For the reaction so2(g) + no2(g) so3(g) + no(g), the equilibrium constant kc is 18.0 at 1200âºc. if 1.0 mole of so2and 2.0 moles of no2 are placed in a 20.0-l container, what concentration of so3 will be present at equilibrium? a. 0.048 mol/l b. 0.11 mol/l c. 0.95 mol/l d. 2.22 mol/l e. 18 mol/l

 when:
[SO2] =  1mol/20L = 0.05 M
[NO2] = 2mol / 20L = 0.1 M
by using ICE table:
               SO2(g) +  NO2(g) ↔  SO3(g) + NO(g)
initial      0.05         0.1                  0            0
change  -X              -X                   +X         +X
Equ       (0.05-X)    (0.1-X)              +X          +X

when Kc= [SO3][NO]/[SO2][NO2]
by substitution:
18  = X^2/ (0.05-X) (0.1-X) by solving this equation for X
∴X(1,2) = -1/34√1.17+27/340 , 1/34√1.17+27/340
we can neglect the negative one 
∴X = 0.11 mol/L

You need to divided twenty four by a dozen them by the temperature it is getting cook by

Olá. porque a reação na temperatura é muito rapida a do copo : )