Utilizando los datos del apéndice C, calcule ΔH^∘, ΔS^∘ y ΔG^∘ a 298 K para cada una de las sig

Utilizando los datos del apéndice C, calcule ΔH^∘, ΔS^∘ y...

Utilizando los datos del apéndice $\mathrm{C}$, calcule $\Delta H^{\circ}, \Delta S^{\circ}$ y $\Delta G^{\circ}$ a $298 \mathrm{~K}$ para cada una de las siguientes reacciones. En cada caso, demuestre que $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$. a) $\mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g})$ b) $\mathrm{C}(\mathrm{s}$, grafito $)+2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{CCl}_4(\mathrm{~g})$ c) $2 \mathrm{PCl}_3(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{POCl}_3(\mathrm{~g})$ d) $2 \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

Utilizando los datos del apéndice $\mathrm{C}$, calcule $\Delta H^{\circ}, \Delta S^{\circ}$ y $\Delta G^{\circ}$ a $298 \mathrm{~K}$ para cada una de las siguientes reacciones. En cada caso, demuestre que $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$.
a) $\mathrm{H}_2(\mathrm{~g})+\mathrm{F}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{HF}(\mathrm{g})$
b) $\mathrm{C}(\mathrm{s}$, grafito $)+2 \mathrm{Cl}_2(\mathrm{~g}) \longrightarrow \mathrm{CCl}_4(\mathrm{~g})$
c) $2 \mathrm{PCl}_3(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow 2 \mathrm{POCl}_3(\mathrm{~g})$
d) $2 \mathrm{CH}_3 \mathrm{OH}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g}) \longrightarrow \mathrm{C}_2 \mathrm{H}_6(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$

Key Concepts

Gibbs Free Energy Equation

This fundamental relationship, expressed as ?G° = ?H° - T?S°, links enthalpy, entropy, and temperature, providing a comprehensive picture of reaction energetics. It demonstrates how the interplay of heat transfer and changes in system disorder governs the spontaneity of a reaction.

Standard Gibbs Free Energy Change (?G°)

?G° quantifies the maximum reversible work that can be performed by a reaction at constant temperature and pressure under standard conditions. It is a key indicator of reaction spontaneity, where negative values signify spontaneous processes, while positive values indicate non-spontaneity.

Standard Enthalpy Change (?H°)

This concept refers to the overall heat absorbed or released during a reaction when conducted under standard conditions. It is calculated from the standard heats of formation or bond energies of the reactants and products, revealing whether a reaction is exothermic (releasing heat) or endothermic (absorbing heat).

Standard Entropy Change (?S°)

This term represents the change in disorder or randomness associated with a chemical reaction under standard conditions. It is determined by the difference in the standard molar entropies of the products and reactants, and it helps assess the degree to which a reaction results in increased or decreased molecular randomness.

Transcript

00:01 Let's calculate the change in enthalpy, the change in entropy, and the change it gives free energy for a series of chemical reactions.

00:10 The equations to be used in each of these cases is shown already.

00:17 So to calculate the enthalpy of reaction under standard conditions, that's what that little zero or not sign is about.

00:26 We do that by summing up, that's what the sigma sign means, the number of moles of a substance times its enthalpy of formation for the product side, and then we do the same thing for the reactant side and subtract the two values.

00:44 We do the same thing for entropy, and we do the same thing for gibbs free energy.

00:51 The n and the m in the equations deal with in both cases number of moles.

01:01 We don't use n in each case just to make sure that you understand that the number on both sides may not be the same.

01:10 Where do you come up with these h's of formation or these s's or these gs of formation, those can be found in a table of values.

01:22 Some that you might not find in a table of values are elements found in their natural state.

01:29 And for those, you should know that the delta h of formation under standard conditions for those elements is zero.

01:45 So you may not find them in the table.

01:48 It's assumed that you know that they're zero.

01:50 Entropies for elements do not have to be zero.

01:59 So you do have to look those up.

02:02 Okay.

02:03 So let's start out with this first equation.

02:07 We have hydrogen in its natural state plus fluorine in its natural state react to become hydrofluor acid.

02:17 And so we're going to use our top of equation to figure out the enthalpy change during this reaction.

02:29 Okay.

02:30 So when we sum things up, what that means is we would add up all of the substances on that particular side of the reaction.

02:36 So if you can see here, we have to look at the product side first.

02:41 So we're going to look at the hf side.

02:43 And we need to know the enthalpy of formation for hf.

02:47 When you go look that up, you'll find it's negative 268 .61 kilojoules per mole.

02:53 And so what we do now is n is the number of moles.

02:57 So we have two moles times the hf, which is negative 268.

03:05 0 .81 kilojoules per mole.

03:10 Now, if we had any other substances, we would add those to this.

03:14 That's what the sigma means.

03:16 But this is the only substance on the product side.

03:18 So we're done with the product side.

03:20 We have a subtraction.

03:22 And now we're going to go take a look at the reactant side.

03:26 Now, the reacticide has one mole of each substance.

03:32 These are elements in their natural state, so the delta h is going to be zero for those.

03:37 But let me just show you how this would be set up.

03:40 There's one mole of h2.

03:42 Now we know that the delta h for this is zero kilojoules per mole.

03:54 But what we have to do now, sigma means to add up, we have to do the same thing with the f2...

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