For the reaction: `N_(2)O_(4)(g) hArr 2NO_(2)(g)`
`(i)" "`In a mixture of `5 mol NO_(2)` and `5` mol `N_(2)O_(4)` and pressure of 20 bar. Calculate the value of `DeltaG` for the reaction. Given `DeltaG_(f)^(@) (NO_(2)) = 50` KJ/mol, `DeltaG_(f)^(@) ((N_(2)O_(4)) =100` KJ/mol and T=298 K. `(ii)` Predict the direction in which the reaction will shift, in order to attain equilibrium
[Given at `T=298 K, 2.303 "RT" = 5.7`KJ/mol.]

Answer

`(i) DeltaG=5.7` KJ/mol `(ii)` backward shifting

Explanation

(a) `(i) DeltaG^(@)` for the reaction
`DeltaG^(@)"reac". = 2DeltaG_(f)^(@) (NO_(2))-DeltaG_(f)^(@)(N_(2)O_(4))`
100 -100=0
Now , `DeltaG=2.303 "RT log"Q_(p) + DeltaG^(@)`
Here `Q_(P) = (P_(N_(2)O_(4))^(@))/(P_(N_(2)O_(4))) = (100)/(10)= 10` atm
So, `DeltaG=2.303 "RT log"Q_(P) +0 =2.303"RT log"Q_(P) = 2.303 "RT log"_(10) 10=2.303 "RT" =5.7` KJ/mol
`(ii)` Since `Q_(P)` is more than `K_(P)`
(Calculate `K_(P)` by putting the value of `DeltaG^(@)` in the equation `DeltaG^(@)=2.303 "RT log" K_(P)` as `DeltaG^(@)=0` that's why `K_(P)` comes as 1.)
Hence , the reaction will proceed in backward direaction.