Chapter 12, Problem 90QRT

(a)

Interpretation Introduction

Interpretation:

The equilibrium concentrations of the atomic form of each element at $\text{1500}\text{K}$ 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.

Explanation of Solution

Using ICE table to calculate x value.

The concentration of ${\text{E}}_{\text{2}}$ is $\text{=}\frac{\text{1}\text{.00}\text{mol}}{\text{1}\text{.0}\text{L}}\text{=1}\text{.0}\text{M}$

    $\begin{array}{l}{\text{E}}_{\text{2(g)}}\underset{}{\rightleftharpoons }{\text{2E}}_{\text{(g)}}\\ \text{initial}\text{1}\text{.0}\text{0}\\ \text{change}\text{-x}\text{+2x}\\ \text{equilibrium}\text{1}\text{.0-x}\text{2x}\end{array}$

At equilibrium,

    ${\text{K}}_{\text{c}}\text{=}\frac{{\text{[E]}}^{\text{2}}}{{\text{[E}}_{\text{2}}\text{]}}\text{=}\frac{{\text{(2x)}}^{\text{2}}}{\text{1}\text{.0-x}}$

X value is calculated as,

    $\begin{array}{c}{\text{(2x)}}^{\text{2}}{\text{=K}}_{\text{2}}\text{(1}\text{.0-x)}\text{=}{\text{K}}_{\text{c}}{\text{-K}}_{\text{c}}\text{x}\\ \text{0}\text{=}{\text{4x}}^{\text{2}}\text{+}{\text{K}}_{\text{c}}{\text{x-K}}_{\text{c}}\\ \text{x}\text{=}\frac{\text{-b±}\sqrt{{\text{b}}^{\text{2}}\text{-4ac}}}{\text{2(4)}}\text{=}\frac{{\text{-K}}_{\text{c}}\text{±}\sqrt{{\text{K}}_{\text{c}}^{\text{2}}{\text{-4(K}}_{\text{2}}{\text{)(-K}}_{\text{c}}\text{)}}}{\text{2(4)}}\\ \text{[E]=}\text{2x}\end{array}$

Taken an example of the Bromine,

    $\begin{array}{c}\text{x}\text{=}\frac{\text{-(8}{\text{.9×10}}^{\text{-2}}\text{)-}\sqrt{\text{(8}{\text{.9×10}}^{\text{-2}}\text{)-4(8}{\text{.9×10}}^{\text{-2}}\text{)(8}{\text{.9×10}}^{\text{-2}}\text{)}}}{\text{2(4)}}\\ \text{=}\frac{\text{1}\text{.1}}{\text{8}}\text{=0}\text{.14}\\ \text{=}\text{2×(0}\text{.14M)}\\ \text{=}\text{0}\text{.28}\text{M}\end{array}$

From the above same method is used to other elements to get equilibrium concentrations of the atomic form of each element at $\text{1500}\text{K}$.

Hence, the equilibrium concentrations of the atomic form of each element at $\text{1500}\text{K}$ are,

    $\begin{array}{l}\text{E}\text{species}{\text{K}}_{\text{c}}\text{x(M)}\text{[E](m)}\\ \text{Br}\text{8}{\text{.9×10}}^{\text{-2}}\text{0}\text{.14}\text{0}\text{.28}\\ \text{Cl}\text{3}{\text{.4×10}}^{\text{-3}}\text{0}\text{.029}\text{0}\text{.057}\\ \text{F}\text{7}\text{.4}\text{0}\text{.72}\text{1}\text{.44}\\ \text{H}\text{3}{\text{.1×10}}^{\text{-10}}\text{8}{\text{.8×10}}^{\text{-6}}\text{1}{\text{.76×10}}^{\text{-5}}\\ \text{N}{\text{1×10}}^{\text{-27}}{\text{2×10}}^{\text{-14}}{\text{4×10}}^{\text{-14}}\\ \text{O}\text{1}{\text{.6×10}}^{\text{-10}}\text{2}{\text{.0×10}}^{\text{-6}}{\text{4×10}}^{\text{-6}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The element which has lowest bond dissociation energy has to be predicted by comparing thermodynamic calculations and Lewis structure.

Explanation of Solution

In generally, a molecule which one have lowest enthalpy value, it is give a more product so the enthalpy vales of given molecules are,

    $\begin{array}{l}{\text{F}}_{\text{2}}\text{-158 kJ}{\text{436 kJ H}}_{\text{2}}\\ {\text{Br}}_{\text{2}}\text{-193 kJ}{\text{498 kJ O}}_{\text{2}}\\ {\text{Cl}}_{\text{2}}\text{-242 kJ }{\text{946 kJ N}}_{\text{2}}\end{array}$

The calculated equilibrium concentrations of the atomic form of each element at $\text{1500}\text{K}$ are,

    $\begin{array}{l}\text{E}\text{species}{\text{K}}_{\text{c}}\text{x(M)}\text{[E](m)}\\ \text{Br}\text{8}{\text{.9×10}}^{\text{-2}}\text{0}\text{.14}\text{0}\text{.28}\\ \text{Cl}\text{3}{\text{.4×10}}^{\text{-3}}\text{0}\text{.029}\text{0}\text{.057}\\ \text{F}\text{7}\text{.4}\text{0}\text{.72}\text{1}\text{.44}\\ \text{H}\text{3}{\text{.1×10}}^{\text{-10}}\text{8}{\text{.8×10}}^{\text{-6}}\text{1}{\text{.76×10}}^{\text{-5}}\\ \text{N}{\text{1×10}}^{\text{-27}}{\text{2×10}}^{\text{-14}}{\text{4×10}}^{\text{-14}}\\ \text{O}\text{1}{\text{.6×10}}^{\text{-10}}\text{2}{\text{.0×10}}^{\text{-6}}{\text{4×10}}^{\text{-6}}\end{array}$

From the above, data the lowest enthalpy containing molecule is yields more product.

From the above, Lewis structures and concentrations of elements, the single bonded compound forms more products.