At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M....

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.100 M and [NO]=0.500 M.

N₂(g) + O₂ (g) <----> 2NO(g)

If more NO is added, bringing its concentration to 0.800 M, what will the final concentration of NO be after equilibrium is re-established?

_____M

Solutions

Expert Solution

The balanced equilibrium reaction is:

N2(g) + O2(g) ---> 2NO(g)

Kc = [NO]² / [N₂] [O₂] = (0.5 )² / 0.1 X 0.1 = 25

Now the new concentration of NO in the system is 0.8 M

Let us see the changes through ICE table

N2(g) + O2(g) ---> 2NO(g)

initial 0.1 0.1 0.8

Change +x +x -2x

Equilibrium 0.1+x 0.1 +x 0.8-2x

Kc = [NO]² / [N₂] [O₂] = 25

25 = (0.8-2x)² / (0.1+x)(0.1+x)

5 = 0.8-2x / 0.1+x

0.5 + 5x = 0.8 -2x

7x = 0.3

x = 0.0429 M

Therefore [NO] after equilibrium = 0.8-2x = 0.8- 2(0.0429) = 0.7142 M

queen_honey_blossom answered 1 year ago

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