Chapter 17, Problem 103A

Interpretation Introduction

Interpretation:

The equilibrium constant expression of a reversible chemical equation is to be described.

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 103A

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

Explanation of Solution

Consider an example of a reversible reaction which is given below

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

$\left[{\text{H}}^{\text{2}}\right]\text{ and }\left[{\text{I}}^{\text{2}}\right]$ are the concentration of products and $\left[\text{HI}\right]$ is the concentration of reactant.

According to Law Mass of Action,

$\begin{array}{l}\text{rate}\text{of}\text{forward}\text{reaction}\text{µ}{\left[\text{HI}\right]}^{\text{2}}\\ {\text{=K}}_{\text{f}}{\left[\text{HI}\right]}^{\text{2}}\end{array}$

.where ${\text{K}}_{\text{f}}$ is the rate constant for forward reaction.

Similarly for backward reaction

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

$\begin{array}{l}\text{rate}\text{of}\text{backward}\text{reaction}\propto \left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]\\ {\text{=K}}_{\text{b}}\left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]\end{array}$

At equilibrium are of forward reaction is equal to rate of backward reaction

$\begin{array}{l}\text{rate}\text{of}\text{forward}\text{reaction}\text{=rate}\text{of}\text{backward}\text{reaction}\\ {\text{K}}_{\text{f}}{\left[\text{HI}\right]}^{\text{2}}{\text{=K}}_{\text{b}}\left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]\\ \frac{{\text{K}}_{\text{f}}}{{\text{K}}_{\text{b}}}\text{=}\frac{\left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]}{{\left[\text{HI}\right]}^{\text{2}}}\\ \text{K=}\frac{\left[{\text{H}}^{\text{2}}\right]\left[{\text{I}}^{\text{2}}\right]}{{\left[\text{HI}\right]}^{\text{2}}}\end{array}$

Where K is called the equilibrium constant.

Conclusion

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