Chapter 17.1, Problem 11SSC

Interpretation Introduction

Interpretation:

The value of chemical equivalent equilibrium constant (Keq) is to be calculated.

Concept introduction:

The equilibrium constant is the ratio of molar concentration of products to the reactants. It is represented by 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 molecules.

Answer to Problem 11SSC

The value of equilibrium constant $\left(\text{Keq}\right)$ is $\text{2}\text{.241}$.

Explanation of Solution

Given data: the given reaction is

${\text{PCl}}_{\text{5}}\left(\text{g}\right)\rightleftharpoons {\text{PCl}}_{\text{3}}\left(\text{g}\right){\text{ + Cl}}_{\text{2}}\left(\text{g}\right)$

$\left[{\text{PCl}}_{\text{3}}\right]\text{ = 0}\text{.550}$ mol/L

$\left[{\text{PCl}}_{\text{5}}\right]\text{ = 0}\text{.135}$ mol/L

$\left[{\text{Cl}}_{\text{2}}\right]\text{ = 0}\text{.550}$ mol/L

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 $\text{Keq}$.

$\text{Keq = }\frac{\left[{\text{PCl}}_{\text{3}}\right]\left[{\text{Cl}}_{\text{2}}\right]}{\left[{\text{PCl}}_{\text{5}}\right]}$

$\text{=}\frac{\text{0}\text{.550 × 0}\text{.550}}{\text{0}\text{.135}}$

$\text{ = 2}\text{.241}$

The equilibrium constant is $\text{2}\text{.2}\text{.41}$.

Conclusion

The value of equilibrium constant $\left(\text{Keq}\right)$ is $\text{2}\text{.241}\text{.}$