Question

Calculating equilibrium concentrations when the net reaction proceeds in reverse Consider mixture C, which will cause the net reaction to proceed in reverse. Concentration (M)initial:change:equilibrium:[XY]0.200+x0.200+x←net⇌[X]0.300−x0.300−x+[Y]0.300−x0.300−x The change in concentration, x, is positive for the reactants because they are produced and negative for the products because they are consumed. Part C Based on a Kc value of 0.170 and the data table given, what are the equilibrium concentrations of XY, X, and Y, respectively? Express the molar concentrations numerically.

Answer 1

PLEASE mention that given Kc is for reverse reaction or for given reaction , in case it is for given reaction then just take inverse of 0.170 = 5.88 then solve as it is. you will get x=0.083 M, which is half of what i calculated, then solve as usual,

YOU WILL GET [x]=[y]= 0.3- 0.083=0.217 M instead of 0.14 M

AND [XY]= 0.200+ x= 0.283 M instead of 0.374 M

now use whichever best fits but procedure is same

ini hat 0-300 О, Зоо 0.2。0 change as - egcli at equi li br/um , equilibrium 4nston4 is gim la- 3-х ) ( b . 3-х) Io 16348 So 97 = C 0.374 M M 0.14 M