Sulfur dioxide reacts with strontium oxide as follows: SO 2 (g) + SrO(g) → SrSO 3 (s) (a) Without using thermochemical data, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.

Hello everyone. So in this video we're being asked to not do any calculations. And we're trying to determine the delta G. For the reaction of F. E. O. With C. 02. And we're trying to see if it's more or less negative than our delta H. Value. Let's go ahead and get started. Then we're gonna go ahead and use the equation to solve for our delta G. That is going to be that the delta H. Um is attracted by the temperature, multiplied by our delta S. Value. So from our chemical equation which is right over here we can see that products contain fewer moles of gas. And therefore we can conclude that our delta S. Will be negative if we have that our delta S. Is negative. That makes our negative T delta S. Value positive. And therefore if we have this being negative and this being positive and we look at this chemical or that chemical reaction. But this equation, we can conclude that the delta G. Is less negative than our delta H. And therefore our answer to this problem is that our delta G. Is less negative then our delta H. So without any calculations we have solved for this problem right here, thank you all so much for watching

Ch.19 - Chemical Thermodynamics

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Brown 15th Edition

Ch.19 - Chemical Thermodynamics

Problem 62a

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Chapter 19, Problem 62a

Sulfur dioxide reacts with strontium oxide as follows: SO₂(g) + SrO(g) → SrSO₃(s) (a) Without using thermochemical data, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.

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Identify the type of reaction: The reaction between sulfur dioxide (SO_2) and strontium oxide (SrO) to form strontium sulfite (SrSO_3) is a combination reaction, where two reactants form a single product.

Consider the entropy change (ΔS°): In this reaction, two gaseous molecules (SO_2 and SrO) combine to form a solid (SrSO_3). This results in a decrease in entropy because gases have higher entropy than solids.

Relate ΔG° and ΔH° using the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. Since ΔS° is negative (entropy decreases), the term -TΔS° becomes positive, which affects the relationship between ΔG° and ΔH°.

Predict the sign of ΔG° relative to ΔH°: Since -TΔS° is positive, ΔG° will be less negative than ΔH° if ΔH° is negative (exothermic reaction). If ΔH° is positive (endothermic reaction), ΔG° could be more positive or less negative than ΔH° depending on the magnitude of TΔS°.

Conclude the prediction: Without specific thermochemical data, the decrease in entropy suggests that ΔG° is likely less negative than ΔH° for this reaction, assuming it is exothermic.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Gibbs Free Energy (ΔG)

Gibbs Free Energy (ΔG) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. It combines the system's enthalpy (ΔH) and entropy (ΔS) to determine spontaneity. A negative ΔG indicates a spontaneous reaction, while a positive ΔG suggests non-spontaneity.

Enthalpy (ΔH)

Enthalpy (ΔH) is a measure of the total heat content of a system, reflecting the energy required to create the system and the energy associated with the pressure and volume of the system. In chemical reactions, ΔH can be either positive (endothermic) or negative (exothermic), influencing the overall energy change during the reaction. Understanding ΔH is crucial for predicting reaction behavior.

Entropy (ΔS)

Entropy (ΔS) is a measure of the disorder or randomness in a system. In chemical reactions, an increase in entropy typically favors spontaneity, as systems tend to evolve towards greater disorder. The relationship between ΔG, ΔH, and ΔS is given by the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin, highlighting how entropy influences the Gibbs Free Energy.

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Related Practice

Textbook Question

For a certain chemical reaction, ΔH° = -35.4 kJ and ΔS° = -85.5 J/K. (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system?

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Textbook Question

Use data in Appendix C to calculate ΔH°, ΔS°, and ΔG° at 25 °C for each of the following reactions.

c. 2 P(s) + 10 HF(g) → 2 PF₅(g) + 5 H₂(g)

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Textbook Question

Using data from Appendix C, calculate ΔG° for the following reactions. Indicate whether each reaction is spontaneous at 298 K under standard conditions. (a) 2 SO₂(g) + O₂(g) → 2 SO₃(g)

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Textbook Question

Classify each of the following reactions as one of the four possible types summarized in Table 19.3: (i) spontaneous at all temperatures; (ii) not spontaneous at any temperature; (iii) spontaneous at low T but not spontaneous at high T; (iv) spontaneous at high T but not spontaneous at low T. (c) N₂F₄(g) ⟶ 2 NF₂(g) ΔH° = 85 kJ; ΔS° = 198 J/K

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Textbook Question

From the values given for ΔH° and ΔS°, calculate ΔG° for each of the following reactions at 298 K. If the reaction is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous? a. 2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g) ΔH° = −844 kJ; ΔS° = −165 J/K

A certain constant-pressure reaction is barely nonspontaneous at 45 °C. The entropy change for the reaction is 72 J/K. Estimate ΔH.

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Sulfur dioxide reacts with strontium oxide as follows: SO2(g) + SrO(g) → SrSO3(s) (a) Without using thermochemical data, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.

Verified Solution

Key Concepts

Gibbs Free Energy (ΔG)

Enthalpy (ΔH)

Entropy (ΔS)

Sulfur dioxide reacts with strontium oxide as follows: SO₂(g) + SrO(g) → SrSO₃(s) (a) Without using thermochemical data, predict whether ΔG° for this reaction is more negative or less negative than ΔH°.