Problem 62
Question
Sulfur dioxide reacts with strontium oxide as follows: $$ \mathrm{SO}_{2}(g)+\mathrm{SrO}(g) \longrightarrow \mathrm{SrSO}_{3}(s) $$ (a) Without using thermochemical data, predict whether \(\Delta G^{\circ}\) for this reaction is more negative or less negative than \(\Delta H^{\circ} .\) (b) If you had only standard enthalpy data for this reaction, how would you estimate the value of \(\Delta G^{\circ}\) at \(298 \mathrm{K},\) using data from Appendix Con other substances.
Step-by-Step Solution
Verified Answer
(a) ΔG° is more negative than ΔH° for the given reaction, due to a decrease in entropy resulting from a net decrease in the number of gaseous particles.
(b) To estimate ΔG° at 298K using only standard enthalpy data, estimate values for ΔH° and ΔS° using similar substances' values from Appendix c, then plug these values into the Gibbs free energy equation: \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\). This will result in a rough estimation of ΔG° for the reaction at 298K.
1Step 1: Examine the Reaction
Observe the given reaction: $$\mathrm{SO}_{2}(g)+\mathrm{SrO}(g) \longrightarrow \mathrm{SrSO}_{3}(s)$$ Here, one gaseous substance (\(SO_2\)) reacts with another gaseous substance (\(SrO\)) to form a solid (\(SrSO_3\)).
2Step 2: Relate ΔG° and ΔH° using the Gibbs Free Energy Equation
The relationship between ΔG° and ΔH° can be described by Gibbs free energy equation: $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ From this equation, we will analyze the entropy change for the reaction to determine whether ΔG° is more or less negative than ΔH°.
3Step 3: Analyze Entropy Change for the Reaction
In general, when a gas turns into a solid, the entropy (\(\Delta S\)) of the system decreases. The reaction has a net decrease in the number of gaseous particles, which also contributes to a decrease in entropy. Because ΔS (\(\Delta S^\circ\)) is negative, and \(T\Delta S^\circ\) contributes to the reduction in Gibbs free energy (ΔG), we can conclude the following:
a) ΔG° is more negative than ΔH° for the given reaction.
4Step 4: Estimating ΔG° Using Standard Enthalpy Data
To estimate ΔG° at 298K using only standard enthalpy data, follow these steps:
1. Estimate the value for \(\Delta H^\circ\) by looking at the enthalpy values of similar substances (from Appendix C or other sources).
2. Calculate \(\Delta S^\circ\) using standard entropy values of similar substances, taking into account that, for this reaction, the entropy change is negative.
3. Substitute the estimated values of \(\Delta H^\circ\) and \(\Delta S^\circ\) into the Gibbs free energy equation and solve for ΔG° at 298K, using the ambient temperature (\(T = 298\: K\)). This will provide a rough estimation of \(\Delta G^\circ\) for the given reaction.
Key Concepts
Chemical Reaction PredictionThermochemical DataEntropy ChangeEnthalpy Data Estimation
Chemical Reaction Prediction
Predicting the direction and extent of a chemical reaction is a critical aspect of chemistry. The Gibbs free energy equation, centrally important in this regard, is expressed as \( centiDelta G^circ = centiDelta H^circ - TcentiDelta S^circ \), where \( centiDelta G^circ \) represents the change in free energy, \( centiDelta H^circ \) is the change in enthalpy, \( centiDelta S^circ \) is the change in entropy, and T is the absolute temperature in Kelvins. For the reaction of sulfur dioxide with strontium oxide to form strontium sulfite, we intuitively assess entropy changes based on physical states and particle count. As gases have higher entropy than solids, the formation of a solid from gaseous reactants (as seen here) suggests a decrease in entropy.
Next, when we apply the Gibbs equation, given that entropy change (centiDelta S^ncirc) is negative due to decreased disorder, and since free energy is a measure of a system's ability to do work, the negative entropy change makes \( centiDelta G^circ \) more negative, considering the temperature T is positive. This means the reaction is thermodynamically favored since free energy is being released. This prediction is valuable when direct measurements are not available, offering a way to anticipate the feasibility of a reaction under standard conditions.
Next, when we apply the Gibbs equation, given that entropy change (centiDelta S^ncirc) is negative due to decreased disorder, and since free energy is a measure of a system's ability to do work, the negative entropy change makes \( centiDelta G^circ \) more negative, considering the temperature T is positive. This means the reaction is thermodynamically favored since free energy is being released. This prediction is valuable when direct measurements are not available, offering a way to anticipate the feasibility of a reaction under standard conditions.
