Chapter 13, Problem 13.45SP

Interpretation Introduction

Interpretation:

The equilibrium constantfor given reaction has to becalculated.

Concept introduction:

Equilibrium constant$\left({\text{K}}_{\text{p}}\right)$:

In gas phase reactions, partial pressure is used to write equilibrium equation than molar concentration. Equilibrium constant $\left({\text{K}}_{\text{p}}\right)$ is defined as ratio of partial pressure of products to partial pressure of reactants. Each partial pressure term is raised to a power, which is same as the coefficients in the chemical reaction

Consider the reaction where A reacts to give B.

$\text{aA}\rightleftharpoons \text{bB}$

$\begin{array}{l}\text{Rate of forward reaction = Rate of reverse reaction}\\ {\text{k}}_{\text{f}}{\text{P}}_{\text{A}}{}^{\text{a}}{\text{=k}}_{\text{r}}{\text{P}}_{\text{B}}{}^{\text{a}}\end{array}$

On rearranging,

$\frac{{\text{P}}_{\text{A}}{}^{\text{a}}}{{\text{P}}_{\text{B}}{}^{\text{b}}}\text{=}\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{r}}}{\text{=K}}_{\text{p}}$

Where,

${\text{k}}_{\text{f}}$ is the rate constant of the forward reaction.

${\text{k}}_{\text{r}}$ is the rate constant of the reverse reaction.

${\text{K}}_{\text{p}}$ is the equilibrium constant.

The pressure of each species of ideal gas and its molar concentration are directly proportional to each other.

$\begin{array}{l}{\text{P}}_{\text{A}}{\text{V = n}}_{\text{A}}\text{RT}\\ \\ {\text{P}}_{\text{A}}\text{ = }\frac{{\text{n}}_{\text{A}}}{\text{V}}\text{RT}\text{=}\text{[A]RT}\end{array}$

In same way ${\text{P}}_{\text{B}}\text{ =}\text{[B]RT}$. Thus, equilibrium constant ${\text{K}}_{\text{p}}$ is given as

$\begin{array}{l}{\text{K}}_{\text{p}}\text{=}\frac{{\text{[B]}}^{\text{b}}}{{\text{[A]}}^{\text{a}}}\text{×}{\text{(RT)}}^{\text{b-a}}\\ \\ {\text{K}}_{\text{p}}{\text{ = K}}_{\text{c}}{\text{(RT)}}^{\text{Δn}}\end{array}$

Where,

R is gas constant

T is absolute temperature

$\text{Δn}$ is difference in moles of products to reactants in gas phase.