Given the following data: NO2(g) ⟶NO(g)+O(g) Δ H=233 kJ 2 O3(g) ⟶ 3 O2(g) Δ H=-427

Given the following data: NO2(g) ⟶NO(g)+O(g) Δ H=233...

Given the following data: $$ \begin{array}{ll}{\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)} & {\Delta H=233 \mathrm{kJ}} \\ {2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)} & {\Delta H=-427 \mathrm{kJ}}\end{array} $$ $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H=-199 \mathrm{kJ} $$ Calculate the bond energy for the $\mathrm{O}_{2}$ bond, that is, calculate $\Delta H$ for: $$ \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \qquad \Delta H=? $$

Given the following data:
$$
\begin{array}{ll}{\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)} & {\Delta H=233 \mathrm{kJ}} \\ {2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)} & {\Delta H=-427 \mathrm{kJ}}\end{array}
$$
$$
\mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H=-199 \mathrm{kJ}
$$
Calculate the bond energy for the $\mathrm{O}_{2}$ bond, that is, calculate $\Delta H$ for:
$$
\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \qquad \Delta H=?
$$

Key Concepts

Bond Energy

Bond energy is the amount of energy required to break a particular bond in one mole of gaseous molecules. It represents the strength of a chemical bond and is often expressed in kilojoules per mole. In thermochemical calculations, knowing bond energies allows one to estimate the enthalpy changes of reactions by relating the energy required to break bonds in reactants to the energy released upon forming bonds in products.

Hess's Law

Hess's Law states that the total enthalpy change of a chemical reaction is the same regardless of the route by which the reaction occurs, because enthalpy is a state function. This principle allows one to combine or reverse multiple reactions with known enthalpy changes to calculate the enthalpy change of a different reaction, such as determining an unknown bond energy by appropriately summing or subtracting the values from related reactions.

Enthalpy

Enthalpy is a thermodynamic quantity equivalent to the total heat content of a system. It is particularly useful in chemical thermodynamics because rather than tracking all individual energy changes, changes in enthalpy (?H) can express the net heat transfer at constant pressure. This concept is crucial when applying Hess's Law to combine several reactions, as it ensures that the algebraic sum of the overall enthalpy changes correctly corresponds to the overall energy change in a reaction.

Transcript

00:01 In this video, we are going to review how to find delta h for the breakdown of the oxygen molecule, the o2 molecule, into two oxygen atoms.

00:13 So i went ahead, and this is a hesse's law equation again.

00:17 These are pretty common in this unit.

00:19 And we're given three equations, a, b, and c, listed 1, 2, 3, or in sequence in the book.

00:26 So i went ahead and reviewed how i added those equations, such that i, end up with o2 on the left leading to two oxidant atoms on the right.

00:35 And then we subsequently add up all of the enthalpys for each individual intermediate reaction for reactions a through c.

00:46 So i'm going to go ahead and walk you through that here.

00:49 We have a and c are the first two equations i'm going to add together.

00:54 And that is because there are some common species here that we can cancel out if we add these two together, we will end up with 03 on the left, breaks down into o and o2.

01:08 So on the next page, i have that broken down.

01:11 O3 has breaks down into o2 and o gas.

01:16 By the way, i did leave out the subscripts because they're all gaseous molecules and it just makes the screen less busy.

01:22 So you can focus on the compounds.

01:25 From here, what i did is i went ahead and times that new equation times two.

01:32 So i doubled all the coefficients.

01:34 So what we end up with is three, or i'm sorry, two ozone atoms or two ozone molecules.

01:42 That's the 203 on the left.

01:45 Then we have two o2 plus two oxygen atoms.

01:49 The last thing we're going to do is we are going to take the second equation, that's b, and we're going to flip it.

01:57 So we end up with three o2 molecules on the left, with two o3 molecules on the right.

02:03 And the last thing we do is i'm going to go ahead and separate these is we are going to cancel out the common species here.

02:15 So we're going to cancel out the ozone molecules...

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