Thermochemical Data
Thermochemical data represent the heat aspects of chemical reactions and include values such as enthalpy (\( centiDelta H^ncirc \)), entropy (\(centiDelta S^ncirc \)), and specific heat capacities. These parameters are essential in thermodynamics for estimating reaction energies and predicting the spontaneity of reactions through the Gibbs free energy equation. The relationship of these data with temperature, a key factor in Gibbs equation, helps chemists predict whether a reaction will occur spontaneously at a given temperature.
In the context of the sulfur dioxide and strontium oxide reaction example, accurate thermochemical data allows for precise calculations. If only enthalpy data is available, assumptions and estimates can be used for entropy and free energy, but this may reduce precision. Furthermore, databases and appendices, like Appendix C mentioned in the exercise, provide standard enthalpy values for many substances, allowing for reasonable enthalpy estimations when direct values for reactants or products are unavailable.
In the context of the sulfur dioxide and strontium oxide reaction example, accurate thermochemical data allows for precise calculations. If only enthalpy data is available, assumptions and estimates can be used for entropy and free energy, but this may reduce precision. Furthermore, databases and appendices, like Appendix C mentioned in the exercise, provide standard enthalpy values for many substances, allowing for reasonable enthalpy estimations when direct values for reactants or products are unavailable.
Entropy Change
Entropy is a measure of the disorder or randomness of a system. A fundamental principle of entropy dictates that for any spontaneous process, the total entropy of the system and its surroundings always increases. In the context of chemical reactions, an increase in gas particles generally signifies an increase in entropy, while a change from gas to a solid infers a decrease in entropy.
The reaction between sulfur dioxide and strontium oxide to form strontium sulfite exemplifies an entropy decrease, as there's a net loss of gaseous molecules upon product formation. This negative entropy change contributes to the calculation of the Gibbs free energy, playing a crucial role in predicting the spontaneous nature of the reaction. When specific standard entropy data (\(centiDelta S^ncirc \)) is not readily available, approximations can be made by comparing with substances of similar structure and state. This helps chemists and students understand not only how a reaction can proceed but also the molecular-level changes accompanying the reaction.
The reaction between sulfur dioxide and strontium oxide to form strontium sulfite exemplifies an entropy decrease, as there's a net loss of gaseous molecules upon product formation. This negative entropy change contributes to the calculation of the Gibbs free energy, playing a crucial role in predicting the spontaneous nature of the reaction. When specific standard entropy data (\(centiDelta S^ncirc \)) is not readily available, approximations can be made by comparing with substances of similar structure and state. This helps chemists and students understand not only how a reaction can proceed but also the molecular-level changes accompanying the reaction.
Enthalpy Data Estimation
Enthalpy, represented by \( centiDelta H^ncirc \) in thermochemical equations, indicates the total heat content of a system and is associated with the breaking and forming of bonds during a chemical reaction. Estimating enthalpy change can be necessary when experimental data is unavailable for specific reactions.
An effective approach to estimate \( centiDelta H^circ \) involves utilizing known enthalpy values of similar substances and considering the stoichiometry of the reaction in question. For instance, if we have the standard enthalpy data for sulfur dioxide and strontium oxide but lack it for strontium sulfite, we may draw from analogous compounds to estimate the enthalpy of formation for SrSO3. This estimation, along with the entropy changes, can then be plugged into the Gibbs free energy equation to derive an approximate value of \( centiDelta G^circ \), providing insight into the reaction's spontaneity at a given temperature, like 298 Kelvin in our exercise example.
An effective approach to estimate \( centiDelta H^circ \) involves utilizing known enthalpy values of similar substances and considering the stoichiometry of the reaction in question. For instance, if we have the standard enthalpy data for sulfur dioxide and strontium oxide but lack it for strontium sulfite, we may draw from analogous compounds to estimate the enthalpy of formation for SrSO3. This estimation, along with the entropy changes, can then be plugged into the Gibbs free energy equation to derive an approximate value of \( centiDelta G^circ \), providing insight into the reaction's spontaneity at a given temperature, like 298 Kelvin in our exercise example.
Other exercises in this chapter
Problem 60
Using data from Appendix \(\mathrm{C}\) , calculate \(\Delta G^{\circ}\) for the following reactions. Indicate whether each reaction is spontaneous at 298 \(\ma
View solution Problem 61
Octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is a liquid hydrocarbon at room temperature that is a constituent of gasoline. (a) Write a balanced equat
View solution Problem 63
Classify each of the following reactions as one of the four possible types summarized in Table \(19.3 :\) (i) spontanous at all temperatures; (ii) not spontaneo
View solution Problem 64
From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at 298 \(\mathrm{K}\
View solution