A student ran the following reaction in the laboratory at 68

A student ran the following reaction in the laboratory at 682 K: H2(g) + I2(g) 2HI(g) When she introduced 0.219 moles of H2(g) and 0.244 moles of I2(g) into a 1.00 liter container, she found the equilibrium concentration of HI(g) to be 0.364 M. Calculate the equilibrium constant, Kc, she obtained for this reaction. Kc =

Solution

since volume is 1 L,
initial concentration of H2 = 0.219 M
initial concentration of I2 = 0.244 M
ICE Table:

                    [H2]                [I2]                [HI]              

initial             0.219               0.244               0                 

change              -1x                 -1x                 +2x               

equilibrium         0.219-1x            0.244-1x            +2x               

Given at equilibrium,
[HI] = 0.364
+2x = 0.364
x = 0.182
Equilibrium constant expression is
Kc = [HI]^2/[H2][I2]
Kc = (+2x)^2/(0.219-1x)(0.244-1x)
Kc = (+2*0.182)^2/(0.219-1*0.182)(0.244-1*0.182)
Kc = 57.8
Answer: 57.8