Discutez de la possibilité thermodynamique (ou spontanéité)...

Discutez de la possibilité thermodynamique (ou spontanéité) des réactions suivantes, en considérant la valeur de la variation d'enthalpie libre $\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\mathrm{o}}$ pour chacune d'elles supposées s'effectuer avec l'ensemble des constituants pris dans leur état de référence. a) $\mathrm{CH}_4+2 \mathrm{O}_2=\mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ (1) b) $2 \mathrm{NH}_3=\mathrm{N}_2+3 \mathrm{H}_2$ c) $2 \mathrm{~N}_2 \mathrm{O}=2 \mathrm{~N}_2+\mathrm{O}_2$

Discutez de la possibilité thermodynamique (ou spontanéité) des réactions suivantes, en considérant la valeur de la variation d'enthalpie libre $\Delta_{\mathrm{r}} G_{\mathrm{m}}^{\mathrm{o}}$ pour chacune d'elles supposées s'effectuer avec l'ensemble des constituants pris dans leur état de référence.
a) $\mathrm{CH}_4+2 \mathrm{O}_2=\mathrm{CO}_2+2 \mathrm{H}_2 \mathrm{O}$ (1)
b) $2 \mathrm{NH}_3=\mathrm{N}_2+3 \mathrm{H}_2$
c) $2 \mathrm{~N}_2 \mathrm{O}=2 \mathrm{~N}_2+\mathrm{O}_2$

Key Concepts

Gibbs Free Energy

Gibbs free energy is a thermodynamic potential that indicates the maximum reversible work obtainable from a system at constant temperature and pressure. It combines both enthalpic and entropic contributions (?G = ?H - T?S) and determines spontaneity; a negative ?G signifies a reaction that can proceed spontaneously under the given conditions.

Thermodynamic Spontaneity

Thermodynamic spontaneity refers to the natural tendency of a process to occur without the need for external energy input. In chemical reactions, this concept is directly linked to the Gibbs free energy change, where a negative value indicates that the reaction is thermodynamically favorable.

Standard State Conditions

Standard state conditions provide a reference framework for comparing the thermodynamic properties of substances. Under these conditions, all reactants and products are in their most stable form at a set pressure and temperature, simplifying the evaluation of reaction spontaneity through standard Gibbs free energy changes.

Enthalpy and Entropy Contributions

Enthalpy (heat content) and entropy (degree of disorder) are the two primary factors affecting the Gibbs free energy of a reaction. The balance between these contributions, captured by the equation ?G = ?H - T?S, explains why some reactions release energy while others may be driven by an increase in disorder, thus impacting their overall spontaneity.

Transcript

00:01 So for problem 66, we're given these to three reactions, and we have to find whether these the three reactions are spontaneous at low, high, or all temperatures.

00:11 So to do that, we're going to have to find the delta s and the delta h of all of the reactions.

00:19 So before we do that, the equation we're going to be referencing is going to be the formula for the free energy change, which is delta g is equal to delta h minus a, temperature times delta s so a reaction is spontaneous when delta g is negative and we're going to be using this formula to determine that for these reactions so for part a our only product is magnesium oxide and we have two moles of this so it's an h value is negative 601 .7 so for delta h it's going to be 2 times negative 601 .7.

01:07 And below that, we're also going to find the delta s.

01:10 So it's going to be two times the s value, which is 26 .94.

01:18 So 26 .94 minus our first reactant, which is two moles of solid magnesium.

01:28 And its h value is zero.

01:31 So it's going to be minus a zero.

01:33 Down here, the s value is going to be 32 .68, so minus 2 times 32 .68, minus the other reactant, which is molecule oxygen, which also has an h value of 0.

01:59 Yes, so it's going to be minus 0, and down here it's going to be minus, since there's only one mole of this, so it's going to be plus times 1.

02:09 So minus 205 .138.

02:13 So minus 205 .138.

02:18 This would give us a delta h value of negative 1 ,203 .4.

02:28 And a delta s value of negative 216 .618.

02:35 So we only have to care about the signs of these values.

02:40 So the delta h value is negative and the delta s value is also negative.

02:46 So when the delta h value is negative and the delta s value is negative, then here it's going to cancel out the minus sign.

02:55 So it's going to be negative delta h plus temperature times delta s.

03:02 So in order for the formula to produce a negative delta g value, the second term must be small.

03:12 Smaller than the delta h value, which means the temperature must be small.

03:19 And thus for a, the temperature must only be spontaneous at a low temperature, so it's going to be option one.

03:31 For b, it's basically the same procedure.

03:33 You find delta s and delta h.

03:35 So our first product is carbon dioxide, and we have two moles of it.

03:41 So it's going to be 2 times the h value of carbon dioxide, which is negative 393 .509.

03:51 So negative 393 .509.

03:56 Down here, it's going to be 2 times the s value, which is 213 .74...

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