Chapter 17, Problem 71A

Interpretation Introduction

Interpretation:

The concentration of hydrogen iodide is to be calculated.

Concept introduction:

The equilibrium constant is the ratio of molar concentration of products to the reactants. It is represented by ${\text{K}}_{\text{eq}}$.

The formula is given as: there is a chemical reaction

$\text{ aA+Bb}\rightleftharpoons \text{cC+dD }$

$\text{Keq= }\frac{{\left[\text{C}\right]}^{\text{C}}{\left[\text{D}\right]}^{\text{d }}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

Where, $\left[\text{C}\right]$ and $\left[\text{D}\right]$ = molar concentration of products

$\left[\text{A}\right]$ and $\left[\text{B}\right]$ = molar concentration of reactants

a, b, c, and d are number of molecules.

Answer to Problem 71A

The value of equilibrium constant $\left(\text{Keq}\right)$ is $\text{3}{\text{.24×10}}^{\text{-3}}\text{mol/L}$.

Explanation of Solution

Given information: the given reaction is

$\text{2HI}\left(\text{g}\right)\rightleftharpoons {\text{H}}^{\text{2}}\left(\text{g}\right){\text{ + I}}^{\text{2}}\left(\text{g}\right)$

$\left[{\text{H}}^{\text{2}}\right]\text{ = 2}{\text{.44×10}}^{\text{-3}}$ mol/L

$\left[{\text{I}}^{\text{2}}\right]\text{ = 7}{\text{.18×10}}^{\text{-5}}$ mol/L

${\text{K}}_{\text{eq}}\text{=1}{\text{.67×10}}^{\text{-2}}$

According to the Le Chatelier’s principle, the equilibrium constant is the ratio of molar concentration of products to the molar concentration of reactants. It is represented by $\text{Keq}$.

$\begin{array}{l}\text{Keq = }\frac{\left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]}{{\left[\text{HI}\right]}^{\text{2}}}\\ {\left[\text{HI}\right]}^{\text{2}}\text{=}\frac{\left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]}{\text{Keq}}\\ \text{=}\frac{\text{2}{\text{.44×10}}^{\text{-3}}\text{×7}{\text{.18×10}}^{\text{-5}}}{\text{1}{\text{.67×10}}^{\text{-2}}}\\ \text{=}\frac{\text{17}{\text{.52×10}}^{\text{-6}}}{\text{1}\text{.67}}\\ \text{=10}{\text{.5×10}}^{\text{-6}}\end{array}$

Or, $\begin{array}{l}\left[\text{HI}\right]\text{=}\sqrt{\text{10}{\text{.5×10}}^{\text{-6}}}\\ \text{=3}{\text{.24×10}}^{\text{-3}}\text{M}\end{array}$

The concentration of hydrogen iodide is $\text{3}{\text{.24×10}}^{\text{-3}}\text{M}$.

Conclusion

The concentration of hydrogen iodide is $\text{3}{\text{.24×10}}^{\text{-3}}\text{M}$.