Question
A 0.25 mol sample of n204 dissociates and comes to equilibrium in a 1.5 L flask at 100 degrees C. The reaction is N2O4 > 2 NO2. The Kc at 100 degrees C is 0.36. What are the equilibrium concentrations of NO2 and N2O4?
Answers
0.25 mol/1.5L = 0.167M which I would round to 0.17. (Note: you may wish to carry an extra place and round the answer at the end to two significant figures. Follow your profs advice on this.)
........N2OP4 ==> 2NO2
I.......0.167.....0
C.......-x........2x
E.....0.167-x....2x
Kc = (NO2)^2/(N2O4)
0.36 = (2x)^2/(0.167-x)
Solve for x and 2x.
........N2OP4 ==> 2NO2
I.......0.167.....0
C.......-x........2x
E.....0.167-x....2x
Kc = (NO2)^2/(N2O4)
0.36 = (2x)^2/(0.167-x)
Solve for x and 2x.