Question
The reaction N2O4(g) -->2NO2(g) has Kc = 0.133 at 25 °C. What is the NO2 concentration in a 5.00 liter flask if it contains 0.250 mol of N2O4 at equilibrium
I know I set something equal to Kc?
I know I set something equal to Kc?
Answers
DrBob222
(NO2) at equilibrium = 0.250 mols/5 L = 0.05
Kc = (NO2)^2/(N2O4) = 0.133
(NO2)2 = x^2
(N2O4 = 0.05
Solve for x
Kc = (NO2)^2/(N2O4) = 0.133
(NO2)2 = x^2
(N2O4 = 0.05
Solve for x