[biomed77]

i am stuck on a question if someone could help i would appreciate it....
at 360 Celcius the value of Kc for the reaction  H2 + I2  <----  2HI is 9
                                                                               ----->
A sample of HI was introduces into an evacuated vessel. when equilibrium was attained the concentration of HI was 1.2 mol dm-3. Using the value above calculate the concentration of H2 and I2 at equilibrium.
i set up a table saing that initial value for HI = 1.2 for H2=0 for I2=0
then the change for HI=-2x  for H2=+x and for I2= +x
equilibrium for HI= 1.2-2x  for H2=x  and for I2 = x
then Kc= [H2]*[I2] / [HI]^2=  X*X / [ 1.2-2X]^2=  X^2/ [1.2-2X]^2
Then Kc in a square root= x/ 1.2-2x   so sqare root of 9 = 3  =  x/ 1.2-2x
i know that seems complicated but its the best i can do...so what do i do next? how do i calculate that??

[melstiza]

looks good to me, except it should be products/reactants. so flip the denominator and numerator and simply solve for x.

[biomed77]

do i just multiply 3 by 1.2= 3.6? i dont understand what do i do with the -2x

can someone help? thanks