Chapter 17, Problem 48A

Interpretation Introduction

Interpretation:

Whether the two reaction with the chemical equilibrium constant at a same temperature are in equilibrium or not needs 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

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

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

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.

Answer to Problem 48A

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

$\text{ A + 2B }\underset{}{\rightleftharpoons }\text{C }$

$\text{Keq= }\frac{{\left[\text{C}\right]}^{}}{\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}}$

Where, [C] = molar concentration of products

[A] and [B] = molar concentration of reactants

The chemical equilibrium constant for both cases is in equilibrium whose value is equal to $\text{3}{\text{.628 M}}^{\text{-2}}$.

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 this is given below,

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

The given chemical reaction iS

$\text{ A + 2B}\underset{}{\rightleftharpoons }\text{ C }$

The equilibrium constant for first case is given as

$\text{Keq}$

$\left(\text{1}\right)\text{ = }\frac{{\left[\text{C}\right]}^{}}{\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}}$

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

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

$\begin{array}{l}\text{Keq }\left(\text{1}\right)\text{ = }\frac{\text{0}\text{.700}}{{\left(\text{0}\text{.62}\right)}^{\text{2}}\text{×}\left(\text{0}\text{.500}\right)}\\ \text{=}\frac{\text{0}\text{.7}}{\left(\text{0}\text{.193}\right)}\\ \text{=3}{\text{.628M}}^{\text{-2}}\end{array}$

Now the equilibrium constant for second case:

$\begin{array}{l}\text{Keq }\left(\text{2}\right)\text{ = }\frac{\text{0}\text{.25}}{\left(\text{0}\text{.25}\right)\text{×}{\left(\text{0}\text{.525}\right)}^{\text{2}}}\\ \text{=3}{\text{.628M}}^{\text{-2}}\end{array}$

Hence, the chemical equilibrium constant for the both cases is in equilibrium.

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

$\text{ A + 2B }\underset{}{\rightleftharpoons }\text{C }$

$\text{Keq= }\frac{{\left[\text{C}\right]}^{}}{\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}}$

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

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

The chemical equilibrium constant for both cases is in equilibrium whose value is equal to $\text{ 3}{\text{.628 M}}^{\text{-2}}$.