Challenge Two enthalpy of formation equations, a and b,...

Challenge Two enthalpy of formation equations, a and b, combine to form the equation for the reaction of nitrogen oxide and oxygen. The product of the reaction is nitrogen dioxide: $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) \Delta H_{\mathrm{mn}}^{\circ}=-58.1 \mathrm{kJ}$ $$\begin{array}{l}{\text { a. } \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) \Delta H_{\mathrm{f}}^{\circ}=91.3 \mathrm{kJ}} \\ {\text { b. } \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) \Delta H_{\mathrm{f}}^{\circ}=?} \\ {\text { What is } \Delta H_{\mathrm{f}}^{\circ} \text { for Equation } \mathrm{b} ?}\end{array}$$

Challenge Two enthalpy of formation equations, a and b, combine to form the equation
for the reaction of nitrogen oxide and oxygen. The product of the reaction is nitrogen
dioxide: $\mathrm{NO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) \Delta H_{\mathrm{mn}}^{\circ}=-58.1 \mathrm{kJ}$
$$\begin{array}{l}{\text { a. } \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}(\mathrm{g}) \Delta H_{\mathrm{f}}^{\circ}=91.3 \mathrm{kJ}} \\ {\text { b. } \frac{1}{2} \mathrm{N}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{NO}_{2}(\mathrm{g}) \Delta H_{\mathrm{f}}^{\circ}=?} \\ {\text { What is } \Delta H_{\mathrm{f}}^{\circ} \text { for Equation } \mathrm{b} ?}\end{array}$$

Key Concepts

Enthalpy of Formation

This is the heat change that accompanies the formation of one mole of a compound from its elements in their most stable forms at standard conditions. It is used as a reference to calculate the energy changes in formation reactions and overall reaction enthalpies.

Hess's Law

A central concept stating that the total enthalpy change of a reaction is path independent, meaning that regardless of the intermediate steps taken, the overall enthalpy change remains constant. This principle allows us to add known reaction enthalpies together to find unknown values.

Reaction Enthalpy

This is the net heat absorbed or released during a chemical reaction occurring at constant pressure. It is calculated by taking the difference between the enthalpies of the products and those of the reactants, often using standard enthalpy of formation values.

Standard State

Standard state refers to the conditions under which thermodynamic properties are measured, commonly at 1 atm pressure and a specified temperature (usually 298 K). It provides the baseline for defining standard enthalpy changes and formation values.

Transcript

00:02 In this problem, we have a reaction, no, which is a gas, reacts with 1 .5 .02, which is also a gas, and it produces n .o .2.

00:23 And the heat of reaction for this is a negative 58 .1 kilojoules.

00:38 So this was given to us in the problem.

00:39 They also give us the two reactions that produce this equation.

00:46 It's a two -step process.

00:48 One reaction is written as 1 half n2, which is a gas, plus 1 .5 .02, which is also a gas, produced n0.

01:02 And that has a delta h of formation, a heat of formation of 91 .3 kilojoules.

01:15 And the other reaction we have, so this was reaction a, the other reaction we have is one half n2, reacts with oxygen to produce no2.

01:36 And we do not know what the heat of formation of this one is.

01:40 This is what we're trying to find.

01:43 But what we need to do is we need to look at putting these two reactions together to equal our original reaction.

01:51 Well, we see that we want no as a reactant.

01:59 So i am going to flip reaction a.

02:02 I'm going to reverse it.

02:04 So when i reverse reaction a, i get no produces 1 half and 2 plus 1 half 02.

02:21 That means i also need to reverse the sign for the heat of formation.

02:29 Of it being endothermic, it is now going to be exothermic.

02:37 All right, let's look what this allows us to do now.

02:39 We can take equation b and this reversed equation that i just wrote, and we can combine them.

02:48 And when we combine them, we see that we get the one -half, let me get a color in here, the one -half n -2 is going to cancel...

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