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At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M. N2(g) + 02(g) <-> 2NO(g)...Asked by Anonymous
At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.500 M.
N2 + O2= 2NO
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?
N2 + O2= 2NO
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?
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Answered by
DrBob222
Substitute the equilibrium concentrations into the Kc expression and solve for Kc. Use this below.
...........N2 + O2 ==> 2NO
I.......0.200..0.200....0.800
C.........+x...+x.......-2x
E.....0.200+x..0.200+x...0.800-2x
Substitute the E line into Kc expression along with the numberical value of Kc determined above and solve for x, then evaluate each of the values in the E line above. Post your work if you get stuck.
...........N2 + O2 ==> 2NO
I.......0.200..0.200....0.800
C.........+x...+x.......-2x
E.....0.200+x..0.200+x...0.800-2x
Substitute the E line into Kc expression along with the numberical value of Kc determined above and solve for x, then evaluate each of the values in the E line above. Post your work if you get stuck.
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