Equilibrium question and about the Reaction Quotient?

Question

The reaction:
2 NO2(g) « N2O4(g)

has an equilibrium constant, Kc, of 170 at 298K. Analysis of this system at 298K reveals that 4.90E-1 mol of NO2 and 4.02E-1 mol of N2O4 are present in a 41.0-L flask. Determine the reaction quotient, Q, for this mixture. Enter your answer in scientific notation. Tell me how to do it also.

Answer 1

[ N2O4 ] =0.402 / 41 = 0.00980 M

[ NO2 ] = 0.490 / 41 = 0.0119 M

Q = 0.00980 / ( 0.0119 )^2 = 69.2 < Kc

The reaction has not reached the equilibrium and it will shift on the right

Answer 2

Q = [N2O4]/[NO2]^2

Just substitute the concentrations.