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Question: In a closed tube \(HI\left( g \right)\) is heated at \({440^ \circ }C\) up to establishment of equil...

In a closed tube HI(g)HI\left( g \right) is heated at 440∘C{440^ \circ }C up to establishment of equilibrium. If it dissociates into H2(g){H_2}\left( g \right) and I2(g){I_2}\left( g \right) up to 22%22\% , the dissociation constant is.
A)0.2820.282
B) 0.07960.0796
C) 0.01990.0199
D) 1.991.99

Explanation

Solution

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,
AxBy⇌xA+yB{A_x}{B_y} \rightleftharpoons xA + yB
The dissociation constant of the reaction is calculate by using the below formula,
Kd=[A]x[B]y[AxBy]{K_d} = \dfrac{{{{\left[ A \right]}^x}{{\left[ B \right]}^y}}}{{\left[ {{A_x}{B_y}} \right]}}
The equilibrium concentrations of A, B are [A]\left[ A \right],[B]\left[ B \right], and [AxBy]\left[ {{A_x}{B_y}} \right].

Complete step by step answer:
First, write the dissociation reaction,
2HI⇆kH2+I22HI\overset k \leftrightarrows {H_2} + {I_2}
Let us take the initial concentration of the hydrogen iodide as 100.100.
The concentration of hydrogen iodide at equilibrium 100−22=78100 - 22 = 78.
We know that the initial concentration of the hydrogen and iodide is 22.22.
K=[H2][I2][HI]2K = \dfrac{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}}{{{{\left[ {HI} \right]}^2}}}
Substituting the values we get,
⇒\Rightarrow K=22×22782K = \dfrac{{22 \times 22}}{{{{78}^2}}}
On simplifying we get,
⇒\Rightarrow K=0.0796K = 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 pHpH of the reaction,
Example:
We can write the dissociation equation of the reaction.
HA+H2O→H3O++A−HA + {H_2}O\xrightarrow{{}}{H_3}{O^ + } + {A^ - }
The constant Ka{K_a} of the solution is4×10−24 \times {10^{ - 2}}.
The dissociation constant of the reaction Ka{K_a} is written as,
Ka=[H3O+][A−][HA]{K_a} = \dfrac{{\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right]}}{{\left[ {HA} \right]}}
Let us imagine the concentration of [H3O+][A−]\left[ {{H_3}{O^ + }} \right]\left[ {{A^ - }} \right] as x.
⇒\Rightarrow 4×10−7=x20.08−x4 \times {10^{ - 7}} = \dfrac{{{x^2}}}{{0.08 - x}}
⇒x2=4×10−7×0.08\Rightarrow {x^2} = 4 \times {10^{ - 7}} \times 0.08
⇒x=1.78×10−4\Rightarrow x = 1.78 \times {10^{ - 4}}
The concentration of Hydrogen is 1.78×10−41.78 \times {10^{ - 4}}
We can calculate the pHpH of the solution is,
pH=−log⁡[H+]=3.75pH = - \log \left[ {{H^ + }} \right] = 3.75
The pHpH of the solution is 3.753.75.