If the equilibrium constant for the reaction
`A_(2)+B_(2) hArr 2AB`
is K, then the backward reaction,
`AB hArr 1/2 A_(2)+1/2 B_(2)`
the equilibrium constant is `1//K`.

If equilibrium constant for the reaction A_(2)+B_(2) hArr 2AB is k, then for the backward reaction AB hArr 1//2 A_(2)+1//2B_(2) the equilibrium constant k' is 1//K .

If equilibrium constant for the reaction A_(2)+B_(2) hArr 2AB is k, then for the backward reaction AB hArr 1//2 A_(2)+1//2B_(2) the equilibrium constant k' is 1//K .

If equilibrium constant for the reactions. A_2 +B_2 hArr 2AB, is K , then the backward reactions. AB hArr (1)/(2) A_2 + (1)/(2) B_2. its value is 1/K . Is it true or false? If false then write the correct constant.

If the equilibrium constant for the reaction 2AB hArr A_(2) + B_(2) is 49 what is the value of equilibrium constant for AB hArr 1/2 A_(2) + 1/2 B_(2) ?

The equilibrium constant for a reaction A+2B hArr 2C is 40 . The equilibrium constant for reaction C hArr B+1//2 A is

The equilibrium constant for a reaction A+2B hArr 2C is 40 . The equilibrium constant for reaction C hArr B+1//2 A is

The equilibrium constant for the reversible reaction N_(2)+3H_(2) hArr 2NH_(3) is K and for the reaction (1)/(2) N_(2)+(3)/(2) H_(2) hArr NH_(3) , the equilibrium constant is K',.K and K' will be related as

If the equilibrium constant for the reaction 2AB hArr A_(2)+B_(2) is 49, what is the value of equilibrium constant for ABhArr (1)/(2) A_(2)+(1)/(2)B_(2)

If equilibrium constant for reaction 2AB = A_(2) + B_(2) . Is 49, then the equilibrium constant for reaction AB ltimplies 1/2_(A_(2)) + 1/2_(B_(2)) , will be: