Posted by Jim on Saturday, July 3, 2010 at 3:00pm.
Consider the following equation:
N2O4(g)+ 2NO2(g) Kc= 5.8 x 10^-3
If the initial concentration of N2O4(g)= 0.040 M and the initial concentration of NO2(g) is 0 M, what is the equilibrium of concentration of N2O4(g)?
- chemistry - DrBob222, Saturday, July 3, 2010 at 3:51pm
Did you make a typo, typing + instead of -->
- chemistry - Jim, Saturday, July 3, 2010 at 3:58pm
Yes, I meant to right -->...I apologize for typing the question incorrectly.
- chemistry - DrBob222, Saturday, July 3, 2010 at 4:00pm
If so, then
N2O4 ==> 2NO2
Set up ICE chart.
N2O4 = 0.040
NO2 = O
NO2 = 2x
N2O4 = -x
NO2 = 2x
N2O4 = 0.040-x
Set up Keq expression of
Keq = (NO2)^2/(N2O4), plug in the equilibrium values from above and solve for x. Finally, 0.040-x will be the N2O4 at equilibrium.
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