The standard enthalpy of formation for N2 H4(g) is 95.4 kJ / mol. Use this and the data in Exer

The standard enthalpy of formation for N2 H4(g) is 95.4 kJ...

The standard enthalpy of formation for $\mathrm{N}_{2} \mathrm{H}_{4}(g)$ is $95.4 \mathrm{~kJ} / \mathrm{mol}$. Use this and the data in Exercise 76 to estimate the $\mathrm{N}-\mathrm{N}$ single bond energy. Compare this with the value in Table $8.4$.

Key Concepts

Hess’s Law

Hess’s law states that the total enthalpy change for a reaction is the same, no matter how the reaction occurs, because enthalpy is a state function. This principle allows for the estimation of bond energies by breaking complex reactions into simpler steps whose enthalpy changes are known.

Thermochemical Calculations

These calculations involve determining the energy changes in chemical reactions using tabulated standard enthalpies, enthalpy of formation data, and bond energies. They form the basis for predicting reaction energetics and understanding the energy required to break or form chemical bonds.

Enthalpy of Formation

This is the heat change that occurs when one mole of a compound is formed from its elements in their standard states. It provides a fundamental measure of the energy content of a compound relative to its constituent elements and is used to calculate reaction enthalpies using Hess’s law.

Bond Energy (Bond Dissociation Energy)

Bond energy is the amount of energy required to break a bond between two atoms in a molecule in the gas phase. It represents a measure of bond strength and is crucial for understanding reaction energetics and stability of various compounds.

Transcript

00:02 Okay, today we're looking at calculating a bond energy from standard enthalpyes of formation.

00:15 So first we need to figure out what reaction is going on here.

00:23 So we have n2h4 being formed, and it's coming from n2 and h.

00:40 These are the values that we have from the previous problem.

00:44 So my nitrogen levels are balanced with my hydrogens or not, so i'll need two hydrogens.

00:52 And what we're calculating is this single nitrogen -nitrogen bond here in the middle of this structure.

01:07 So what we'll be breaking or what we'll be doing is we have this equation up here.

01:15 We know the delta h formation is going to be equal to bonds broken minus bonds.

01:20 Formed and that nitrogen -nitrogen bond is going to be one of the bonds formed.

01:25 So we're going to estimate how much energy that bond has by solving for it.

01:31 So first we're only making one and two di - nitrogen tetroxide, so 95 .4 is our enthalpy of formation.

01:46 And bonds broken, we're breaking one and two bond, and we're breaking two, h2 bonds and this is going to be subtracted by the bonds formed.

02:06 So we're forming 4 nh bonds, which we estimated was around 202 from the previous problem and we're also forming 1 n -n bond which is what we're solving for.

02:29 So i'm going to use x there.

02:32 Okay, so now let's simplify in order to solve.

02:37 So 95 .4 equals 472 .7 plus 216 times 2.

02:46 So this comes out to 904 .7...

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