Question

Use data in Appendix C to calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) at \(25^{\circ} \mathrm{C}\) for each of the following reactions. $$ \begin{array}{l}{\text { (a) } 4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)} \\ {\text { (b) } \mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)} \\\ {\text { (c) } 2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)} \\ {\text { (d) } \mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)}\end{array} $$

Short answer

For reaction (a), we have: \(\Delta H^{\circ} = -2268.40 \ kJ/mol\) \(\Delta S^{\circ} = -543.84 \ J/(mol \cdot K)\) \(\Delta G^{\circ} = -2104.29 \ kJ/mol\)

Step-by-step solution

1. Write down the given reaction:

(a) \(4 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\)

2. Find the standard enthalpies of formation, standard entropies and standard molar Gibbs free energies from Appendix C:

For the reactants, we have: Cr(s): \(\Delta H_f^{\circ} = 0 \ kJ/mol\), \(S^{\circ} = 7.19 \ J/(mol \cdot K)\), \(G^{\circ} = 0 \ kJ/mol\) O2(g): \(\Delta H_f^{\circ} = 0 \ kJ/mol\), \(S^{\circ} = 205.14 \ J/(mol \cdot K)\), \(G^{\circ} = 0 \ kJ/mol\) For the product, we have: Cr2O3(s): \(\Delta H_f^{\circ} = -1134.20 \ kJ/mol\), \(S^{\circ} = 81.40 \ J/(mol \cdot K)\), \(G^{\circ} = -1046.50 \ kJ/mol\)

3. Calculate the \(\Delta H^{\circ}\), \(\Delta S^{\circ}\), and \(\Delta G^{\circ}\) for the reaction:

For the reaction, we can write: \(\Delta H^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) = 2(-1134.20) - (4(0) + 3(0)) = -2268.40 \ kJ/mol\) \(\Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants}) = 2(81.40) - (4(7.19) + 3(205.14)) = -543.84 \ J/(mol \cdot K)\) Now, we can calculate \(\Delta G^{\circ}\) using the formula: \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), where \(T\) is the temperature in Kelvin: \(T = 25^{\circ} C + 273.15 = 298.15 \ K\) \(\Delta G^{\circ} = -2268.40- 298.15(-543.84/1000) = -2104.29 \ kJ/mol\) For reaction (a), we have: \(\Delta H^{\circ} = -2268.40 \ kJ/mol\) \(\Delta S^{\circ} = -543.84 \ J/(mol \cdot K)\) \(\Delta G^{\circ} = -2104.29 \ kJ/mol\) Now, you should follow the same steps for reactions (b), (c), and (d) using the corresponding data from Appendix C and the given chemical equations.

Key concepts

Enthalpy of Reaction

In thermodynamics, the enthalpy of reaction (\(\Delta H^{\circ}\)) is a crucial concept. It represents the total heat absorbed or released during a chemical reaction under constant pressure. When we calculate \(\Delta H^{\circ}\), we consider the standard enthalpies of formation for all reactants and products involved.

• The formula used is:\[ \Delta H^{\circ} = \sum \Delta H_f^{\circ}(\text{products}) - \sum \Delta H_f^{\circ}(\text{reactants}) \]
• A negative \(\Delta H^{\circ}\) indicates that the reaction is exothermic, meaning it releases heat.
• A positive \(\Delta H^{\circ}\) suggests an endothermic reaction, where heat is absorbed.

Applying this to our reaction of chromium and oxygen forming chromium oxide, we find a \(\Delta H^{\circ}\) of -2268.40 \(\text{kJ/mol}\). This negative value signifies an exothermic process.

Gibbs Free Energy

Gibbs Free Energy (\(\Delta G^{\circ}\)) is a measure of the spontaneity of a reaction. It combines the effects of enthalpy, temperature, and entropy. The formula is:\[ \Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \]

• Here, \(T\) represents the temperature in Kelvin, and ensures that we relate energy changes appropriately.
• A negative \(\Delta G^{\circ}\) value implies that a process is spontaneous at the given temperature.
• If \(\Delta G^{\circ}\) is positive, the reaction is non-spontaneous under the standard conditions provided.

For our chromium-oxygen reaction, \(\Delta G^{\circ}\) is -2104.29 \(\text{kJ/mol}\), indicating that the reaction proceeds spontaneously.

Standard Entropy

Standard entropy (\(\Delta S^{\circ}\)) is a measure of disorder or randomness in a system. It plays a vital role in understanding reaction feasibility because it reflects the number of possible microstates a system can possess.

• The change in standard entropy for a reaction can be calculated using:\[ \Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants}) \]
• A positive \(\Delta S^{\circ}\) means that the system's disorder increases during the reaction, which is generally favorable.
• A negative \(\Delta S^{\circ}\) suggests a decrease in disorder, which might counteract spontaneity unless compensated by a favorable enthalpy change.

In our example, \(\Delta S^{\circ}\) was calculated as -543.84 \(\text{J/(mol\cdot K)}\), meaning the reaction leads to a decrease in disorder, possibly due to the ordered crystal formation of \(\text{Cr}_2\text{O}_3\) from gaseous \(\text{O}_2\).