Question

When 1.326 mol of gaseous HI is sealed in a 1.00-L flask at 225°C, it decomposes until the equilibrium amount of I2 present is 0.241 mol: 2HI(g) ⇔ H2(g) + I2(g) Use these data to calculate Keq for this reaction at 225°C.

Answer 1

ICE Table:

                    [HI]                [H2]                [I2]              

initial             1.326               0                   0                 

change              -2x                 +1x                 +1x               

equilibrium         1.326-2x            +1x                 +1x               

Given at equilibrium,
[I2] = 0.241
+1x = 0.241
x = 0.241
Equilibrium constant expression is
Kc = [H2][I2]/[HI]^2
Kc = (+1x)(+1x)/(1.326-2x)^2
Kc = (+1*0.241)(+1*0.241)/(1.326-2*0.241)^2
Kc = 0.0815
Answer: 0.0815