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

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

Text Solution

Verified by Experts

We know that
pV=nRT
Moles of `N_2O_4 =13.8/92`=0.15 (Molar mass of `N_2O_4`=92)
R=0.083 bar L `mol^(-1) K^(-1)`, T=400 K, V=1L
`therefore p=(nRT)/V=(0.15xx0.083xx400)/1`=4.98 bar
`{:(,N_2O_4hArr,2NO_2),("Initial conc.",4.98,0),("At equi.",4.98-x,2x):}`
Total pressure = 4.98-x+2x=9.15 bar
x=9.15-4.98=4.17bar
`p_(N_2O_4)` = 4.98-4.17=0.81 bar
`p_(NO_2)` = 4.17 x 2 = 8.34 bar
`K_p=(p_(NO_2))^2/(p_(N_2O_4))=(8.34)^2/0.81`=85.87
Now, `K_c=K_p(1/"RT")^(Deltan_g)`
`Deltan_g`=2-1=1
`K_c=85.87xx1/(0.083xx400)^1`
=2.586=2.6
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