are written for each equilibrium reaction, read the text Kc and Kp equilibrium constants.

The relationship between these constants is established by the following formulas:

K_ç = K_P. (R. T)ⁿ and K_P = K_ç. (R. T)^-n

But how were these formulas arrived at?

Well, let's consider the following generic reaction where the lowercase letters are the coefficients of the equation and the uppercase letters are the substances (reagents and products), all of which are gaseous:

a A + b B ↔ c C + d D

For such a reaction, the expressions of the equilibrium constants Kc and Kp are, respectively, given by:

K_ç = [Ç]^ç. [D]^d K_P = (Praça)^ç. (pD)^d
[THE]^The. [B]^B (pA)^The. (pB)^B

So let's use the Clapeyron equation or gas state equation:

P. V = n. A. T

p = no. A. T
V

The concentration in quantity of matter (in mol/L) of the substances can be calculated by n/V. So, we can make the following substitution in the formula above:

p = [substance]. A. T

Using this formula for each of the reactants and products of the reaction in question, we have:

P_THE = [A]. A. T p_B = [B]. A. T p_Ç = [C]. A. T p_D = [D]. A. T
[A] = __P_THE_[B] = __P_B_[C] = __P_Ç_[D] = __P_D_
A. T R. T R. T R. T

Thus, we can substitute these concentrations in the Kc expression shown above:

But as we've seen, (Praça)^ç. (pD)^d is exactly the same as Kp. Therefore, we have:
(pA)^The. (pB)^B

K_ç = K_P. (R. T)^{(a + b) – (c +d)}

Note that (a + b) – (c +d) is the same as: “sum of the coefficients of the reactants – sum of the coefficients of the products”. So, we can simplify even more like this:

(a + b) – (c +d) = ∆n

So, we come to the formulas that relate Kc and Kp:

K_ç = K_P. (R. T)^∆no or K_P = K_ç. (R. T)^-∆no

Let's look at some chemical equilibrium reactions and how to determine these expressions for them.

Important note:∆n involves only the coefficients of substances that are in the gaseous state.

N_2(g) + 3 H_2(g) ↔ 2 NH_3(g)
K_ç = K_P. (R. T)^{(4 – 2)}
K_ç = K_P. (R. T)²

3 O_3(g) ↔ 2 O_2(g)
K_ç = K_P. (R. T)^{(3 - 2)}
K_ç = K_P. (R. T)¹
K_ç = K_P. A. T

H_2(g) + I_2(g) ↔ 2 HI_(g)
K_ç = K_P. (R. T)^{(2 – 2)}
K_ç = K_P. (R. T)⁰
K_ç = K_P

CO_(g) + NO_2(g) ↔ CO_2(g)+ NO_(g)
K_ç = K_P. (R. T)^{(2 – 2)}
K_ç = K_P. (R. T)⁰
K_ç = K_P

2 SO_3(g) ↔ 2 SO_2(g) + O_2(g)
K_ç = K_P. (R. T)^{(2 – 3)}
K_ç = K_P. (R. T)^-1

2 NO_2(g) ↔ N₂O_4(g)
K_ç = K_P. (R. T)^{(2 – 1)}
K_ç = K_P. (R. T)¹
K_ç = K_P. A. T

HCl_(here) + AgNO_3(aq) ↔ AgCl_(s) + HNO_3(aq)
Kc = not defined – has no gases.

Ç_(s) + O_2(g) ↔ CO_2(g)
K_ç = K_P. (R. T)^{(1- 1 )}
K_ç = K_P. (R. T)⁰
K_ç = K_P

Note that, in this case, the coefficient of C_(s) did not participate.

By Jennifer Fogaça
Graduated in Chemistry

Source: Brazil School - https://brasilescola.uol.com.br/quimica/relacao-entre-constantes-equilibrio-kc-kp.htm