One mole of` X_(2)`is mixxed with one mole of `Y_(2)` in a flask of valume 1 litre . If at equilibrium 0.5 mole of `Y_(2)` are obtained. Then find out `K_(P)` for reaction `X_(2(g))+Y_(2(g))hArr2XY_(g)`

One mole of X_(2) is mixed with one mole of Y_(2) in a flask of volume 1 litre . If at equilibrium 0.5 mole of Y_(2) are obtained. Then find out K_(P) for reaction X_(2(g))+Y_(2(g))hArr2XY_(g)

Initially 0.8 mole of PCl_(5) and 0.2 mole of PCl_(3) are mixed in one litre vessel . At equilibrium 0.4 mole of PCl_(3) is present . The value of K_(C) for the reaction PCl_(5(g)) hArr PCl_(3(g)) + Cl_(2(g)) is

One mole of N_(2) (g) is mixed with 2 moles of H_(2)(g) in a 4 litre vessel If 50% of N_(2) (g) is converted to NH_(3) (g) by the following reaction : N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) What will the value of K_(c) for the following equilibrium ? NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)

One mole of N_(2) (g) is mixed with 2 moles of H_(2)(g) in a 4 litre vessel If 50% of N_(2) (g) is converted to NH_(3) (g) by the following reaction : N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) What will the value of K_(c) for the following equilibrium ? NH_(3)(g)hArr(1)/(2)N_(2)(g)+(3)/(2)H_(2)(g)

1 mole of 'A' 1.5 mole of 'B' and 2 mole of 'C' are taken in a vessel of volume one litre. At equilibrium concentration of C is 0.5 mole /L .Equilibrium constant for the reaction , A_((g))+B_((g) hArr C_((g)) is

1 mole of 'A' 1.5 mole of 'B' and 2 mole of 'C' are taken in a vessel of volume one litre. At equilibrium concentration of C is 0.5 mole /L .Equilibrium constant for the reaction , A_((g))+B_((g) hArr C_((g)) is

One mole of N_(2) and 3 moles of H_(2) are mixed in a litre flask. If 50% N_(2) is converted into ammonia by the reaction. N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g) then the total number of moles of gas at equilibrium is :