At 35°c, kc = 1.6 multiplied by10-5 for the following reaction
2 nocl(g) reverse reaction arrow 2 no(g)+ cl2(g)
calculate the concentrations of all species at equilibrium if
2.0 mol no and 1.0 mol of cl2 are placed in a 1.0 l flask

Answer : The equilibrium concentrations of all species are, 0.05 M, 0.043 M and 0.975 M respectively.

Explanation : Given,

Moles of   = 2 mole

Moles of   = 1 mole

Volume of solution = 1 L

Initial concentration of = 2 M

Initial concentration of = 1 M

The given balanced equilibrium reaction is,

Initial conc.          2 M            1 M            0

At eqm. conc.    (2-2x) M   (1-x) M         (2x) M

The expression for equilibrium constant for this reaction will be,

The for reverse reaction =

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

By solving the term 'x', we get :

x = 0.975

Thus, the concentrations of at equilibrium are :

Concentration of = (2-2x) M  = (2 - 2 × 0.975) M = 0.05 M

Concentration of = (1-x) M = 1 - 0.975 = 0.043 M

Concentration of = x M = 0.975 M

Therefore, the equilibrium concentrations of all species are, 0.05 M, 0.043 M and 0.975 M respectively.

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