An equilibrium mixture at `300 K` contains `N_(2)O_(4)` and `NO_(2)` at `0.28` and `1.1 atm`, respectively. If the volume of container is doubles, calculate the new equilibrium pressure of two gases.

Explanation

Step 1 . Calculate of `K_(p)`
` {:(,N_(2)O_(4),hArr,2 NO_(2)),("Equilibrium",0*28 "atm",,1*1"atm"):}`
`K_(p) = (p_(NO_(2))^(2))/(p_(N_(2)O_(4)))= (1*1 "atm")^(2)/(0*28 "atm") =4*32 "atm"`
Step 2 . Calsulation of new equilibrium pressures . On doubling the volume , pressure will decrease to half . Hence, equilibrium will shift to the side accompanied by increase in the number of moles , i.e., forward direction . This means that pressure of `N_(2)O_(4) ` will decrease while that of `NO_(2)` will increase . Suppose decrease in pressure of `N_(2)O_(4) = p` . Then
` {: (,N_(2)O_(4),hArr,2NO_(2)), ("Intial pressures ",0*28//2 "atm",,1*1//2 "atm"), ("New eqm. pressures ",(0*28/2-p)"atm",,(1*1/2 +2 p)"atm" ),(,=(0*14 -p)"atm",,=(0*55 +2p) "atm"):} ` .
` 0* 3025 + 4p^(2) + 2*2 p = 0* 6048 - 4* 32 p `
` 4p^(2) + 6*52 p - 0*3023 = 0 `
` p = (-6* 52 pm sqrt(42 *51 + 4*84 ))/8 = 0*045 "atm" ` (minus value is neglected )
For quadratic equation ` ax^(2) + bx + c = 0 : x = (-b pm sqrt(b^(2) - 4ac))/ (2a)`
` :. " New equilibrium pressures ")`
` p_(N_2O_(4)) = 0 *14 - 0* 045 = 0* 095 "atm" `
` p_(NO_(2)) = 0* 55 + 2 xx 0* 045 = 0* 64 " atm"` `