Chapter 21, Problem 121SCQ

(a)

Interpretation Introduction

Interpretation: Draw resonance structures of ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ and it has to be explained that why one $N-O$ bond distance is $114.2pm$, whereas the other two bonds are longer and nearby equal to each other.

Concept introduction:

Nitrogen forms various types of oxides on reaction with oxygen. Some of the common oxides are ${\text{NO}}_{\text{2}}$, $\text{NO}$, ${\text{N}}_{\text{2}}\text{O}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$. The oxide ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ is relatively unstable and decomposes to give $\text{NO}$ and ${\text{NO}}_{\text{2}}$. All these compounds are in gaseous state.

Resonance structures: when a single Lewis structure does not adequately represent a substance, the true structure is intermediate between two or more structures which are called resonance structures and these structures are created by moving electrons, not atoms.

Bond length is the distance between the nuclei in a bond and it is related to the sum of the covalent radii at the bonded atoms.

Bond length is inversely proportional to the bond order.

(b)

Interpretation Introduction

Interpretation: The entropy change and equlibrium constant for the decomposition $N_2{O}_3$ has to be determined.

Concept introduction:

Nitrogen forms various types of oxides on reaction with oxygen. Some of the common oxides are ${\text{NO}}_{\text{2}}$, $\text{NO}$, ${\text{N}}_{\text{2}}\text{O}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$. The oxide ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ is relatively unstable and decomposes to give $\text{NO}$ and ${\text{NO}}_{\text{2}}$. All these compounds are in gaseous state.

Equation for Standard Gibb’s energy change is,

$\begin{array}{l}{\text{Δ}}_{\text{r}}{\text{G}}^{\text{o}}{\text{=Δ}}_{\text{r}}{\text{H}}^{\text{o}}{\text{-TΔ}}_{\text{r}}{\text{S}}^{\text{o}}\\ \\ {\text{Δ}}_{\text{r}}{\text{H}}^{\text{o}}\text{is}\text{the}\text{standard}\text{enthalpy}\text{change}\\ {\text{Δ}}_{\text{r}}{\text{S}}^{\text{o}}\text{is}\text{the}\text{standard}\text{entropy}\text{change}\end{array}$

$\Delta G^{\circ }$ is also related to the equilibrium constant $K$ by the equation,

${\Delta }_{\text{r}}{G}^{\circ }=-RT\ln K_{\text{p}}$

The rearranged expression is,

$K_{\text{p}}=e^{-\frac{{\Delta }_{\text{r}}{G}^{\circ }}{RT}}$

(c)

Interpretation Introduction

Interpretation: The enthalpy of formation of ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ has to be determined.

Concept introduction:

Nitrogen forms various types of oxides on reaction with oxygen. Some of the common oxides are ${\text{NO}}_{\text{2}}$, $\text{NO}$, ${\text{N}}_{\text{2}}\text{O}$ and ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$. The oxide ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ is relatively unstable and decomposes to give $\text{NO}$ and ${\text{NO}}_{\text{2}}$. All these compounds are in gaseous state.

Equation for Standard enthalpy change ${\text{Δ}}_f{\text{H}}^{\text{o}}$ of a reaction can be calculted using the equation,

${\text{Δ}}_{\text{r}}{H}^{\text{°}}=\sum n{\Delta }_f{H}^{\text{°}}\left(\text{products}\right)-\sum n{\Delta }_f{H}^{\text{°}}\left(\text{reactants}\right)$