At 298K, for attainment of equilibrium of the reaction `N_(2)O_(4)(g)hArr 2NO_(2)(g)`, 5 mol of each of the constituents is taken. Due to this, total pressure of the mixture turns 20 atm. If `DeltaG_(f)^(0)(N_(2)O_(4))=100kJ*mol^(-1)` and `DeltaG_(f)^(0)(NO_(2))=50K(J*mol^(-1))` then- (1) Give the value of `DeltaG` of the reaction? (2) In which direction will the reaction proceed more to attain equilibrium?

Explanation

For the reaction, `DeltaG^(0)=2DeltaG_(f)^(0)(NO_(2))-DeltaG_(f)^(0)(N_(2)O_(4))`
`=(2xx50-100)=0kJ*mol^(-1)`
Total number of moles in the reaction mixture=5+5=10
So, `P_(N_(2)O_(4))=(5)/(10)xx20=10 atm&p_(NO_(2))=(5)/(10)xx20=10atm`
Therefore, `Q_(p)` of the reaction`=((p_(NO_(2)))^(2))/(p_(N_(2)O_(4)))=((10)^(2))/(10)=10`
(1) We know, `DeltaG=DeltaG^(0)+RTlnQ_(p)`
`thereforeDeltaG=0+8.314xx298ln10=5.706kJ" "[becauseDeltaG^(0)=0]`
(2) Again, `DeltaG^(0)=-RTlnK_(p)`,
So, `0=-RTlnK_(p)" "[because DeltaG^(0)=0]`
or, `K_(p)=1`
Since, `Q_(p) gtK_(p)`, the reaction will proceed more towards the left to attain equilibrium.