A sample of air consisting of `N_(2)` and `O_(2)` was heated to `2500 K` until the equilibrium
`N_(2)(g)+O_(2)(g)hArr2NO(g)`
was established with an equlibrium constant, `K_(c )=2.1xx10^(-3)`. At equilibrium, the `"mole"%` of `NO` was `1.8`. Eatimate the initial composition of air in mole fraction of `N_(2)` and `O_(2)`.

A sample of air consisting mainly of N_2 and O_2 is heated at 2500K till the given equilibrium attains- N_2(g) + O_2 (g) hArr 2NO(g) . If the value of equilibrium constant (K_c) at this temperature is 2.1xx10^(-3) and the percentage of number of moles of NO(g) in the above equilibrium mixture is 1.8 %, then what will be the mole fractions of N_2 and O_2 in the sample of air used'?

In the equilibrium constant for N_(2)(g) + O_(2)(g)hArr2NO(g) is K, the equilibrium constant for (1)/(2) N_(2)(g) + (1)/(2)O_(2)(g)hArrNO(g) will be:

In the equilibrium constant for N_(2)(g) + O_(2)(g)hArr2NO(g) is K, the equilibrium constant for (1)/(2) N_(2)(g) + (1)/(2)O_(2)(g)hArrNO(g) will be:

If the equilibrium constant for N_(2(g))+O_(2(g))hArr2NO_((g)) is K, the equilibrium constant for 1/2N_(2(g))+1/2O_(2(g))hArrNO_((g)) will be