Calculate the standard molar enthalpy of formation of NO(g) from the following data: N2(g)+

Calculate the standard molar enthalpy of formation of NO(g)...

Calculate the standard molar enthalpy of formation of $\mathrm{NO}(g)$ from the following data: $$\begin{array}{ll}{\mathrm{N}_{2}(g)+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g)} & {\Delta H_{298}^{\mathrm{o}}=66.4 \mathrm{kJ}} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g)} & {\Delta H_{298}^{\circ}=-114.1 \mathrm{kJ}}\end{array}$$

Calculate the standard molar enthalpy of formation of $\mathrm{NO}(g)$ from the following data:
$$\begin{array}{ll}{\mathrm{N}_{2}(g)+2 \mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g)} & {\Delta H_{298}^{\mathrm{o}}=66.4 \mathrm{kJ}} \\ {2 \mathrm{NO}(g)+\mathrm{O}_{2} \longrightarrow 2 \mathrm{NO}_{2}(g)} & {\Delta H_{298}^{\circ}=-114.1 \mathrm{kJ}}\end{array}$$

Key Concepts

Standard Molar Enthalpy of Formation

This is the change in enthalpy when one mole of a compound is formed from its elements in their standard states under standard conditions (usually 1 atm and 298 K). It is a fundamental property used to determine the energy changes in chemical reactions, and it serves as a reference point for calculating reaction enthalpies in thermochemistry.

Hess's Law

Hess's Law states that the total enthalpy change of a reaction is the same regardless of the pathway taken, provided the initial and final conditions are identical. This concept allows us to add, subtract, or reverse chemical equations with known enthalpy changes to calculate the enthalpy change for a reaction of interest, making it a powerful tool in thermochemistry.

Reaction Manipulation

Reaction manipulation involves the mathematical adjustment of chemical equations, such as reversing, scaling, or combining reactions. This technique is crucial when applying Hess's Law, as it allows one to construct a target reaction from known reactions by appropriately combining their enthalpy changes.

Balanced Chemical Equations

Balanced chemical equations ensure that matter and energy are conserved, reflecting the correct stoichiometry of reactants and products in a chemical reaction. Accurate balancing is essential for thermochemical calculations because the enthalpy changes depend on the amounts of substances reacting or being produced.

Transcript

00:01 So today we're trying to calculate the enthalpy of formation of no gas.

00:05 So it's important to know what the definition of enthalpy of formation is, but we are going to be creating one mole of products, so no, and it's going to be comprised of its individual elements.

00:20 So it's comprised of nitrogen, which is naturally found in the world's n2, and o2, which is how oxygen is normally.

00:30 Found.

00:32 Now the interesting thing is with hess's law and enthalpya formation, we're only producing one mole, and that means we are going to actually end up with a fractional coefficient in front of each of our reactants.

00:48 This might seem strange, but this is kind of just the convention that we use so that we can scale things.

00:55 So if we're trying to create three moles of no gas, we can multiply this by 3 or any number, etc.

01:01 So we're going to use these two equations that we've been given in the problem to find the enthalpy for this reaction.

01:08 So delta h is what we don't know.

01:12 My first step is to see that we need to have no end up in our products and we can see that no really only shows up in the second equation.

01:22 And because it's in the reactance of this first equation, we're going to have to flip it.

01:27 We can also see that the coefficient in front of no is 2.

01:32 In order to get to our 1no, we're going to have to divide everything by 2.

01:37 So first i'm going to flip it.

01:40 So it's going to become 2no2 becomes 2n2.

01:47 When i flip it, we know that delta h should change sign.

01:51 So it becomes positive, 114 .1 kilojoules...

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