Figure 17.16 shows how Δf G^o for four gases varies with temperature. (a) Which compounds become

Figure 17.16 shows how Δf G^o for four gases varies with...

Figure 17.16 shows how $\Delta_{\mathrm{f}} G^{\mathrm{o}}$ for four gases varies with temperature. (a) Which compounds become less thermodynamically favoured as the temperature increases from 300 to $1500 \mathrm{K} ?$ (b) For which compounds is the formation from the constituent elements favourable between 300 and $1500 \mathrm{K} ?(\mathrm{c})$ For which compound is $\Delta_{\mathrm{f}} G^{\mathrm{o}}$ approximately temperature-independent between 300 and $800 \mathrm{K} ?(\mathrm{d}) \mathrm{What}$ can you say about the relative thermodynamic stabilities of $\mathrm{SO}_{2}(\mathrm{g})$ and $\mathrm{CO}(\mathrm{g})$ with respect to their constituent elements as a function of temperature? (e) Write an equation for the reaction to which $\Delta_{\mathrm{f}} G^{\mathrm{o}}\left(\mathrm{GeCl}_{4}, \mathrm{g}\right)$ refers.

Figure 17.16 shows how $\Delta_{\mathrm{f}} G^{\mathrm{o}}$ for four gases varies with temperature. (a) Which compounds become less thermodynamically favoured as the temperature increases from 300 to $1500 \mathrm{K} ?$ (b) For which compounds is the formation from the constituent elements favourable between 300 and $1500 \mathrm{K} ?(\mathrm{c})$ For which compound is $\Delta_{\mathrm{f}} G^{\mathrm{o}}$
approximately temperature-independent between 300 and $800 \mathrm{K} ?(\mathrm{d}) \mathrm{What}$ can you say about the relative thermodynamic stabilities of $\mathrm{SO}_{2}(\mathrm{g})$ and
$\mathrm{CO}(\mathrm{g})$ with respect to their constituent elements as a function of temperature? (e) Write an equation for the reaction to which $\Delta_{\mathrm{f}} G^{\mathrm{o}}\left(\mathrm{GeCl}_{4}, \mathrm{g}\right)$ refers.

Key Concepts

Reaction Equations in Thermodynamics

Writing the chemical reaction equations associated with thermodynamic data is essential for clarity and proper interpretation. It outlines the precise process by which a compound is formed from its constituent elements and identifies the standard states of the elements. This clarity is necessary for understanding the source of the reported thermodynamic values and for comparing the stabilities of different compounds.

Thermodynamic Favorability

Thermodynamic favorability is determined by the sign and magnitude of the Gibbs free energy change. When the standard Gibbs free energy of formation is negative, the formation reaction is thermodynamically favored, meaning that the formation of the compound from its elements releases energy. Evaluating favorability across a temperature range helps in understanding which reactions may be spontaneous under different conditions.

Thermodynamic Stability

Thermodynamic stability refers to the relative free energy of a compound compared to its potential decomposition into constituent elements. A compound with a highly negative standard Gibbs free energy of formation is typically more stable because it requires additional energy input to decompose back into the elements. Evaluating stability as a function of temperature helps determine how the persistence of a compound may change under different thermal conditions.

Standard Gibbs Free Energy of Formation

This concept refers to the change in Gibbs free energy when one mole of a compound is formed from its constituent elements under standard conditions (usually 1 atm and a specified temperature, often 298 K). It is a critical parameter used to assess the thermodynamic favorability of a formation reaction, as negative values indicate that the formation is spontaneous and thermodynamically favored.

Temperature Dependence of Gibbs Free Energy

Gibbs free energy is not a static value but rather depends on temperature. Changes in temperature affect both the enthalpy and entropy contributions to the Gibbs free energy. This temperature dependence can determine how the spontaneity of a reaction shifts with temperature, providing insights into whether a compound becomes more or less thermodynamically favored at higher or lower temperatures.

Transcript

00:01 For each one of the four reactions that we are given in this problem, we need to first determine, without doing any calculations, what the signs are for the change in entropy of the system and for the surroundings.

00:15 We then need to determine whether or not the reaction is spontaneous at various temperature conditions.

00:23 For each of the reactions, we are provided the overall stoichiometry, which will help us determine delta s of the system.

00:32 Which in this case is delta s of the reaction.

00:37 And then we can use a top equation in order to help us determine the sign of delta s surroundings because we have values for delta h of the reaction.

00:49 So starting with part a, this is the overall reaction that we are provided.

00:54 We begin by determining what the sign is for delta s of the system, which again is delta s of the reaction since this is a chemical reaction.

01:03 We see that this is a gas phase reaction.

01:06 All of the species participating in the reaction are gases.

01:11 So we need to count up the total number of moles of gas on each side of the reaction in order to help us determine delta s of the reaction.

01:19 On the reactant side, we have six total moles of gas and on the product side we have seven.

01:23 So as a result of the chemical reaction, we increase the total number of moles of gas from six to seven.

01:30 Whenever we increase the number of moles of gas that we have, we increase the amount of disorder, and therefore we increase the entropy.

01:39 So that means that delta s of the reaction or of the system will have a positive change.

01:49 And then for delta s of the surroundings, we are given a value for delta h of this reaction, and we see that it has a negative sign.

02:00 When we go to calculate delta s of the surroundings, we take a negative in front of that delta h of reaction.

02:08 And since delta h of reaction for the reaction in part a is negative, those negatives will cancel out.

02:14 So the numerator will be a positive value.

02:16 Since the temperature is measured in kelvin, that's always going to be positive.

02:20 So we have a positive number divided by another positive number.

02:22 So that means that delta s of the surroundings will also have a positive value.

02:34 So since both delta s of the system and the surroundings are positive, when we add them together, that will come out to a positive value for delta s of the universe at all temperatures, because both of them are positive.

02:47 And whenever a delta s of the universe is positive, that means that the reaction is spontaneous.

02:53 So that means we can conclude that this reaction is spontaneous at all temperatures.

03:05 Moving on to the reaction that we are given in part b, we see that we have, have 1 plus 1, so 2 total moles of gas on the reactant side, and 2 total moles of gas on the product side.

03:15 However, if we examine the actual gases themselves, we can distinguish what the sign of delta s of the reaction should be.

03:28 We have, on the reactant's side, one mole of two different gases, and on the product side we have 2 moles of 1 gas...

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