At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M [N2]=[O2]=0.100 M and [ NO ] = 0.600

At equilibrium, the concentrations in this system were found to be [ N 2 ] = [ O 2 ] = 0.100 M [N2]=[O2]=0.100 M and [ NO ] = 0.600 M . [NO]=0.600 M. N 2 ( g ) + O 2 ( g ) − ⇀ ↽ − 2 NO ( g ) N2(g)+O2(g)↽−−⇀2NO(g) If more NO NO is added, bringing its concentration to 0.900 M, 0.900 M, what will the final concentration of NO NO be after equilibrium is re‑established?

Answers

The final concentration of NO be after equilibrium is re‑established is 0.825 M.

Explanation:

The given balanced chemical equation is as follows.

      N_{2}(g) + O_{2}(g) rightleftharpoons 2NO(g)

Now, equilibrium constant for this reaction will be as follows.

            K_{c} = frac{[NO]^{2}}{[N_{2}][O_{2}]}

It is given that concentrations at the equilibrium are:

   [N_{2}] = [O_{2}] = 0.1 M and [NO] = 0.6 M

Therefore, the value of K_{c} is as follows.

          K_{c} = frac{(0.6)^{2}}{(0.1)(0.1)}

                     = 36.0

NO concentration of 0.9 M is added to the system. So,

              N_{2}(g) + O_{2}(g) rightleftharpoons 2NO(g)

Initial:      0.1        0.1          0.9

Change:    x         x             2x

Equibm:(0.1 + x)  (0.1 + x)   (0.9 - 2x)

Now, we will find the value of x as follows.

         36.0 = frac{(0.9 - 2x)^{2}}{(0.1 + x)^{2}}

          x = 0.0375

Therefore, final concentration of NO after the equilibrium that is re-established is as follows.

                     0.9 - 2x

                  = 0.9 - (2 times 0.0375)

                  = 0.9 - 0.075

                  = 0.825 M

Therefore, we can conclude that the final concentration of NO be after equilibrium is re‑established is 0.825 M.

Idon’t know sorry i don’t know the answer i’m s sorry
Acovalent bond is formed between the two hydrogen bonds

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