The reaction below is carried out at a different temperature at which Kc=0.055. This time, however, the reaction mixture starts with only the product, [NO]=0.0100M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium.

Thee given question is incomplete. the complete question is:

The reaction below is carried out at a different temperature at which Kc = 0.055. This time, however, the reaction mixture starts with only the product, [NO] = 0.0100 M, and no reactants. Find the equilibrium concentrations of N2, O2, and NO at equilibrium. The equation is N2(g) + O2(g) <--> 2NO(g)

Concentration of at equilibrium = 0.001 M

Concentration of = 0.0045 M

Concentration of  = 0.0045 M

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stoichiometric coefficients.

Initial concentration of = 0.0100 M

The given balanced equilibrium reaction is,

Initial conc.            0 M           0 M         0.0100 M

At eqm. conc.     (x) M          (x) M       (0.0100-2x) M

The expression for equilibrium constant for this reaction will be,

Now put all the given values in this expression, we get :

By solving the term 'x', we get :

x = 0.0045

Thus, the concentrations of at equilibrium are :

Concentration of at equilibrium = (0.0100-2x) M  =

Concentration of = (x) M = 0.0045 M

Concentration of = (x) M = 0.0045 M

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be