Question

At 60 and 1 atm of n2o4 is 50 % dissociated into no2 then kp is

Answer 1

N2O4-->2NO2..let initial conc..1, 0 resp.. after dissociation, 1- € ( € is diss constant) nd 2€ resp.. where 1-€=0.5 given in qstn.. so 2€ is 1, ie, NO2 is 1.. Kp= 1²/0.5=2

Answer 2

Answer: 1.33 atm

Explanation:

$N_2O_4(g)\rightleftharpoons 2NO_2$

Initial conc.   1              0    

as $N_2O_4$ is 50% dissociated.

At eqm. conc.   (1-0.5=0.5)     $0.5\times 2=1.0$

Total moles at eqm = 0.5+1.0 = 1.5

The expression for equilibrium constant for this reaction will be,

$K_p=\frac{(p_{NO_2)^2}}{p_{N_2O_4}}$

$p_{NO_2}=\chi_{NO_2}\times P_T$

$p_{NO_2}$ = partial pressure of $NO_2$

$\chi_{NO_2}$ = mole fraction of $NO_2$ =

$\frac{\text{Moles of }NO_2}{\text{Total moles}}=\frac{0.5}{1.5}$

$P_{T}$ = total pressure = 1 atm

$p_{N_2O_4}=\chi_{N_2O_4}\times P_T$

$p_{NO_2}$ = partial pressure of $N_2O_4$

$\chi_{NO_2}$ = mole fraction of $N_2O_4$ =

$\frac{\text{Moles of }N_2O_4}{\text{Total moles}}=\frac{1.0}{1.5}$

$P_{T}$ = total pressure = 1 atm

Putting values in above equation, we get:

$K_p=\frac{(\frac{0.5}{1.5}\times 1atm)^2}{\frac{1.0}{1.5}\times 1atm}$

$K_p=1.33atm$

Thus equilibrium constant is 1.33 atm