Chapter 12, Problem 89QRT

(a)

Interpretation Introduction

Interpretation:

The sign of ${\text{Δ}}_{\text{r}}\text{H}°$ for given reaction positive or negative has to be given.

  ${\text{H}}_{\text{2}}{}_{\text{(g)}}{\text{+Br}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{2HBr}}_{\text{(g)}}$

(b)

Interpretation Introduction

Interpretation:

The value of ${\text{K}}_{\text{c}}$ given reaction has to be given.

  $\frac{1}{2}{\text{H}}_{\text{2}}{}_{\text{(g)}}\text{+}\frac{1}{2}{\text{Br}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{HBr}}_{\text{(g)}}$

(c)

Interpretation Introduction

Interpretation:

The percentage decomposition of $\text{HBr}$ into ${\text{H}}_{\text{2}}$ and ${\text{Br}}_{\text{2}}$ at given equilibrium condition 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,

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

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

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