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.]

Explanation

(i) For the given reaction,
`DeltaG^(@) = Delta_(f) G^(@) ` ( Products) `- Delta_(f)G^(@) `( Reactants)
`= 2 xx 50 - 100 = 0`
`DeltaG^(@) = - 2.303 RT log K_(p)` ,brgt `:.log K_(p ) = 0, i.e., K_(p) =1`
Initially, as equal moles of `N_(2) O_(4)` and `NO_(2)` are taken,
`p_(N_(2)O_(4))= p_(NO_(2))= 10` bar ( `:'` total pressure `= 20 ` bar )
`:. Q_(p)` ( Initial) `= ( p_(NO_(2))^(2) )/p_(N_(2)O_(4)) = ((" 10 bar")^(2))/(1"10 bar" ) = 10 `bar
Initial free energy of the reaction will be
`DeltaG= DeltaG^(@) + 2.303 RT log Q_(p) `
`= 0 + ( 2.303 ) * 8.314 JK^(-1) mol^(-1) ) ( 298 K ) ( log 10)`
` = 5705.8 Jmol^(-1) = 5.706 kJ mol^(-1)`
(ii) As the initial Gibbs free energy change of the reaction is positive, so the reverse reaction will take place.