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 \(\mathrm{C}\) on other substances.

Short answer

(a) The standard Gibbs free energy change (∆G°) is less negative than the standard enthalpy change (∆H°). (b) To estimate the value of ∆G° at 298 K using only ∆H° data, we can look at similar compounds and assume an approximate value for the entropy change (∆S°) and use the relation ∆G° = ∆H° - T∆S° to calculate the estimated value. However, this method has limitations and is not very accurate.

Step-by-step solution

(Step 1: Understand the relation between ∆G°, ∆H°, and ∆S°)

(The standard Gibbs free energy change (∆G°) in a reaction can be expressed as: ∆G° = ∆H° - T∆S° Here, T is the absolute temperature and ∆S° is the standard entropy change. Based on the given reaction, we need to predict if the entropy change (∆S°) will be positive or negative. As we proceed from reactants to products, sulfur dioxide gas and strontium oxide gas react to form a solid, strontium sulfite. In this process, we can assume the entropy change to be negative as going from gaseous to solid state decreases the freedom of movement of particles.)

(Step 2: Predict the sign of the Gibbs free energy change (∆G°))

(Since the entropy change (∆S°) is negative and the temperature (T) is always positive, the term T∆S° is negative. Considering the equation for ∆G°: ∆G° = ∆H° - T∆S° Thus, ∆H° is more negative than ∆G°, implying that ∆G° is less negative than ∆H°.)

(Step 3: Estimate the value of ∆G° at 298 K using only ∆H° data)

(To estimate the value of ∆G° at 298 K, we can use the equation: ∆G° = ∆H° - T∆S° We are given that we only have standard enthalpy data (∆H°). We can assume the entropy change for similar compounds and use that to estimate the value of ∆G° at 298 K. For example, we can look up the standard entropy change for the formation of a similar group 2 metal sulfite and use that value. However, this estimation method has limitations and is not very accurate.) In summary: (a) The standard Gibbs free energy change (∆G°) is less negative than the standard enthalpy change (∆H°). (b) To estimate the value of ∆G° at 298 K using only ∆H° data, we can look at similar compounds and assume an approximate value for the entropy change (∆S°) and use the relation ∆G° = ∆H° - T∆S° to calculate the estimated value. However, this method has limitations and is not very accurate.

Key concepts

Entropy Change

Entropy (\(\Delta S^{\circ}\)) refers to the measure of disorder or randomness in a system. When considering chemical reactions, it's important to understand how entropy changes. In the reaction between sulfur dioxide (\(\text{SO}_2\)) gas and strontium oxide (\(\text{SrO}\)) gas to form solid strontium sulfite (\(\text{SrSO}_3\)), entropy decreases. This is because gases, with their free-moving particles, have higher entropy compared to solids where particles are fixed in a structure.- Moving from gaseous reactants to a solid product reduces the number of microstates.- A decrease in microstates results in a negative entropy change.Entropy change is a crucial factor in calculating the Gibbs Free Energy (\(\Delta G^{\circ}\)), using the formula:\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]This tells us how spontaneous a reaction might be.

Enthalpy Change

Enthalpy (\(\Delta H^{\circ}\)) represents the heat change at constant pressure during a reaction. In many chemical processes, the enthalpy change helps us understand whether the reaction absorbs or releases heat.- Endothermic reactions absorb heat and have a positive \(\Delta H^{\circ}\).- Exothermic reactions release heat and have a negative \(\Delta H^{\circ}\).For the reaction of sulfur dioxide with strontium oxide, predicting the sign of \(\Delta H^{\circ}\) relies on whether the bond formation releases more energy than is required to break the bonds in the reactants.In our case:- The formation of \(\text{SrSO}_3\) likely releases energy, indicating a negative \(\Delta H^{\circ}\).Understanding \(\Delta H^{\circ}\) is essential because it is directly used in the Gibbs Free Energy formula to help predict reaction spontaneity.

Thermodynamic Predictions

Thermodynamic predictions are made using Gibbs Free Energy (\(\Delta G^{\circ}\)), which tells us whether a reaction will occur spontaneously.- A negative \(\Delta G^{\circ}\) indicates a spontaneous process.- A positive \(\Delta G^{\circ}\) means the process is non-spontaneous.For the reaction:\[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\]If \(\Delta H^{\circ}\) is more negative than \(T\Delta S^{\circ}\), \(\Delta G^{\circ}\) becomes more negative, suggesting spontaneity. Students can utilize known data from similar compounds to predict \(\Delta G^{\circ}\) at different temperatures, like 298 K. Using approximate \(\Delta S^{\circ}\) values from similar reactions enables this estimation.Remember, while useful, these predictions have limitations:- They depend closely on the accuracy of assumed or available values.- Conditions like pressure and concentration can influence results.Thus, combining \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for effective prediction helps understand a reaction's pathway and feasibility.