Question

In a closed tube $HI\left( g \right)$ is heated at ${440^ \circ }C$ up to establishment of equilibrium. If it dissociates into ${H_2}\left( g \right)$ and ${I_2}\left( g \right)$ up to $22\% $, the dissociation constant is.
A)$0.282$
B) $0.0796$
C) $0.0199$
D) $1.99$

Answer 1

Hint:We know that a dissociation constant is an equilibrium constant which measures the tendency of a bigger object to separate reversibly into smaller components; the equilibrium constant also can be called an ionization constant.
For a general reaction can be written as,
${A_x}{B_y} \rightleftharpoons xA + yB$
The dissociation constant of the reaction is calculate by using the below formula,
${K_d} = \dfrac{{{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}{{\left[ {{A_x}{B_y}} \right]}}$
The equilibrium concentrations of A, B are \[\left[ A \right]\],\[\left[ B \right]\], and \[\left[ {{A_x}{B_y}} \right]\].

Complete step by step answer:
First, write the dissociation reaction,
$2HI\overset k \leftrightarrows {H_2} + {I_2}$
Let us take the initial concentration of the hydrogen iodide as $100.$
The concentration of hydrogen iodide at equilibrium $100 - 22 = 78$.
We know that the initial concentration of the hydrogen and iodide is $22.$
$K = \dfrac{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}{{{{\left[ {HI} \right]}^2}}}$
Substituting the values we get,
$ \Rightarrow $$K = \dfrac{{22 \times 22}}{{{{78}^2}}}$
On simplifying we get,
$ \Rightarrow $$K = 0.0796$

Therefore, the option A is correct.

Note: We must remember that the concept of the equilibrium constant is applied in various fields of chemistry and pharmacology. In protein-ligand binding the equilibrium constant describes the affinity between a protein and a ligand. A little equilibrium constant indicates a more tightly bound the ligand. Within the case of antibody-antigen binding the inverted equilibrium constant is employed and is named affinity constant.
Using the equilibrium constant we can calculate the $pH$ of the reaction,
Example:
We can write the dissociation equation of the reaction.
$HA + {H_2}O\xrightarrow{{}}{H_3}{O^ + } + {A^ - }$
The constant ${K_a}$ of the solution is$4 \times {10^{ - 2}}$.
The dissociation constant of the reaction ${K_a}$ is written as,
${K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}$
Let us imagine the concentration of \[\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]\] as x.
$ \Rightarrow $$4 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.08 - x}}$
$ \Rightarrow {x^2} = 4 \times {10^{ - 7}} \times 0.08$
$ \Rightarrow x = 1.78 \times {10^{ - 4}}$
The concentration of Hydrogen is $1.78 \times {10^{ - 4}}$
We can calculate the $pH$ of the solution is,
$pH = - \log \left[ {{H^ + }} \right] = 3.75$
The $pH$ of the solution is $3.75$.