Question

Consider the reaction: 2HI(g) ⇄ H2(g) + I2(g). It is found that, when equilibrium is reached at a certain temperature, HI is 35.4 percent dissociated. Calculate the equilibrium constant Kc for the reaction at this temperature?

Answer 1

Answer:

Kc = 0.075

Explanation:

The dissociation (α) is the initial quantity that ionized divided by the total dissolved. So, let's calling x the ionized quantity, and M the initial one:

α = x/M

x = M*α

x = 0.354M

For the stoichiometry of the reaction (2:1:1), the concentration of H₂ and I₂ must be half of the acid. So the equilibrium table must be:

2HI(g) ⇄   H₂(g) +    I₂(g)

M               0             0               Initial

-0.354M  +0.177M  +0.177M       Reacts

0.646M     0.177M   0.177M        Equilibrium

The equilibrium constant Kc is the multiplication of the products' concentrations (elevated by their coefficients) divided by the multiplication of the reactants' concentrations (elevated by their coefficients):

$Kc = \frac{[H2]*[I2]}{[HI]^2}$

$Kc = \frac{0.177M*0.177M}{(0.646M)^2}$

$Kc = \frac{0.03133M^2}{0.41732M^2}$

Kc = 0.075