Question

In the reaction N2O4 => 2 NO2 (g), equilibrium is reached at a temperature at which P NO2 = 3 P (NO4) ^ 1/2. What must be the value of Kp at this temperature?

Answer 1

The equation for the Kp is the multiplication of the products exponentiated by its stoichiometric coefficients divided by the same for the reactants.

In this case, we have a single product, NO₂ (g), with stoichiometry 2, and a single reactant, N₂O₄ (g), with stoichiometry 1. So, the equation for Kp, is:

`K_p=\frac{(p_{NO_(2)})^2}{(p_{N_(2)O_(4)})^1}`

Since we know that the equilibrium was reached where:

`p_(NO_2)=3(p_(N_2O_4))^{(1)/(2)}`

So, we can substitute that into the Kp equation to get:

`K_p=((p_(NO_2))^2)/((p_(N_2O_4))^1)=\frac{(3(p_(N_2O_4))^{(1)/(2)}_{})^2}{p_(N_2O_4)^{}}=\frac{3^2(p_(N_2O_4))^{(1)/(2)\cdot2}}{p_(N_2O_4)^{}}=(9p_(N_2O_4))/(p_(N_2O_4))=9`

So, the Kp for this temperature must be 9.