# 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?

Jul 21, 2018

Assuming no ${\text{NO}}_{2}$ existed prior, and defining $P$ as the EQUILIBRIUM total pressure,

$\alpha = \sqrt{{K}_{p} / \left({K}_{p} + 4 P\right)}$

For example, if $P = \text{1 atm}$, 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 ${\text{NO}}_{2} \left(g\right)$ previously was in the container, the ICE table would simply be:

${\text{N"_2"O"_4(g) " "rightleftharpoons" " 2"NO}}_{2} \left(g\right)$,

$\text{I"" "P_(N_2O_4,i)" "" "" "" } 0$
$\text{C"" "-x" "" "" "" } + 2 x$
$\text{E"" "P_(N_2O_4,i)-x" "" } 2 x$

where ${P}_{{N}_{2} {O}_{4} , i}$ indicates the initial partial pressure of ${\text{N"_2"O}}_{4} \left(g\right)$. $x$ 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, $\alpha {P}_{{N}_{2} {O}_{4} , i}$ is the dissociated fraction of ${\text{N"_2"O}}_{4}$.

Thus, instead of ${P}_{{N}_{2} {O}_{4} , i} - x$, we write:

${P}_{{N}_{2} {O}_{4} , i} - x = \left(1 - \alpha\right) {P}_{{N}_{2} {O}_{4} , i}$

Similarly,

$2 x = 2 \alpha {P}_{{N}_{2} {O}_{4} , i}$

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

${K}_{p} = {P}_{N {O}_{2}}^{2} / \left({P}_{{N}_{2} {O}_{4}}\right)$

$= {\left(2 \alpha {P}_{{N}_{2} {O}_{4} , i}\right)}^{2} / \left(\left(1 - \alpha\right) {P}_{{N}_{2} {O}_{4} , i}\right)$

$= \left(\frac{4 {\alpha}^{2}}{1 - \alpha}\right) {P}_{{N}_{2} {O}_{4} , i}$

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

${P}_{{N}_{2} {O}_{4} , e q} + {P}_{N {O}_{2} , e q} = P$

$= \left(1 - \alpha\right) {P}_{{N}_{2} {O}_{4} , i} + 2 \alpha {P}_{{N}_{2} {O}_{4} , i}$

$= \left(1 + \alpha\right) {P}_{{N}_{2} {O}_{4} , i}$

So,

${P}_{{N}_{2} {O}_{4} , i} = \frac{P}{1 + \alpha}$

and we proceed to get:

${K}_{p} = \left(\frac{4 {\alpha}^{2}}{1 - \alpha}\right) \frac{P}{1 + \alpha}$

$= \left(\frac{4 {\alpha}^{2}}{1 - {\alpha}^{2}}\right) P$

Now we are asked to solve for $\alpha$.

${K}_{p} \left(1 - {\alpha}^{2}\right) = 4 {\alpha}^{2} P$

${K}_{p} - {K}_{p} {\alpha}^{2} = 4 {\alpha}^{2} P$

Rearrange to solve for $\alpha$:

${K}_{p} = {K}_{p} {\alpha}^{2} + 4 P {\alpha}^{2}$

$= \left({K}_{p} + 4 P\right) {\alpha}^{2}$

Therefore, knowing that $\alpha > 0$,

$\textcolor{b l u e}{\alpha = \sqrt{{K}_{p} / \left({K}_{p} + 4 P\right)}}$