5 moles of `N_(2)` and 10 moles of `H_(2)` are made to react in one liter vessel. At equilibrium 4 moles of `NH_(3)` is formed. The value of `K_(C)` is
(A) "0.83
(B) 0.083
(C) "8.3
(D) 0.125

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

Initially, 0.8 mole of PCl_(5) and 0.2 mol of PCl_(30 are mixed in one litre vessel. At equilibrium, 0.4 mol of PCl_(3) is present. The value of K_(c ) for the reaction PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g) would be

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

1.1 mole of A mixed with 2.2 mole of B and the mixture is kept in a 1 litre at the equilibrium 0.2 mole of C is formed, then the value of K_(c) will be:

One mole of A and 2 moles of B are allowed to react in a 0.5 lit flask. What is the value of K if at equilibirum, 0.4 moles of C is formed in the reaction A+2B Leftrightarrow C+2D

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

1 "mole of" N_(2) "and" 3 "moles of " H_(2) are placed in 1L vessel. Find the concentration of NH_(3) at equilibrium, if the equilibrium constant (K_(c)) at 400K "is" (4)/(27)

If, from 10 moles NH_(3) and 5 moles of H_(2)SO_(4), all the H-atoms are removed in order to form H_(2) gas, then find the number moles of H_(2) formed.