Kc = 3.07 x 10-4 at 24°C for 2NOBr(g) ↔ 2NO(g) + Br2(g). If the initial concentration of NOBr = 0.878 M, what is the equilibrium concentration (in M to 4 decimal places) of NO?

The equilibrium concentration of NO is 0.02124 M.

Explanation:

Given that,

Initial concentration of NOBr = 0.878 M

Temperature = 24°C

We know that,

The balance equation is

Initial concentration is,

Concentration is,

Equilibrium concentration

We need to calculate the value of x

Using formula of concentration

Put the value into the formula

We need to calculate the equilibrium concentration of NO

Using formula of concentration of NO

Put the value of x

Hence, The equilibrium concentration of NO is 0.02124 M.

i believe the answer is three water molecules