For the dissociation reaction
`N_(2)OhArr2NO_(2)(g),`
the equilibrium constant `K_(P)` is `0.120` atm at `298 K` and total pressure of system is `2` atm. Calculate the degree of dissociation of `N_(2)O_(4)`.

For the dissociation reaction N_(2)O_(4)(g)hArr2NO_(2)(g), the equilibrium constant K_(P) is 0.120 atm at 298 K and total pressure of system is 2 atm. Calculate the degree of dissociation of N_(2)O_(4) .

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 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 dissociation reaction N_(2)O_(4) (g)hArr 2NO_(2)(g) , the degree of dissociation (alpha) interms of K_(p) and total equilibrium pressure P is:

At 30^(@)C, K_(p) for the dissociation reaction : SO_(2)Cl_(2)(g) hArr SO_(2)(g)+Cl_(2)(g) is 2.9xx10^(-2) atm. If the total pressure is 1 atm, the degree of dissociation of SO_(2)Cl_(2) is : (assume 1-alpha^(2)=1 ).

The equilibrium constant K_(p) for the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) at 497^(@)C is found to be 636 mm Hg . If the pressure of the gas mixture is 182 mm , calculate the presentage dissociation of N_(2)O_(4) . At what pressure will it be dissociated?

The equilibrium constant K_(p) for the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) at 497^(@)C is found to be 636 mm Hg . If the pressure of the gas mixture is 182 mm , calculate the presentage dissociation of N_(2)O_(4) . At what pressure will it be dissociated?

For the reaction N_2O_4 harr 2NO_(2(g)) , the degree of dissociation at equilibrium is 0.2 at 1 atm. Then K_p will be

For the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant K_(p) will be

For the reaction N_(2)O_(4)(g)hArr2NO_(2)(g) , the degree of dissociation at equilibrium is 0.2 at 1 atm pressure. The equilibrium constant K_(p) will be