Déterminez les variations d'enthalpie Δr Hm^o et d'entropie Δr Sm^o des réactions suivantes, à 2

Déterminez les variations d'enthalpie Δr Hm^o et d'entropie...

Déterminez les variations d'enthalpie $\Delta_{\mathrm{r}} H_{\mathrm{m}}^{\mathrm{o}}$ et d'entropie $\Delta_{\mathrm{r}} S_{\mathrm{m}}^{\mathrm{o}}$ des réactions suivantes, à $298 \mathrm{~K}$. À partir de ces valeurs, discutez les conditions de température dans lesquelles ces réactions peuvent être spontanées (en admettant que $\Delta_{\mathrm{r}} H_{\mathrm{m}}^{\mathrm{o}}$ et $\Delta_{\mathrm{r}} S_{\mathrm{m}}^{\mathrm{o}}$ ne varient pas avec la température). a) $2 \mathrm{Al}+3 / 2 \mathrm{O}_2=\mathrm{Al}_2 \mathrm{O}_3$ b) $\mathrm{Fe}_2 \mathrm{O}_3=2 \mathrm{Fe}+3 / 2 \mathrm{O}_2$ c) $\mathrm{Fe}_2 \mathrm{O}_3+2 \mathrm{Al}=\mathrm{Fe}+\mathrm{Al}_2 \mathrm{O}_3$ d) $3 \mathrm{O}_2=2 \mathrm{O}_3$ (Les données nécessaires se trouvent dans l'Annexe I).

Déterminez les variations d'enthalpie $\Delta_{\mathrm{r}} H_{\mathrm{m}}^{\mathrm{o}}$ et d'entropie $\Delta_{\mathrm{r}} S_{\mathrm{m}}^{\mathrm{o}}$ des réactions suivantes, à $298 \mathrm{~K}$. À partir de ces valeurs, discutez les conditions de température dans lesquelles ces réactions peuvent être spontanées (en admettant que $\Delta_{\mathrm{r}} H_{\mathrm{m}}^{\mathrm{o}}$ et $\Delta_{\mathrm{r}} S_{\mathrm{m}}^{\mathrm{o}}$ ne varient pas avec la température).
a) $2 \mathrm{Al}+3 / 2 \mathrm{O}_2=\mathrm{Al}_2 \mathrm{O}_3$
b) $\mathrm{Fe}_2 \mathrm{O}_3=2 \mathrm{Fe}+3 / 2 \mathrm{O}_2$
c) $\mathrm{Fe}_2 \mathrm{O}_3+2 \mathrm{Al}=\mathrm{Fe}+\mathrm{Al}_2 \mathrm{O}_3$
d) $3 \mathrm{O}_2=2 \mathrm{O}_3$
(Les données nécessaires se trouvent dans l'Annexe I).

Key Concepts

Gibbs Free Energy

Gibbs free energy (?G) is a thermodynamic potential that combines enthalpy and entropy to predict reaction spontaneity under constant temperature and pressure conditions using the equation ?G = ?H - T?S. A negative ?G denotes a spontaneous process, while a positive ?G implies non-spontaneity, making it a central concept in evaluating the feasibility and equilibrium of chemical reactions.

Temperature Dependence of Spontaneity

The interplay between enthalpy and entropy changes determines how temperature influences reaction spontaneity. When both ?H and ?S are negative, spontaneous behavior is more likely at lower temperatures; when both are positive, higher temperatures tend to favor spontaneity; and when ?H and ?S have opposite signs, a specific temperature threshold (T = ?H/?S) can be identified, above or below which the reaction becomes spontaneous. This concept is crucial for controlling reaction conditions and optimizing industrial processes.

Entropy Change

Entropy change (?S) quantifies the change in disorder or randomness in a system undergoing a reaction. It is determined by subtracting the sum of the standard molar entropies of the reactants from that of the products. A positive entropy change suggests increased disorder, which can favor spontaneity, whereas a negative change indicates a decrease in disorder.

Enthalpy Change

Enthalpy change (?H) represents the heat exchanged at constant pressure during a reaction. It is calculated as the difference between the sum of the standard enthalpies of formation of the products and that of the reactants. This concept is essential for understanding whether a reaction is endothermic or exothermic, and it often serves as an indicator of the energy efficiency and feasibility of chemical processes.

Transcript

00:01 In this problem, we need to use the thermodynamic data in order to answer it because we need to calculate delta h, delta s, and delta g from the tables that are provided at the back of the book.

00:17 So it takes a lot of time to look these up.

00:21 You have to be very careful that you're using the exact compound in the correct state of matter.

00:29 So our first reaction is two chromium solid plus three oxygens.

00:45 That's a gas.

00:48 This gives us two cro3 solid.

00:58 The order of the chart gives you your delta h, your delta g, and then your delta s.

01:08 So you have to look up each of these.

01:10 Remember, you have to do products minus reactants.

01:13 And we're going to multiply by the coefficient in front of our chemical formula in the balanced equation.

01:20 So for delta h, i'm doing two times the value for the cr -03.

01:27 So that's two times a negative 1139 .7 minus two times the value for chromium.

01:38 That's a zero and minus three times the value for oxygen, which is also a zero.

01:44 This gives me negative 2279 .4, and this is in kilojoules.

01:58 The delta g, again, we're doing in reverse order, so we have negative 1058 .1, minus 2 times 0, minus 3 times 0, gives me negative 211 .2 .2 .000.

02:22 Is also in kilojoules.

02:25 Your delta s for this problem, 2 times 81 .2 minus 2 times 0 .0.

02:39 Whoops, sorry, i skip to the wrong thing.

02:46 Sorry, it's minus, it's two times 23 .6 minus 3 times 205 .0.

02:59 When you calculate all of that, you should get a negative 499 .8.

03:08 This is in joules per kelvin.

03:13 So it has a different unit than the other two do.

03:17 All right.

03:19 That is the first part of this question.

03:25 The next one is baco3 solid yields, b .a .o.

03:39 Solid.

03:41 Plus co2 gas.