In the reaction equilibrium
`N_(2)O_(4) hArr 2NO_(2)(g)`
When `5` mol of each is taken and the temperature is kept at `298 K`, the total pressure was found to be `20` bar.
Given : `Delta_(f)G_(n_(2)O_(4))^(ɵ)=100 kJ, Delta_(f)G_(NO_(2))^(ɵ)=50 KJ`
a. Find `DeltaG` of the reaction at `298 K`.
b. Find the direction of the reaction.

(a) In the following equilibrium N_(2)O_(4)(g)hArr2NO_(2)(g) when 5 moles of each are taken, the temperature is kept at 298K the total pressure was found to be 20 bar. Given that DeltaG_(f)^(@)(N_(2)O_(4))=100kJ, DeltaG_(f)^(@)(NO_(2))=50kJ (i) Find DeltaG of the reaction. (ii) The direction of th reaction in which the equilibrium shifts. (b) A graph is plotted for a real gas which follows van der Waals' equation with pV_(m) taken on Y -axis and p on X -axis. Find the intercept of the ine where V_(m) is molar volume.

For the equilibrium reaction 2NO_2(g) hArr N_2O_4(g) + 60.0 kJ the increase in temperature

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.]

For the following equilibrium reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) , NO_(2) is 50% of the total volume at a given temperature. Hence, vapour density of the equilibrium mixture is :

For the dissociation reaction N_(2)O_($) (g)hArr 2NO_(2)(g) , the degree of dissociation (alpha) interms of K_(p) and total equilibrium pressure P is:

For the equilibrium, N_(2)O_(4)hArr2NO_(2) , (G_(N_(2)O_(4))^(@))_(298)=100kJ//mol and (G_(NO_(2))^(@))_(298)=50kJ//mol . ( a ) When 5 mol/litre of each is taken, calculate the value of DeltaG for the reaction at 298 K . ( b ) Find the direction of reaction.