Question

In: Chemistry

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


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