When sulphur is heated at 900 K, `S_(8)` is converted to `S_(2)`. What will be the equilibrim constant for the reaction if initial pressure of 1 atm falls by `25%` at equilibrium ?

When sulphur is heated at 800 K, the initial pressure of 1 atm decreases by 20% at equilibrium K_(p) of the reaction S_(8)(g)rArr4S_(2)(g) is :-

When sulphur is heated at 800 K, the initial pressure of 1 atm decreases by 20% at equilibrium K_(p) of the reaction S_(8)(g)rArr4S_(2)(g) is :-

When sulphur in the form of S_(8) is heated at 900 K , the initial pressure of 1 atm falls by 10% at equilibrium. This is because of conversion of some S_(8) to S_(2) . Find the value of equilibrium constant for this reaction.

When sulphur in the form of S_(8) is heated at 900 K , the initial pressure of 1 atm falls by 10% at equilibrium. This is because of conversion of some S_(8) to S_(2) . Find the value of equilibrium constant for this reaction.

A sample of pure NO_2 gas heated to 1000 K decomposes : 2NO_2(g) hArr 2NO(g)+O_(2)(g) The equilibrium constant K_P is 100 atm. Analysis shows that the partial pressure of O_2 is 0.25 atm. At equilibrium.The partial pressure of NO_2 at equilibrium is:

When sulphur in the form of S_8 is heated at 900k the initial pressure of 1 atm falls by 29% at equilibrium . This is because of conversion of some S_8 "to" S_2 . Find the value of equilibrium constant for this reaction. S_8(g) hArr 4S_2(g)

The equilibrium constant K_(p) of the reaction: 2SO_(2)+O_(2) hArr 2SO_(3) is 900 atm^(-1) at 800 K . A mixture constaining SO_(3) and O_(2) having initial pressure of 1 atm and 2 atm respectively, is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at 800 K at equilibrium.

The equilibrium constant K_(p) of the reaction: 2SO_(2)+O_(2) hArr 2SO_(3) is 900 atm^(-1) at 800 K . A mixture constaining SO_(3) and O_(2) having initial pressure of 1 atm and 2 atm respectively, is heated at constant volume to equilibriate. Calculate the partial pressure of each gas at 800 K at equilibrium.