Chapter 17.5, Problem 17.8AFP

Interpretation Introduction

Interpretation:

The equilibrium partial pressure of $\text{NO}$ in the air we breathe has to be calculated.

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,

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

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.

Relation between ${\text{K}}_{\text{c}}\text{ and K}{\text{p}}_{}$:

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

Where,

     R is gas constant

    T is absolute temperature

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