Chapter 17, Problem 38A

Interpretation Introduction

Interpretation:

The equilibrium constant expression is to be explained.

Concept introduction:

According to law of mass action, 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

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

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

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 moles.

Answer to Problem 38A

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

considered 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 moles.

Explanation of Solution

According to Law of mass action, the equilibrium constant is the ratio of molar concentration of products to the molar concentration of reactants. It is represented by $\text{Keq}$.

Considered a reversible reaction which is given below

$\text{A + B}\underset{}{\rightleftharpoons }\text{ C + D }$

According to law of mass action,

the rate of forward reaction $\propto \left[\text{A}\right]\text{ ×}\left[\text{B}\right]$

${\text{= k}}_{\text{f}}\left[\text{A}\right]\text{×}\left[\text{B}\right]$

Where is ${\text{k}}_{\text{f}}$ is rate constant for forward reaction

And, the rate of backward reaction $\propto \left[\text{C}\right]\text{×}\left[\text{D}\right]$

${\text{= k}}_{\text{b}}\left[\text{C}\right]\text{×}\left[\text{D}\right]$

Where, k_b is rate constant for backward reaction.

Now at equilibrium position, both rates are equal.

So,

${\text{k}}_{\text{f}}\left[\text{A}\right]\left[\text{B}\right]{\text{ = k}}_{\text{b}}\left[\text{C}\right]\left[\text{D}\right]$

$\frac{{\text{k}}_{\text{f}}}{\text{Kb}}\text{=}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

${\text{k =}}_{}\frac{\left[\text{C}\right]\left[\text{D}\right]}{\left[\text{A}\right]\left[\text{B}\right]}$

Conclusion

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

Considered a chemical reaction

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

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

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 moles.