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13.8 g of N(2)O(4) was placed in 1 L rea...

13.8 g of `N_(2)O_(4)` was placed in 1 L reaction vessel at 400K and allowed to attain equilibrium :`N_(2)O_(4) (g) hArr 2NO_(2)(g).`
the total pressure at equilibrium was found to be 9.15 bar. Calculate `K_(c),K_(p)` and partial pressure at equilibrium .

Text Solution

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`"No. of moles of " N_(2) O_(4) = (13.8)/(92) " mol " = 0.15 " mol " " " (" Molar mass of " N_(2)O_(4) = 92 g " mol "^(-1))`
`PV=nRT or P = (nRT)/(V) or P = ((0.15 "mol") xx (0.083 " bar L mol"^(-1) K^(-1)) xx 400 K)/(1 L)= 4.98 " bar "`
`{:(,N_(2)O(g),hArr, 2NO_(2)),("Intital pressure",4.98"bar",,0),("Equilbibrium pressure",(4.98-p),,2_(P)):}`
`:. (P_(N_(2)O_(4)) eq = 4.98 -4.17 =0.81 "bar" , (P_(NO_(2)) eq =2x 4.17 =8.34 "bar"`
`K_(p) = p_(NO_(2))^(2)//P_(N_(2)O_(4) = (8.34)^(2)//0.81=85.86,`
`k_(p) =K_(c) (RT)^(Deltan) :. 85.87 = K_(c) (0.083xx400)^(1)`
`or " "K_(c) = (85.87)/(0.083xx 400) = 2.586 =2.6`
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