A mixture of $NO_2(g)$ and $N_2O_4(g)$ is introduced into a container.
Equilibrium is established:
$N_2O_4(g)$ $2NO_2(g)$, with $K_p=4,90\;atm$
At equilibrium the mixture has a total pressure of $1,45\;atm$ .
Calculate the partial pressures of the two gases.
$K_p$ $=$ $\frac{p_{N_2O_4}}{p_{NO_2}^2}$
Let $x$ be the partial pressure of $NO_2$.
Then the partial pressure of $N_2O_4$ will be $1,45$ $-$ $x$ and we have:
$4,90$ $=$ $\frac{1,45-x}{x^2}$
$x$ $=$ $0,45$ atm or $x$ $=$ $-0,66\; atm $
(to be excluded, pressure is always positive)
$p_{NO_2}$ $=$ $0,45\; atm$ ;
$p_{N_2O_4}$ $ =$ $1,45$ $-$ $0,45$ $=$ $1,00\;atm$
$2NO_2(g)$, with $K_p=4,90\;atm$
At equilibrium the mixture has a total pressure of $1,45\;atm$ .
Calculate the partial pressures of the two gases.