At 273 K and 1atm , 10 litre of `N_(2)O_(4)` decompose to `NO_(4)` decompoes to ` NO_(2)` according to equation
`N_(2)O_(4)(g)hArr2NO_(@)(G)`
What is degree of dissociation `(alpha)` when the original volume is 25% less then that os existing volume?

At 273K and 1 atm a L of N_(2)O_(4) decomposes to NO_(2) according to equation N_(2)O_(4)(g) hArr 2NO_(2)(g) . To what extent ha the decomposition proceeded when the original volume is 25% less than that of exisiting volume?

At 273 K one atm, 'a' litre of N_(2)O_(4) decomposes to NO_(2) as : N_(2)O_(4)(g) hArr 2NO_(2)(g) . To what extent has the decomposition proceeded when he original volume is 25% less then that of existing volume ? [Report your answer up to decimal places.]

If in the reaction, N_(2)O_(4)(g)hArr2NO_(2)(g), alpha is the degree of dissociation of N_(2)O_(4) , then the number of moles at equilibrium will be

If in the reaction, N_(2)O_(4)(g)hArr2NO_(2)(g), alpha is the degree of dissociation of N_(2)O_(4) , then the number of moles at equilibrium will be

If in reaction, N_(2)O_(4)hArr2NO_(2) , alpha is degree of dissociation of N_(2)O_(4) , then the number of molecules at equilibrium will be:

For the reaction, N_(2)O_(4)(g)hArr 2NO_(2)(g) the degree of dissociation at equilibrium is 0.l4 at a pressure of 1 atm. The value of K_(p) is

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

Vapour density of the equilibrium mixture of NO_(2) "and" N_(2)O_(4) is found to be 38.33 . For the equilibrium N_(2)O_(4)(g)hArr2NO_(2)(g) . Calculatedegree of dissociation.