The degree of dissociation of `N_(2)O_(4)(1)` obeying the equilibrium, `N_(2)O_(4) (g) hArr 2NO_(2)(g)`, is approximately related to the pressure at equilibrium by `:`
The degree of dissociation of SO_(3) at equilibrium pressure is: K_(p) for 2SO_(3) (g)hArr2SO_(2)(g) + O_(2) (g)
The degree of dissociation 'alpha' of the reaction N_(2)O_(4)(g)hArr 2NO_(2)(g) Can be related of K_(p) as [Given : Total pressure at equilibrium = P]
For the equilibrium reaction 2NO_2(g) hArr N_2O_4(g) + 60.0 kJ the increase in temperature
Gaseous N_(2)O_(4) dissociates into gaseous NO_(2) according to the reaction : [N_(2)O_(4)(g)hArr2NO_(2)(g)] At 300 K and 1 atm pressure, the degree of dissociation of N_(2)O_(4) is 0.2. If one mole of N_(2)O_(4) gas is contained in a vessel, then the density of the equilibrium mixture is :
In the given reaction N_(2)(g)+O_(2)(g) hArr 2NO(g) , equilibrium means that
For the following equilibrium N_(2)O_(4)(g)hArr 2NO_(2)(g) K_(p) is found to be equal to K_(c) . This is attained when :
For the following gases equilibrium, N_(2)O_(4)(g)hArr2NO_(2)(g) K_(p) is found to be equal to K_(P) . This is attained when:
The temperature at which K_(c) and K_(p) will have the same value for the equilibrium , N_(2) O_(4) (g) hArr 2 NO_(2) (g) is