[FollowTheFez]

The Question
HI gas at a concentration of 2.0mol/L is placed in a flask and deated to 628°C. The following equilibrium is established...

2HI    H₂  =  I₂

Given that at this temperature K_c =0.0380, calculate the concentration of I₂ which will be present at equilibrium.

Note that all products and reactants are gaseous.


My Attempt

I think I've got most of the working correct but I just can't seem to get the correct answer (which is 0.281mol/L).


Initial Concentrations      [HI]=2mol/L   [H₂]=0mol/L   [I₂]=0mol/L

Change in conc. in reaching equilibrium     HI = -2x     H₂ = x     I₂ = x

Equilibrium conc.     [HI] = 2-2x     [H₂] = x     [I₂] = x

Equilibrium expression     K_c = ([H₂][I₂]) / [HI]^2

Substituting into K_c      0.0380 = (x × x) / (2-2x)^2

Square rooting both sides    0.1949 = x / 2-2x

Rearranging and calculating    0.1949(2-2x) = x

0.3898-0.3898x=x
0.3898 = 0.3898x
1=x

Therefore, [I_2] at equilibrium will be 1mol/L. However the textbook answer give it as 0.281mol/L.

[iamback]

0.3898-0.3898x=x
0.3898 = 0.3898x
1=x

Simple math error.

0.3898-0.3898x=x
0.3898 = 0.3898x + x
0.3898 = 1.3898x

x= 0.3898/1.3898
x= 0.281mol/L.