Although nitrogen trifluoride (NF3) is a thermally stable...

Although nitrogen trifluoride $\left(\mathrm{NF}_{3}\right)$ is a thermally stable compound, nitrogen triiodide $\left(\mathrm{NI}_{3}\right)$ is known to be a highly explosive material. $\mathrm{NI}_{3}$ can be synthesized according to the equation $$ \mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g) $$ a. What is the enthalpy of formation for $\mathrm{NI}_{3}(s)$ given the enthalpy of reaction $(-307 \mathrm{~kJ})$ and the enthalpies of formation for $\mathrm{BN}(s)(-254 \mathrm{~kJ} / \mathrm{mol}), \mathrm{IF}(g)(-96 \mathrm{~kJ} / \mathrm{mol})$, and $\mathrm{BF}_{3}(g)(-1136$ $\mathrm{kJ} / \mathrm{mol}$ ? b. It is reported that when the synthesis of $\mathrm{NI}_{3}$ is conducted using 4 mol IF for every 1 mol BN, one of the by-products isolated is $\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .$ What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

Although nitrogen trifluoride $\left(\mathrm{NF}_{3}\right)$ is a thermally stable compound, nitrogen triiodide $\left(\mathrm{NI}_{3}\right)$ is known to be a highly explosive material. $\mathrm{NI}_{3}$ can be synthesized according to the equation
$$
\mathrm{BN}(s)+3 \mathrm{IF}(g) \longrightarrow \mathrm{BF}_{3}(g)+\mathrm{NI}_{3}(g)
$$
a. What is the enthalpy of formation for $\mathrm{NI}_{3}(s)$ given the enthalpy of reaction $(-307 \mathrm{~kJ})$ and the enthalpies of formation for $\mathrm{BN}(s)(-254 \mathrm{~kJ} / \mathrm{mol}), \mathrm{IF}(g)(-96 \mathrm{~kJ} / \mathrm{mol})$, and $\mathrm{BF}_{3}(g)(-1136$
$\mathrm{kJ} / \mathrm{mol}$ ?
b. It is reported that when the synthesis of $\mathrm{NI}_{3}$ is conducted using 4 mol IF for every 1 mol BN, one of the by-products isolated is $\left[\mathrm{IF}_{2}\right]^{+}\left[\mathrm{BF}_{4}\right]^{-} .$ What are the molecular geometries of the species in this by-product? What are the hybridizations of the central atoms in each species in the by-product?

Key Concepts

Molecular Geometry (VSEPR Theory)

Molecular geometry is the three-dimensional arrangement of atoms within a molecule. VSEPR (Valence Shell Electron Pair Repulsion) theory provides a model for predicting this arrangement by minimizing electron pair repulsions around a central atom, which in turn determines properties like bond angles and overall shape.

Reaction Stoichiometry in Thermochemistry

Reaction stoichiometry involves the use of the balanced chemical equation to determine the relative amounts of reactants and products. In thermochemical calculations, the stoichiometric coefficients are crucial for relating the enthalpy changes of individual species to the overall reaction enthalpy.

Hybridization

Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals that are used to form bonds in molecules. The type of hybridization adopted by the central atom is directly related to the observed molecular geometry, influencing the shape and bond properties of the molecule.

Standard Enthalpy of Formation

This is the heat change that occurs when one mole of a compound is formed from its constituent elements in their standard states. It is a fundamental thermochemical property used to compare the stability of different compounds and to calculate reaction enthalpies.

Hess's Law

Hess's Law states that the total enthalpy change for a reaction is the same regardless of the pathway taken, as enthalpy is a state function. This principle allows the calculation of unknown enthalpy values by combining known enthalpy changes from other reactions.

Transcript

00:01 Please consider nitrogen triiodide.

00:07 And our first order of business is to calculate the enthalpya formation for the nitrogen triadide given the information you see right here.

00:17 Let me begin.

00:20 Pretty simple.

00:25 So i'm going to have my will be equal to some of my heats of formation for my products minus the sum.

00:54 Of my heats of formation of my reactants.

01:01 Okay, so let me start this.

01:08 This will be equal to.

01:10 So first my products for my bf3, i have one mole times my bf3 is negative 1136 kilojoules per mole.

01:29 And i forgot a very very important thing.

01:31 This is given.

01:33 So this is negative 307 kilojoules.

01:42 We're going to add to this.

01:44 This is my n -i -3.

01:47 That's my unknown.

02:08 And then we're going to subtract from that.

02:10 I'm going to move this down the line.

02:14 My first, my b -n will be, did i forget to balance this whole thing? no, i should be good.

02:24 I only have one coefficient.

02:26 Excellent.

02:28 And this will be one mole times my b -n is negative 254 kilojoules per mole.

02:39 And then we'll add to that three moles times my if, which is negative 96 kilojoules per mole.

02:54 Keep track of your signs very carefully here.

02:56 For instance, this sign will be turning into a minus, obviously, and then plus...

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