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

Under S.T.P. 1 mol of N_(2) and 3 mol of H_(2) will form on complete reaction

In a closed vessel of 1 dm^(3) capacity, 1 mol N_(2)(g) and 2 mol H_(2)(g) interact to produce 0.8 mol NH_(3)(g) in the equilibrium mixture. What is the concentration of H_(2)(g) in the equilibrium mixture?

1 mol of N_2 and 4 mol of H_2 are allowed to react in a vessel and after reaction, H_2O is added. Aqueous solution required 1 mol of HCI for neutralization. Mol fraction of H_2 in the mixture after reaction is :

1 mol of N_2 and 4 mol of H_2 are allowed to react in a vessel and after reaction, H_2O is added. Aqueous solution required 1 mol of HCI for neutralization. Mol fraction of H_2 in the mixture after reaction is :

1 mole of N_(2) and 3 mole of H_(2) are mixed in a closed vessel of 1 dm^(3) capacity. At equilibrium if the vessel contains a total of 2.4 moles, calculate equilibrium constant K_(c)" for "N_(2)+3H_(2)hArr2NH_(3) .

2 mol of N_(2) is mixed with 6 mol of H_(2) in a closed vessel of 1L capacity. If 50% of N_(2) is converted into NH_(3) at equilibrium, the value of K_(C ) for the reaction N_(2)(g0+3H_(2)(g)hArr 2NH_(3)(g) is