`K_p` for the reaction `N_2O_4 (g) hArr 2NO_2(g)` is 0.66 at `46^@C`. Calculate the part per cent dissociation of `N_2O_4` at `46^@C` and a total pressure of 0.5 atm. Also calculate the partial pressure of `N_2O_4 and NO_2` at equilibrium .

Explanation

This problem can be solved by two methods.
Method 1`:` Let the number of moles of `N_(2)O_(4)` initially be 1 and `alpha` is the degree of dissociation of `N_(2)O_(4)`.
`N_(2)O_(4) ` ` +` `2NO_(2)`
Initial moles 1 0
Moles at equilibrium `1- alpha` `2 alpha`
Total moles at equilibrium `= 1-alpha + 2 alpha = 1 + alpha`
`p_(N_(2)O_(4)) = ( 1- alpha)/( 1+ alpha) xx P_(T)`
`p_(NO_(2)) = ( 2alpha )/( 1+ alpha) xx P_(T )`
`:. K_(p) = ( p_(NO_(2))^(2))/( p_(N_(2)O_(4))) = ( 4alpha^(2) P_(T))/((1-alpha) ( 1+ alpha)) = ( 4alpha^(2) xx 0.5)/( 1- alpha^(2))`
`alpha = 0.5` i.e., 50% dissociation
Hence, partial pressure of `N_(2)O_(4) = 0.167` atm.
and partial pressure of `NO_(2)` = 0.333 atm.
Method 2`:` Let the partial pressure of `NO_(2)` at equilibrium be p atm, than the partial pressure of `N_(2)O_(4)` at equilibrium will be ( 0.5-p) atm.
`:. K_(p) = ( p^(2))/( ( 0.5-p)) = 0.66`
`p^(2) + 0.66 p - 0.33 = 0 `
On solving p = 0.333 atm.
`:. p_(NO_(2)) = 0.333` atm and `p_(N_(2)O_(4)) = 0.167 `atm.