Question

Consider the equilibrium, N2(g) + 2O2(g) <----> N2O4(g). Calculate the equilibrium constant, Kpif the equilibrium partial pressures are 0.431 atm for N2O4, 0.0867 atm for N2 and 0.00868 atm for O2

1.52 x 10-5

6.60 x 104

1.75 x 10-3

573

Answer 1

Firstly, we'll consider the following reaction:

$aA_{(g)}+bB_{(g)} \longleftrightarrow cC_{(g)}+dD_{(g)}$

The equlibrium constant (Kp) is calculated using the formula below:

$K_p = \dfrac{(p_C)^c\cdot (p_D)^d}{(p_A)^a\cdot(p_B)^b}$

Thus, if we use that  in the given chemical reaction:

$N_2_{(g)}+2O_2_{(g)} \longleftrightarrow N_2O_4_{(g)}\\\\ K_p = \dfrac{(p_C)^c}{(p_A)^a\cdot(p_B)^b}\\\\ K_p = \dfrac{(p_{N_2O_4})^1}{(p_{N_2})^1\cdot(p_{O_2})^2}\\\\ K_p = \dfrac{p_{N_2O_4}}{p_{N_2}\cdot(p_{O_2})^2}\\\\ K_p = \dfrac{0.431}{0.0867\cdot(0.00868)^2}\\\\ K_p \approx 65980.97\\\\ \boxed{K_p\approx 6.6\times 10^4}$

So, the approximate answer is 6.60 × 10⁴.