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.

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^(ɵ) 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.

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?

In the reaction N_(2(g)) + O_(2(g)) hArr 2NO_((g)) , Delta H = + 180 kJ On increasing the temperature the production of NO

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