Question

At a certain temperature only 50% HI is dissociated at equilibrium in the reaction 2HI(g)<—-> H2(g)+I2(g) the equilibrium constant for reaction is10.50.25

Answer 1

Answer: The equilibrium constant for the given reaction is 0.25

Explanation: We are given a chemical reaction, in which 50% of the reactant dissociates.

The reaction follows:

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

At t = 0                        x            0       0

At $t=t_{eq}$         $[(x-\frac{x}{2})=\frac{x}{2}]$       $\frac{x}{4}$        $\frac{x}{4}$

The equilibrium constant, $k_c$ for the reaction is:

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

Putting values in above equation, we get:

$k_c=\frac{(\frac{x}{4})(\frac{x}{4})}{(\frac{x}{2})^2}$

$k_c=0.25$

Answer 2

Answer: The dissociation constant is 0.25.

Explanation: We are given a chemical reaction, in which 50% of the reactant dissociates.

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

Initial                   c            0       0

Eq: $[(c-c\alpha)$   $\frac{c\alpha}{2}]$ $\frac{c\alpha}{2}$      

The equilibrium constant, $k_a$ for the reaction is the ratio of product of concentration of products to the product of concentration of reactants each term raised to their stochiometric coefficients.

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

Putting values in above equation, we get:

$k_a=\frac{(\frac{c\alpha}{2})(\frac{c\alpha}{2})}{(c-c\alpha)^2}$

$k_a=\frac{(\alpha)^2}{4\times (1-\alpha)^2}$

Putting value of $\alpha=0.5$ we get

$k_a=0.25$