1.

` 13*8 g " of "N_(2)O_(4)` was placed in 1 L reaction vessel in 1 L reaction vessel at 400 K and allowed to attain equilibrium : `N_(2)O_(4) (g) hArr 2 NO(g).` The total pressure at equilibrium was found to be` 9* 15` bar. Calculate `K_(c) , K_(p)` and partial pressures at equilibrium.

Answer» `13*8 "g" N_(2)O_(4) = (13*8)/92" mol"=0*15" mol" "` `13*8 N_(2)O_(4) = (13*8)/92" mol"=0*15" mol " "(Molar mass of "N_(2)O_(4) = 92 g " mol"^(-1))`
PV= nRT
` :. P xx 1 L = 0* 15 mol xx 0* 083 " bar"L mol^(-1) K^(-1) xx 400 K or P = 4*98 " bar"`
` {:(,N_(2)O_(4)(g),hArr,2NO_(2)),("Initial pressures",4*98 "bar",,0),(" At equilibrium",(4*98-p),,2p):}`
` :. 4*98 - p+ 2 p = 9*15 " bar" or p= 4*17 " bar"`
`:. (p_(N_(2)O_(4)))eq = 4* 98 - 4*17 = 0*81 "bar", (p_(NO_(2)))eq = 2 xx 4*17 = 8*34 "bar"`
`K_(p) = p_(NO_(2))^(2)//_(P_(N_(2)O_(4)))=(8*34)^(2)//)*81=85*87,`
` K_(p) = K_(c) (RT)^(Delta n) :. 85*87 = K_(c) (0*083 xx400)^(1) or K_(c)= 2*586 = 2*6`


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