If `0.5 `mole `H_(2)` is reacted with 0.5 mole `I_(2)` in a ten `-` litre container at `444^(@)C` and at same temperature value of equilibrium constant `K_(C)` is 49, the ratio of `[Hl]` and `[l_(2)]` will be `:`

One mole of PCl_(5) is heated in one litre closed container. If 0.6 mole of chlorine is found at equilibrium, calculate the value of equilibrium constant.

0.5 mol of H_(2) and 0.5 mole of I_(2) react in 10 litre flask at 448^(@)C . The equilibrium brium constant K_(C) is 50 for H_(2)(g)I_(2)(g) hArr 2HI(g) Calcualte mole of I_(2) at equilibrium.

How much PCI_5 must be taken in a 9.2 L vessel to get 0.5 moles of Cl_2 at a particular temperature? The value of equilibrium constant (K_c) at the given temperature is 0.0414.

15 moles of H_(2) and 5.2 moles of I_(2) are mixed are allowed to attain equilibrium at 500^(@)C . At equilibrium the concentration of HI is found to be 10 moles. The equilibrium constant for the formation of HI is

0.5 mol of H_(2) and 0.5 mole of I_(2) react in 10 litre flast at 448^(@)C . The equilibrium constant K_(c) is 50 for H_(2)(g)+I_(2)(g)hArr2HI(g) a. What is the value of K_(p) b. Calculate mole of I_(2) at equilibrium.