13.8 g of `N_2O_4` was placed in a 1 L reaction vessel at 400 K and allowed to attain equilibrium,
`N_2O_(4(g)) hArr 2NO_(2(g))`
The total pressure at equilibrium was found to be 9.15 bar. Calculate `K_c, K_p` and partial pressure at equilibrium.

Explanation

Molecular mass of `N_2O_4`=2(N)+4(O)
=2(14)+4(16) = 92 g `"mol"^(-1)`
Mole of `N_2O_4 = n = "Weight"/"Molecular mass"= "13.8 g"/"92 g mol"^(-1)`
= 0.15
Temperature T=400 K
Gas constant = R= 0.083 bar L `mol^(-1) K^(-1)`
pV=nRT where , p=Partial pressure `N_2O_4`
`therefore p=(nRT)/V`
`therefore p=((0.1 "mol")(0.083 "bar L mol"^(-1) K^(-1))(400 K))/(1 L)`
= 4.98 bar partial pressure of `N_2O_4`
`{:("Equilibrium reaction :",N_2O_(4(g)) hArr, 2NO_(2(g))),("Initial pressure :", "4.98 bar","0 bar"),("Pressure change :","-x bar" , "+2x bar"),("Partial pressure at equilibrium :","(4.98-x)bar","2x bar"):}`
`therefore` Total pressure at equilibrium = Addition of partial pressure
`p_"total"=p_(N_2O_4)+p_(NO_2)`
`therefore` 9.15=(4.98-x)+2x
`therefore` 9.15=4.98 +x
`therefore` x=(9.15-4.98)=4.17 bar
So, partial pressure at equilibrium
`p_(NO_2)`= 2x=2(4.17)=8.34 bar
`p_(N_2O_4)`=(4.98-x)=(4.98-4.17)=0.81 bar
The expression of `K_p`,
`K_p=(p_(NO_2))^2/(p_(N_2O_4))="(8.34 bar)"^2/"0.81 bar"`=85.87 bar
The calculation of `K_c` from `K_p`:
`K_p=K_c(RT)^(Deltan)` So, `K_c=K_p/(RT)^(Deltan_(g))`
where , `K_p`=85.87 bar, R=0.083 L bar `mol^(-1) K^(-1)`
T=400 K , `Deltan_((g))`=(2-1)=+1
`therefore K_c=(85.87)/(400xx0.083)^1=2.5864 approx` 2.586 M