4g `H_(2)` & 127g `I_(2)` are mixed "&" heated in "10" lit closed vessels until equilibrium is reached. If the equilibrium concentration of HI is 0.05M total number of moles present at equilibrium is

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

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

Hydrogen (a moles ) and iodine (b moles ) react to give 2x moles of the HI at equilibrium . The total number of moles at equilibrium is

Equilivalent amounts of H_(2) and I_(2) are heated in a closed vessel till equilibrium is obtained. If 80% of the hydrogen is converted to HI , the K_(c) at this temperature is

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

H_(2)(g) + I_(2)(g)hArr2HI(g) When 46 g of I_(2) and 1 g of H_(2) gas heated at equilibrium at 450^(@)C , the equilibrium mixture contained 1.9 g of I_(2) . How many moles of I_(2) and HI are present at equilibrium ?

2 mole of H_(2) and "1 mole of "I_(2) are heated in a closed 1 litre vessel. At equilibrium, the vessel contains 0.5 mole HI. The degree of dissociation of H_(2) is

One mole of H_(2) and 2 moles of I_(2) are taken initially in a two litre vessel. The number of moles of H_(2) at equilibrium is 0.2. Then the number of moles of I_(2) and HI at equilibrium is