Question

If ${K_c}$ for the formation of HI from ${H_2}$ and ${I_2}$ is 48, then ${K_c}$ for decomposition of 1 mole of HI is _____.
A. $\dfrac{1}{{\sqrt {48} }}$
B. 48
C. 0.144
D. None of these

Answer 1

The dissociation constant for one mole hydrogen iodide is equal to one by under root of dissociation constant of formation of hydrogen iodide from hydrogen and iodine.

Complete step by step answer:
Given,
${K_c}$ for the formation of HI from ${H_2}$ and ${I_2}$ is 48.
Hydrogen and iodine react with each other to form hydrogen iodide.
The reaction between hydrogen and iodine is shown below.
${H_2} + {I_2} \to 2HI$
In this reaction, one mole of hydrogen reacts with one mole of iodine to form two mole of hydrogen iodide.
In a chemical reaction, at equilibrium condition the equilibrium constant of concentration is defined as the ratio of products concentration to the reactants concentration raised to their stoichiometric coefficient. The equilibrium constant of concentration is represented by ${K_c}$.
The ${K_c}$ for the reaction is shown below.
${K_c} = \dfrac{{{{[HI]}^2}}}{{[{H_2}][{I_2}]}} = 48$
The decomposition of 1 mole HI is shown below.
$HI \to \dfrac{1}{2}{H_2} + \dfrac{1}{2}{I_2}$
In this reaction, one mole of hydrogen iodide gives half mole of hydrogen and half mole of iodine.
The ${K_c}$ for the reaction is shown below.
${K_c} = \dfrac{{{{[{H_2}]}^{1/2}}{{[{I_2}]}^{1/2}}}}{{[HI]}}$
The ${K_c}$ is given as shown below.
${K_{c1}} = \dfrac{1}{{\sqrt {{K_c}} }}$
To calculate the dissociation constant for the decomposition of hydrogen iodide, substitute the value in the above equation.
$\Rightarrow {K_{c1}} = \dfrac{1}{{\sqrt {48} }}$
$\Rightarrow {K_{c1}} = 0.144$
Thus the dissociation constant of 1 mol HI is 0.144.
Therefore, the correct options are A and C.

Note: You can see that when one mol of hydrogen and one mole of iodine reacts with each other two mole of hydrogen iodide is formed but on decomposition of one mole hydrogen iodide half mole of hydrogen and oxygen is formed.