Question

t 745 K, the reaction below has an equilibrium constant (Kc) of 5.00 × 102. H2 (g) + I2 (g) ⇌ 2 HI (g) If a mixture of 0.10 mol H2, 0.10 mol I2 and 0.50 mol HI is sealed in a 1.00 L flask and heated to 745 K, what is the concentration of HI(g) when equilibrium is established?

Answer 1

Answer : The concentration of HI (g) at equilibrium is, 0.643 M

Explanation :

The given chemical reaction is:

                        $H_2(g)+I_2(g)\rightarrow 2HI(g)$

Initial conc.    0.10        0.10      0.50

At eqm.        (0.10-x)  (0.10-x)   (0.50+2x)

As we are given:

$K_c=5.00\times 10^2$

The expression for equilibrium constant is:

$K_c=\frac{[HI]^2}{[H_2][I_2]}$

Now put all the given values in this expression, we get:

$5.00\times 10^2=\frac{(0.50+2x)^2}{(0.10-x)\times (0.10-x)}$

x = 0.0713  and x = 0.134

We are neglecting value of x = 0.134 because the equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.0713

The concentration of HI (g) at equilibrium = (0.50+2x) = [0.50+2(0.0713)] = 0.643 M

Thus, the concentration of HI (g) at equilibrium is, 0.643 M