In a container equilibrium `N_2O_4(g) hArr 2NO_2(g)` is attained at `25^@C`.The total equilibrium pressure in container is 380 torr.If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of `N_2O_4` at this temperature wil be :

In a container equilibrium N_(2)O_(4) (g)hArr2NO_(2) (g) is attained at 25^(@)C . The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of N_(2)O_(4) at this temperature will be:

At constant temperature, vapour density of N_2O_4 is found to be 40. Percentage degree of dissociation of N_2O_4 at this constant temperature will be N_2O_4(g)hArr2NO_2(g)

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

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

Examine the chemical equilibrium. 4NH_(3(g)) + 5O_(2(g)) harr 4NO_(g) + 6H_2O_(g) Write the expression for equilibrium constatn (K_c) for the above equilibrium.

N_2(g) + O_2(g) harr 2NO_(g) what is the effect of pressur on the above equilibrium?

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) .