The equilibrium are simultaneoulsy existing in a vessel at `25^@C`
`NO(g) +NO_(2) (g) Leftrightarrow N_2O_3 (g) K_(p_(1)) ("say")`
`2NO_(2) (g) Leftrightarrow N_2O_4 (g), K_(p_(x))="8 atm"^(-1)`
If initially only NO and `NO_2` are present in a 3:5 mole ratio and the total pressure at equilibrium is 5.5 atm with the pressure of `NO_2` at 0.5atm, calculate `K_(p_(1))`.

Explanation

As mole ratio represents pressure ratio in gases, for the following equilibria, we have
Initial pressure `"3p atm (say) 5p atm(say) 0 atm"`
`" NO(g) +NO_(2) (g) Leftrightarrow N_2O_3 (g), kp_1`
Equilibrium pressure `p_(1) ("say") " 0.5 atm "p_2 (say)`
And `2NO_2 (g) Leftrightarrow N_2O_4 (g), " "Kp_2=8`
Equilibrium pressure 0.5 atm 2 atm
`Kp_(2)=(p_(N_2)O_4)/(p_(NO_(2))^(2))=(p_(N_(2))O_(4))/(0.5)^2=8, p_(N_2O_4)=2atm`
Now, out of 3p of `NO, p_2` atm of it converts to `p_2` atm of `N_2O_3`,
`3p-p_2=p_1`.......(1)
As `NO_2` converts to both `N_2O_3 and N_2O_4` out of 5p atm of `NO_2, p_2` atm converts to `p_2 atm of N_2O_3` and 4 atm of it gave 2 atm of `N_2O_4 and 0.5" atm of "NO_2` remained at equilibrium.
`5p-p_2-4=0.5 .....(2)`
Further at equilibrium the total pressure is given by
`p_(NO)+p_(NO_(2))+p_(N_(2)O_(3))+p_(N_(2)O_(4))=5.5`
or `p_(1)+0.5 +p_(2)+2=5.5 ……………..(3)`
Solving eqns. (1), (2) and (3),
`p_(1)=2.5 and p_(2)=0.5`
`Kp_(1)=p_(N_(2)O_(3))/(p_(NO) xx p_(NO_(2)))=(p_(2))/(p_(1) xx 0.5) =(0.5)/(2.5 xx 0.5) =0.40 atm^(-1)`