At `25^(@)C` and `1 atm` pressure, the partial pressure in equilibrium mixture of `N_(2)O_(4)` and `NO_(2)`, are `0.7` and `0.3 atm`, respectively. Calculate the partial pressures of these gases when they are in equilibrium at `25^(@)C` and a total pressure of `10 atm`.

Answer

`p_(N_2O_4)` = 8.92 atm , `p_(NO_2)` =1.08 atm

Explanation

(i)Calculation of `K_p`
When total pressure is 1 atm.
`{:("Pressure at equilibrium",N_2O_4 hArr, 2NO_2),(,0.7,0.3):}`
`K_p=([p_(NO_2)]^2)/p_(N_2O_4)=(0.3)^2/0.7`=0.1286 atm
Let the degree of dissociation of `N_2O_4=alpha`
`{:(,N_2O_4 hArr, 2NO_2),("Moles at equi.", 1-alpha, 2alpha):}`
Total numbers of moles =`1-alpha +2alpha =1+alpha`
`p_(NO_2)=(2alpha)/(1+alpha)xx10` (P=10)
`p_(N_2O_4)=(1-alpha)/(1+alpha)xx10`
`K_p=([p_(NO_2)]^2)/p_(N_2O_4)=((2alpha)/(1+alpha)xx10)^2/((1-alpha)/(1+alpha)xx10)=(4alpha^2)/(1-alpha^2)xx10`
Now, `alpha` is very small so that `1-alpha^2 approx 1`
`therefore 0.1286 =(4alpha)^2/1xx10` or `alpha^2=0.1286/40`=0.0032
`therefore alpha` =0.057
`{:(p_(N_2O_4)=(1-0.057)/(1+0.057)xx10="8.92 atm"),(p_(NO_2)=(2xx0.057)/(1+0.057)xx100="1.08 atm"):}}`