2NO
(g) + Cl
(g) 2NOCl
(g)At 25°C, the total mixture of the three gasses in equilibrium is 1.55 atm. The percentage of the NOCL in the mixture is 77.4%. The mol fraction of the NO is 0.032. Calculate the K
c for the reaction.
So, I assume I would first express myself ... K
c=[NOCl]
2/[NO]
2[Cl
2]
Next I believe you make an "ICE" table
[NO]
2i:
[Cl
2]
i:
[NOCl]
2i:
[NO]
2eq:
[Cl
2]
eq:
[NOCl]
2eq:
Now I just can't seem to figure out how to use the percentage and mol fraction to come up with an answer.
I would imagine that the mol fraction formula, χ
a=mol A/mol total, as well as the mass percent formula, g A/ g total * 100, would play a roll. Also, would K
c be switched to K
p and the square brackets be changed to parentheses due to the pressure?