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 = ((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 = ((p_C)^c)/((p_A)^a\cdot(p_B)^b)\\\\ K_p = ((p_(N_2O_4))^1)/((p_(N_2))^1\cdot(p_(O_2))^2)\\\\ K_p = (p_(N_2O_4))/(p_(N_2)\cdot(p_(O_2))^2)\\\\ K_p = (0.431)/(0.0867\cdot(0.00868)^2)\\\\ K_p \approx 65980.97\\\\ \boxed{K_p\approx 6.6* 10^4}`

So, the approximate answer is 6.60 × 10⁴.