Question

How to do this equilibrium derivation? (I'll post an answer showing the problem)

Answer 1

`N_2O_4(g) <=> 2NO_2(g)`
Assuming that `n_0` is the beginning number of moles of `N_2O_4` and we start with `0` moles of `NO_2`, show that `xi_(eq)`, the extent of reaction, can be expressed as:

`xi_(eq)/n_0 = (K_p/(K_p + 4P))^(1/2)`

where `K_p` is the equilibrium constant for partial pressures, and `P` is the total pressure.

What I tried was:

`K_p = (P_(NO_2))^2/(P_(N_2O_4)) = [((2xi_(eq))P)/(1-xi_(eq))]^2/[((1-xi_(eq))P)/(1-xi_(eq))] = [(2xi_(eq))^2P]/[1-xi_(eq)]`
`K_p - K_pxi_(eq) = 4Pxi_(eq)^2`
`4Pxi_(eq)^2 + K_pxi_(eq) - K_p = 0`

But when I try the quadratic formula on that, I get:

`xi_(eq) = (-K_p pm sqrt(K_p^2-4(4P)(-K_p)))/(8P)`
= `-K_p pm sqrt(K_p^2+16PK_p)/(8P)`

Answer 2

Here's how you'd go about solving this problem.

Start with the ICE table for the equilibrium reaction

`" "N_2O_(4(g)) rightleftharpoons 2NO_(2(g))`
I..........`n_0`..................0
C.......`-epsilon`..............`+2epsilon`
E...`n_0-epsilon`..............`2epsilon`

Express the mole fraction of the two species like this

`chi_(N_2O_4) = (n_0 - epsilon)/(n_0 - epsilon + 2epsilon) = (n_0 - epsilon)/(n_0 + epsilon)`

`chi_(NO_2) = (2epsilon)/(n_0 + epsilon)`

Now just use the definition of `K_p` to solve for `epsilon`

`K_p = P_(NO_2)^(2)/P_(N_2O_4) = ((2epsilon)^(2)/(n_0 + epsilon)^(2) * P^(2))/((n_0 - epsilon)/(n_0 + epsilon) * P)`

`K_p = ((2epsilon)^(2) * P^(cancel(2)))/((n_0 + epsilon)^(cancel(2))) * cancel(n_o + epsilon)/((n_0 - epsilon) * cancel(P))`

`K_p = (4epsilon^(2) * P)/((n_o - epsilon)(n_o + epsilon)) = (4epsilon^(2) * P)/(n_0^(2) - epsilon^(2))`

`K_p(n_0^(2) - epsilon^(2)) = 4epsilon^(2) P`

`4epsilon^(2)P + epsilon^(2)K_p - n_0^(2)K_p = 0`

`epsilon^(2)(4P + K_p) = n_0^(2) K_p | : n_0^(2)`

`epsilon^(2)/n_0^(2) = K_p/((4P + K_p)) |sqrt`

`color(green)(epsilon/n_0 = (K_p/(4P + K_p))^(1/2))`