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?

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.

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)))#