For the equilibrium at `298 K, N_(2)O_(4)(g) hArr 2NO_(2)(g), G_(N_(2)O_(4))^(ɵ)=100 kJ mol^(-1)` and `G_(NO_(2))^(ɵ)=50 kJ mol^(-1)`. If `5` mol of `N_(2)O_(4)` and `2` moles of `NO_(2)` are taken initially in one litre container than which statement are correct.A. reaction proceeds in forward directionB. `K_(c)=1`C. `DeltaG=-0.55 KJ, DeltaG^(ɵ)=0`D. At equilibrium `[N_(2)O_(4)]=4.84 M` and `[NO_(2)]=0.212 M`

For the equilibrium at `298 K, N_(2)O_(4)(g) hArr 2NO_(2)(g), G_(N_(2)

1.	For the equilibrium at `298 K, N_(2)O_(4)(g) hArr 2NO_(2)(g), G_(N_(2)O_(4))^(ɵ)=100 kJ mol^(-1)` and `G_(NO_(2))^(ɵ)=50 kJ mol^(-1)`. If `5` mol of `N_(2)O_(4)` and `2` moles of `NO_(2)` are taken initially in one litre container than which statement are correct.A. reaction proceeds in forward directionB. `K_(c)=1`C. `DeltaG=-0.55 KJ, DeltaG^(ɵ)=0`D. At equilibrium `[N_(2)O_(4)]=4.84 M` and `[NO_(2)]=0.212 M`
Answer» Correct Answer - A::B::C::D `DeltaG=DeltaG^(ɵ)+2.303 RT log Q` `DeltaG=2xxG_(NO_(2))^(ɵ)-G_(N_(2)O_(4))^(ɵ)=2xx50-100=0` `:. DeltaG=0+2.303xx8.314xx10^(-3)xx298 log 22/5` `=0-0.55 kJ` `:. DeltaG=-0.55 kJ`, i.e., reaction proceed in forward direction Also `DeltaG^(ɵ)=0=2.303 RT log K :. K=1` `{:("Now,",N_(2)O_(4),=,2NO_(2)),(,5,,2),(,5-x,,2+2x):}` `:. K_(p)=((P_(NO_(2))))/((P_(N_(2)O_(4))))=1=((2+2x)^(2))/(5-x)` or `x=0.106`

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