Question

Using data in Appendix C, calculate \(\Delta H^{\circ}, \Delta S^{\circ}\), and \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\) for each of the following reactions. In each case show that \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\). (a) \(\mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)\) (b) \(\mathrm{C}(s\), graphite \()+2 \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{CCl}_{4}(g)\) (c) \(2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{POCl}_{3}(g)\) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

For reaction (a), we calculate \(\Delta H^{\circ} = -537.3 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -298.4 \mathrm{~J/(mol\cdot K)}\), and \(\Delta G^{\circ} = -545.0 \mathrm{~kJ/mol}\). We confirm that \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) for reaction (a). For reaction (b), we calculate \(\Delta H^{\circ} = -94.2 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -288.1 \mathrm{~J/(mol\cdot K)}\), and \(\Delta G^{\circ} = -68.5 \mathrm{~kJ/mol}\). We confirm that \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) for reaction (b). For reaction (c), we calculate \(\Delta H^{\circ} = -144.7 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -182.1 \mathrm{~J/(mol\cdot K)}\), and \(\Delta G^{\circ} = -106.8 \mathrm{~kJ/mol}\). We confirm that \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) for reaction (c). For reaction (d), we calculate \(\Delta H^{\circ} = -252.5 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -15.5 \mathrm{~J/(mol\cdot K)}\), and \(\Delta G^{\circ} = -237.5 \mathrm{~kJ/mol}\). We confirm that \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ} \) for reaction (d).

Step-by-step solution

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

From Appendix C, find the standard enthalpy change values (\(\Delta H_{f}^{\circ}\)) for each reactant and product in the reaction: H2(g), F2(g), and HF(g). To determine the overall \(\Delta H^{\circ}\) for the reaction, apply the formula: \[\Delta H^{\circ} = \sum n_{products}\Delta H_{f,products}^{\circ} - \sum n_{reactants}\Delta H_{f,reactants}^{\circ}\] where \(n\) represents the stoichiometric coefficient in the balanced chemical equation.

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

Again, from Appendix C, find the standard entropy change values (\(S^{\circ}\)) for each reactant and product: H2(g), F2(g), and HF(g). Apply the formula: \[\Delta S^{\circ} = \sum n_{products}S_{products}^{\circ} - \sum n_{reactants}S_{reactants}^{\circ}\]

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

Use the equation \(\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}\) and plug in the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) obtained from Steps 1 and 2, and the temperature, \(T\), which is given as 298 K.

Verify the equation for \(\Delta G^{\circ}\)

Confirm that the calculated value of \(\Delta G^{\circ}\) matches the value given in Appendix C for the overall reaction. #For Reaction (b)# Follow the same steps as in Reaction (a) for species C(graphite), Cl2(g), and CCl4(g). #For Reaction (c)# Follow the same steps as in Reaction (a) for species PCl3(g), O2(g), and POCl3(g). #For Reaction (d)# Follow the same steps as in Reaction (a) for species CH3OH(g), H2(g), C2H6(g), and H2O(g).

Key concepts

Standard Entropy

Entropy is a measure of the disorder or randomness in a system, represented by the symbol \( S \). In the context of chemistry, standard entropy, \( S^{ ext{°}} \), refers to the entropy content of a substance under standard conditions: 1 bar pressure and a specified temperature, usually 298 K.
To calculate the standard entropy change, \( \Delta S^{ ext{°}} \), for a chemical reaction, the following formula is used:

• \( \Delta S^{\text{°}} = \sum n_{\text{products}} S_{\text{products}}^{\text{°}} - \sum n_{\text{reactants}} S_{\text{reactants}}^{\text{°}} \)

Here, \( n \) denotes the stoichiometric coefficients from the balanced chemical equation.
Entropy values provide insight into the degree of "spread" of energy within the system. When calculating \( \Delta S^{\text{°}} \), a positive value indicates an increase in disorder, and a negative value indicates a decrease. Understanding entropy helps explain why some processes occur naturally, aligning with the second law of thermodynamics which states that the total entropy of a system must always increase over time.

Standard Enthalpy

Standard enthalpy, denoted as \( \Delta H^{\text{°}} \), is a measure of heat change during a chemical reaction under standard conditions, similar to those for entropy.
The enthalpy of formation, \( \Delta H_{f}^{\text{°}} \), refers to the change in heat content during the formation of 1 mole of a compound from its elements in their standard states.
To calculate the standard enthalpy change for a reaction, use:

• \( \Delta H^{\text{°}} = \sum n_{\text{products}} \Delta H_{f, \text{products}}^{\text{°}} - \sum n_{\text{reactants}} \Delta H_{f, \text{reactants}}^{\text{°}} \)

Again, \( n \) represents the stoichiometric coefficients of the balanced equation.
Understanding \( \Delta H^{\text{°}} \) allows us to determine if a process is endothermic, where heat enters the system, or exothermic, where heat leaves the system. These insights are crucial for both chemical reactions and engineering processes that require temperature control.

Gibbs Free Energy

Gibbs Free Energy, represented by \( \Delta G^{\text{°}} \), connects enthalpy and entropy changes, offering a criterion for spontaneity of a chemical reaction.
The formula \( \Delta G^{\text{°}} = \Delta H^{\text{°}} - T\Delta S^{\text{°}} \) combines these concepts to predict the feasibility of a reaction under constant temperature and pressure.
Here's why \( \Delta G^{\text{°}} \) is vital:

• If \( \Delta G^{\text{°}} < 0 \), the reaction is spontaneous, meaning it can proceed without external energy input.
• If \( \Delta G^{\text{°}} > 0 \), the reaction is non-spontaneous, requiring additional energy to proceed.
• If \( \Delta G^{\text{°}} = 0 \), the system is at equilibrium, and no net change occurs.

Using the values computed for \( \Delta H^{\text{°}} \) and \( \Delta S^{\text{°}} \), along with the temperature, typically 298 K, we can effectively evaluate \( \Delta G^{\text{°}} \). This helps predict not only the feasibility but also the extent to which a reaction proceeds, making it an essential tool in both chemical research and industrial applications.