Chapter 17, Problem 16STP

Interpretation Introduction

Interpretation ;

The concentration of fluorine is to be calculated.

Concept introduction:

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

The formula is given as: there is a chemical reaction

$\text{ aA + bB }\underset{}{\rightleftharpoons }\text{cC + dD }$

$\text{ Keq}$$\text{= }\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 moles of reactants and products.

Answer to Problem 16STP

Correct answer: the concentration of fluorine is $\text{1}{\text{.09×10}}^{\text{-1}}\text{ mol/L}$. So, Option C is correct.

Explanation of Solution

Given information:

The equilibrium constant for the reaction $\text{= 3}{\text{.42×10}}^{\text{-9}}$

The concentration of chlorine $\text{= 0}\text{.563 mol/L}$

The concentration of oxygen $\text{= 1}{\text{.01 mole/L}}_{}$

The concentration of $\left[{\text{ClO}}_{\text{3}}\text{F}\right]\text{ = 1}{\text{.47×10}}^{\text{-5}}$

The given reaction is

${\text{Cl}}_{\text{2}}{\text{+3O}}_{\text{2}}{\text{+F}}_{\text{2}}\underset{}{\rightleftharpoons }{\text{2ClO}}_{\text{3}}\text{F }$

$\text{Keq= }\frac{{\left[{\text{ClO}}_{\text{3}}\text{F}\right]}^{\text{2 }}}{{\left[{\text{O}}_{\text{2}}\right]}^{\text{3}}{\left[{\text{F}}_{\text{2}}\right]}^{}\left[{\text{Cl}}_{\text{2}}\right]}$

$\text{3}{\text{.42×10}}^{\text{-9}}\text{ = }\frac{{\left[\text{1}{\text{.47×10}}^{\text{-5}}\right]}^{\text{2}}}{\left(\text{0}\text{.563}\right)\text{ × }{\left(\text{1}\text{.01}\right)}^{\text{2}}\text{×}\left(\text{ x}\right)}$

$\text{ x = 1}{\text{.09×10}}^{\text{-1}}\text{M}$

The concentration of fluorine is $\text{1}{\text{.09×10}}^{\text{-1}}\text{mol/L}$.