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)?
3 answers
Did you make a typo, typing + instead of -->
Yes, I meant to right -->...I apologize for typing the question incorrectly.
If so, then
N2O4 ==> 2NO2
Set up ICE chart.
initial:
N2O4 = 0.040
NO2 = O
change:
NO2 = 2x
N2O4 = -x
equilibrium:
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.
N2O4 ==> 2NO2
Set up ICE chart.
initial:
N2O4 = 0.040
NO2 = O
change:
NO2 = 2x
N2O4 = -x
equilibrium:
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.