(i) N_(2)O_(4)(g) hArr 2NO_(2)(g) This reaction is carried out at 298 K and 20 bar. 5 moleseach of N_(2)O_(4) and NO(2) are taken initially . Given DeltaG_(N_(2)O_(4))^(@)= 100 kJ mol^(-1) , Delta G_(NO_(2))^(@)= 50 kJ mol^(-1) Find DeltaG for the reaction at 298 K under given condition . (ii) The reaction procees at an initial pressure of 20 bar. Find the direction in which the reaction proceedsto achieve equilibrium.

(i) N_(2)O_(4)(g) hArr 2NO_(2)(g) This reaction is carried out at 298

1.	(i) N_(2)O_(4)(g) hArr 2NO_(2)(g) This reaction is carried out at 298 K and 20 bar. 5 moleseach of N_(2)O_(4) and NO(2) are taken initially . Given DeltaG_(N_(2)O_(4))^(@)= 100 kJ mol^(-1) , Delta G_(NO_(2))^(@)= 50 kJ mol^(-1) Find DeltaG for the reaction at 298 K under given condition . (ii) The reaction procees at an initial pressure of 20 bar. Find the direction in which the reaction proceedsto achieve equilibrium.
Answer» Solution :(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.