4) Hydrogen iodide decomposes according to the equation 2HI(g) –><– H2(g) + I2(g), for which K= 0.0156 at 400C. .550 mol HI was injected into a 2.00 L reaction vessel at 400C. Calculate the concentration of H2 at equilibrium.
A) .275 M
B) .138 M
C) .0550 M
D) .220 M
2 Answers

2HI(g) <===> H2(g) + I2(g)
ICE table to tract concentrations @ equilibrium:
. . . . . . HI. . . . . . . H2. . . . . I2
I. . . . 0.275. . . . . . 0. . . . . . 0
C. . . . x. . . . . . . . .+x. . . . . +x
E. . . 0.2752x. . . . .x. . . . . . x
Kc = [H2]*[I2] / [HI]²
Kc = x*x / (0.275 – 2x)²
Kc = x² / (0.275 – 2x)² …..Take the square root of both sides
0.0156 = x / 0.275 – 2x
x = 0.027
From our ICE table, the concentration of H2 @ equilibrium was x.
Thus, at equilibrium, the concentration of H2 = 0.027 M, none of these answers match exactly, but C is closest.

D 0.220