The factor which changes equilibrium constant of the reaction
`A_(2)(g)+B_(2)(g) rarr 2AB(g)` is

The equilibrium constant (K_(c)) of the reaction A_(2)(g) + B_(2)(g) rarr 2AB(g) is 50. If 1 mol of A_(2) and 2 mol of B_(2) are mixed, the amount of AB at equilibrium would be

The equilibrium constant (K_(c)) of the reaction A_(2)(g) + B_(2)(g) rarr 2AB(g) is 50. If 1 mol of A_(2) and 2 mol of B_(2) are mixed, the amount of AB at equilibrium would be

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The equilibrium constant K_(p) for the reaction H_(2)(g)+I_(2)(g) hArr 2HI(g) changes if:

The answer to each of the folowing questions is a single digit integar, ranging from 0 to 9. If the correct answers to the question numbers A, B, C and D (say) are 4,0,9 and 2 respectively , then the correct darkening of bubbles should be as shown on the side : Equilibrium constant for the reaction A_(3) (g) + 3 B_(2) (g) hArr 3 AB_(2) (g) " is " 64*0 "Then the equilibrium constant for the reaction " 1/3 A_(3) (g) + B_(2) (g) hArr AB_(2)(g) will be

The equilibrium constant of the reaction A_(2)(g)+B_(2)(g) hArr 2AB(g) at 100^(@)C is 50 . If a 1L flask containing 1 mol of A_(2) is connected to a 2 L flask containing 2 mol of B_(2) , how many moles of AB will be formed at 373 K ?

The equilibrium constant of the reaction A_(2)(g)+B_(2)(g) hArr 2AB(g) at 100^(@)C is 50 . If a 1-L flask containing 1 mol of A_(2) is connected to a 2 L flask containing 2 mol of B_(2) , how many moles of AB will be formed at 373 K ?