Question

For the dissociation reaction `N_2O_4(g)rightleftharpoons2NO_2(g)`, the degree of dissociation `(alpha)` in terms of `K_p` and equilibrium total pressure P is?

Answer 1

Assuming no `"NO"_2` existed prior, and defining `P` as the EQUILIBRIUM total pressure,

`alpha = sqrt(K_p/(K_p + 4P))`

For example, if `P = "1 atm"`, and `K_p = 126`, then `alpha = 0.9845`.

That makes sense because `126` indicates the equilibrium is very skewed towards the products.

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For this reaction, assuming no `"NO"_2(g)` previously was in the container, the ICE table would simply be:

`"N"_2"O"_4(g) " "rightleftharpoons" " 2"NO"_2(g)`,

`"I"" "P_(N_2O_4,i)" "" "" "" "0`
`"C"" "-x" "" "" "" "+2x`
`"E"" "P_(N_2O_4,i)-x" "" "2x`

where `P_(N_2O_4,i)` indicates the initial partial pressure of `"N"_2"O"_4(g)`. `x` is the extent of reaction.

This can be rewritten in terms of the fraction of dissociation `alpha`, which is the extent of dissociation in a relative representation. That is, `alphaP_(N_2O_4,i)` is the dissociated fraction of `"N"_2"O"_4`.

Thus, instead of `P_(N_2O_4,i) - x`, we write:

`P_(N_2O_4,i) - x = (1 - alpha)P_(N_2O_4,i)`

Similarly,

`2x = 2alphaP_(N_2O_4,i)`

The equilibrium mass action expression can then be written in terms of partial pressures:

`K_p = P_(NO_2)^2/(P_(N_2O_4))`

`= (2alphaP_(N_2O_4,i))^2/((1 - alpha)P_(N_2O_4,i))`

`= ((4alpha^2)/(1 - alpha)) P_(N_2O_4,i)`

Now, the partial pressure adds up to give the total pressure.

`P_(N_2O_4,eq) + P_(NO_2,eq) = P`

`= (1 - alpha)P_(N_2O_4,i) + 2alphaP_(N_2O_4,i)`

`= (1 + alpha)P_(N_2O_4,i)`

So,

`P_(N_2O_4,i) = P/(1+alpha)`

and we proceed to get:

`K_p = ((4alpha^2)/(1 - alpha)) P/(1 + alpha)`

`= ((4alpha^2)/(1 - alpha^2))P`

Now we are asked to solve for `alpha`.

`K_p(1 - alpha^2) = 4alpha^2P`

`K_p - K_palpha^2 = 4alpha^2P`

Rearrange to solve for `alpha`:

`K_p = K_palpha^2 + 4Palpha^2`

`= (K_p + 4P)alpha^2`

Therefore, knowing that `alpha > 0`,

`color(blue)(alpha = sqrt(K_p/(K_p + 4P)))`