3.1 mol `Fe^(3+)` and 3.2 mol `"SCN"^(Theta)` present in 1 L solution. At equilibrium 3 mol `FeSCN^(2+)` are formed The equilibrium constant `K_(c)` for the reaction `Fe^(3+)+SCN^(Theta)hArrFeSCN^(2+)` will

Write the equilibrium constant expression for the following reactions : Fe^(3+) (aq) + SCN^(-) (aq) hArr FeSCN^(2+) (aq)

1 mol of N_(2) and 2 mol of H_(2) are allowed to react in a 1 dm^(3) vessel. At equilibrium, 0.8 mol of NH_(3) is formed. The concentration of H_(2) in the vessel is

1 mol of N_(2) and 2 mol of H_(2) are allowed to react in a 1 dm^(3) vessel. At equilibrium, 0.8 mol of NH_(3) is formed. The concentration of H_(2) in the vessel is

The equilibrium constant for the reaction : Fe^(3+)(aq)+SCN^(-)(aq)hArr Fe SCN^(2+)(aq) is 140 at 298 K. The equilibrium constant for the reaction : 2Fe^(3+)(aq)+2SCN^(-)(aq)hArr 2Fe SCN^(2+)(aq) is :

A mixture of 1.57 mol of N_(2) , 1.92 mol of H_(2) and 8.13 mol of NH_(3) is introduced into a 20L reaction vessel at 500K . At this temperature, the equilibrium constant, K_(c) for the reaction N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g) is 1.7xx10^(2) . Is the reaction mixture at equilibrium ? If not, what is the direction of the net reaction ?

A mixture of 1.57 mol of N_(2), 1.92 mol of H_(2) and 8.13 mol of NH_(3) is introduced into a 20 L reaction vessel at 500 K . At this temperature, the equilibrium constant K_(c ) for the reaction N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g) is 1.7xx10^(2) . Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?