Asked by Anonymous

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?

Answers

Answered by 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
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