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

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

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

3 moles of NH_3 are allowed to dissociate in a 5 litre vessel and the equilibrium concentration of N_2 is 0.2 mole/lit . Then the total number of moles at equilibrium is

One mole each of hydrogen and iodine are allowed to react at certain temperature till the equilibrium is reached. Calculate the composition of equilibrium mixture if K= 66.5.

4 moles each of SO_(2) and O_(2) gases are allowed to react to form SO_(3) in a closed vessel. At equilibrium 25% of O_(2) is used up. The total number of moles of all the gases at equilibrium is

4 moles each of SO_(2) "and" O_(2) gases are allowed to react to form SO_(3) in a closed vessel. At equilibrium 25% of O_(2) is used up. The total number of moles of all the gases at equilibrium is

4 moles each of SO_(2) "and" O_(2) gases are allowed to react to form SO_(3) in a closed vessel. At equilibrium 25% of O_(2) is used up. The total number of moles of all the gases at equilibrium is

1 mol of hydrogen and 2 mol of iodine are taken initially in a 2L vessel. The number of moles of hydrogen at equilibrium is 0.2 . Then the number of moles of iodine and hydrogen iodide at equilibrium are