H2(g)Â Â Â Â +Â Â I2(g)
<---------------------------> 2HI(g)
3.50Â Â Â Â Â Â Â Â Â Â Â
2.00Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
0 ----------------------> initial
3.50-x        Â
2.00-x                                      Â
2x ----------------------> equilibrium
but equilibrium concentration of I2 is 0.0900 M is
given so
2.00 - x = 0.09
x = 1.91
equilibrium concentrations :
[H2] = 3.50 - x = 3.50 -1.91 = 1.59 M
[I2] = 0.09 M
[HI] = 2x = 2 x 1.91 = 3.82 M
Kc = [HI]^2 / [H2] [I2]
Kc = (3.82)^2 / (1.59) (0.09)
Kc = 102
