Question

Use data in Appendix \(\mathrm{C}\) to calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\) (a) \(2 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CrO}_{3}(s)\) (b) \(\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)\) (c) \(2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)\) (d) \(\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)\)

Short answer

For (a) \(2 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CrO}_{3}(s)\): 1. Using the formula \(\Delta H^{\circ} = [n_{products}\times \Delta H_f^{\circ}(products)] - [n_{reactants}\times \Delta H_f^{\circ}(reactants)]\), we find \(\Delta H^{\circ} = -971.0\, \mathrm{kJ/mol}\). 2. Using the formula \(\Delta S^{\circ} = [n_{products}\times S_m^{\circ}(products)] - [n_{reactants}\times S_m^{\circ}(reactants)]\), we find \(\Delta S^{\circ} = -156.8\, \mathrm{J/mol\cdot K}\). 3. Using the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\cdot \Delta S^{\circ}\) with \(T = 298.15\,\mathrm{K}\), we find \(\Delta G^{\circ} = -1276.1\, \mathrm{kJ/mol}\). 4. The equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\cdot \Delta S^{\circ}\) holds true for reaction (a). Repeat these steps for reactions (b), (c), and (d) to find their respective \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and \(\Delta G^{\circ}\) values and verify that the equation holds true for each.

Step-by-step solution

Calculate \(\Delta H^{\circ}\) for the reaction

To find \(\Delta H^{\circ}\), we will use the following formula: \(\Delta H^{\circ} = [n_{products}\times \Delta H_f^{\circ}(products)] - [n_{reactants}\times \Delta H_f^{\circ}(reactants)]\). We will gather the standard enthalpy of formation values from Appendix C for each species and calculate \(\Delta H^{\circ}\).

Calculate \(\Delta S^{\circ}\) for the reaction

To find \(\Delta S^{\circ}\), we will use the following formula: \(\Delta S^{\circ} = [n_{products}\times S_m^{\circ}(products)] - [n_{reactants}\times S_m^{\circ}(reactants)]\). We will gather the standard molar entropy values from Appendix C for each species and calculate \(\Delta S^{\circ}\).

Calculate \(\Delta G^{\circ}\) using \(T\Delta S^{\circ}\)

Now we will calculate \(\Delta G^{\circ}\) using the following equation: \(\Delta G^{\circ} = \Delta H^{\circ} - T\cdot \Delta S^{\circ}\). The temperature, \(T\), is given as \(25^{\circ}\,\mathrm{C}\), which is equivalent to \(298.15\,\mathrm{K}\). Substitute the values for \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) obtained in the previous steps, and calculate \(\Delta G^{\circ}\).

Verify that \(\Delta G^{\circ}\) holds true for the reaction

Finally, verify that the calculated value of \(\Delta G^{\circ}\) matches with the equation \(\Delta G^{\circ} = \Delta H^{\circ} - T\cdot \Delta S^{\circ}\). If it holds true, then our calculations are consistent. Repeat these steps for reactions (b), (c), and (d).

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation (\r \( \Delta H_f^\circ \) \r) represents the heat change when one mole of a substance is formed from its elements in their standard states. Think of it as the energy required to create a substance from the basic materials that make it up. For example, when we form water (\r \( H_2O \) \r) from hydrogen and oxygen gases, the process releases or absorbs energy, which is the standard enthalpy of formation of water.In thermochemistry calculations, the standard enthalpy of formation is a foundational piece when predicting how much heat a chemical reaction will produce or require. To calculate it, we follow the equation highlighted in the exercise, where the energy of the products and reactants are considered. Knowing this value is crucial for understanding the energetic implications of a reaction under standard conditions—typically defined as 1 atm of pressure and a specified temperature, usually 25°C (298 K).

Standard Molar Entropy

Standard molar entropy (\r \( S_m^\circ \) \r) is a measure of the energy dispersal within a mole of a substance at a constant temperature. Essentially, it gives an indication of how much energy is spread out or how disordered the molecules are within a substance. The higher the entropy, the more disordered the substance is.We calculate the change in standard molar entropy (\r \( \Delta S^\circ \) \r) by taking the difference between the entropy of the products and the reactants. This concept is intertwined with the Second Law of Thermodynamics, which states that entropy of the universe always increases over time. Thus, a positive change in entropy (\r \( \Delta S^\circ > 0 \) \r) suggests a collision that becomes more disorderly, which is often what happens when solids react to form gases. Conversely, a negative change indicates an increase in order, such as when gases condense into a liquid or solid.

Gibbs Free Energy

Gibbs free energy (\r \( G \) \r) is a central concept in chemical thermodynamics, representing the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. It's named after Josiah Willard Gibbs, who introduced the concept in the 1870s. The change in Gibbs free energy (\r \( \Delta G^\circ \) \r) during a reaction indicates whether the process is spontaneous or requires energy to proceed.The equation \r \( \Delta G^\circ = \Delta H^\circ - T\cdot \Delta S^\circ \) \r (where \r \( T \) \r is the temperature in kelvins) expresses this relationship beautifully. A negative \r \( \Delta G^\circ \) \r means the reaction will happen naturally under standard conditions, while a positive value suggests that it will not. This equation takes into account both enthalpy and entropy, revealing the delicate balance between energy released/absorbed and the disorder within the system that determines the fate of a chemical reaction.

Chemical Thermodynamics

Chemical thermodynamics is the study of the interrelation of heat and work with chemical reactions or with physical changes of state within the confines of the laws of thermodynamics. It incorporates all the core concepts we've discussed: standard enthalpy of formation, standard molar entropy, and Gibbs free energy.Thermodynamics paints a big picture of energy transformations in chemical processes, providing insights into the feasibility and extent of reactions. One key principle is the conservation of energy: energy can neither be created nor destroyed, only transformed. Understanding the principles of this field is essential for predicting reaction behavior, designing chemical processes, and even tackling real-world problems like energy production and material synthesis. As we saw in the exercise, the calculations of standard enthalpy, entropy, and Gibbs free energy all feed into the broader thermodynamics framework that governs chemical reactions and their energy profiles.