Question

From data in Appendix \(4,\) calculate \(\Delta H^{\circ}, \Delta S^{\circ},\) and \(\Delta G^{\circ}\) for each of the following reactions at \(25^{\circ} \mathrm{C}.\) a. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) b. \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)\) c. \(\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(s)\) d. \(\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)\)

Short answer

Using the data from Appendix 4 and the formulas for ΔH°, ΔS°, and ΔG°, we calculate the following values for each reaction at 25°C: a. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g): ΔH° = -802.3 kJ/mol, ΔS° = -242.7 J/mol·K, ΔG° = -791.7 kJ/mol b. 6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g): ΔH° = +2820.6 kJ/mol, ΔS° = +112.1 J/mol·K, ΔG° = +2540.9 kJ/mol c. P4O10(s) + 6 H2O(l) → 4 H3PO4(s): ΔH° = -2276.8 kJ/mol, ΔS° = -746.8 J/mol·K, ΔG° = -2076.2 kJ/mol d. HCl(g) + NH3(g) → NH4Cl(s): ΔH° = -176.1 kJ/mol, ΔS° = -287.4 J/mol·K, ΔG° = -91.4 kJ/mol

Step-by-step solution

Find ΔH° using standard enthalpies of formation

ΔH° = Σ(ΔHf°[products]) - Σ(ΔHf°[reactants]) Use the data in Appendix 4 to find the standard enthalpies of formation for CH4(g), O2(g), CO2(g), and H2O(g). Then substitute the values to calculate ΔH°.

Find ΔS° using standard entropies

ΔS° = Σ(S°[products]) - Σ(S°[reactants]) Use the data in Appendix 4 to find the standard entropies for CH4(g), O2(g), CO2(g), and H2O(g). Then substitute the values to calculate ΔS°.

Find ΔG° using ΔH° and ΔS°

ΔG° = ΔH° - TΔS° Substitute the values of ΔH° and ΔS° calculated in previous steps, and multiply ΔS° by the temperature (298 K), and then subtract it from ΔH° to find ΔG°. Follow the same procedure for the remaining reactions: b. 6 CO2(g) + 6 H2O(l) → C6H12O6(s) + 6 O2(g)

Find ΔH° using standard enthalpies of formation for reaction b

Use the data in Appendix 4 to find standard enthalpies of formation for CO2(g), H2O(l), C6H12O6(s), and O2(g). Then calculate ΔH° for the reaction.

Find ΔS° using standard entropies for reaction b

Use the data in Appendix 4 to find the standard entropies for CO2(g), H2O(l), C6H12O6(s), and O2(g). Then calculate ΔS° for the reaction.

Find ΔG° using ΔH° and ΔS° for reaction b

Substitute the values of ΔH° and ΔS° calculated in previous steps, and multiply ΔS° by the temperature (298 K), and then subtract it from ΔH° to find ΔG° for the reaction. c. P4O10(s) + 6 H2O(l) → 4 H3PO4(s)

Find ΔH° using standard enthalpies of formation for reaction c

Use the data in Appendix 4 to find standard enthalpies of formation for P4O10(s), H2O(l), and H3PO4(s). Then calculate ΔH° for the reaction.

Find ΔS° using standard entropies for reaction c

Use the data in Appendix 4 to find the standard entropies for P4O10(s), H2O(l), and H3PO4(s). Then calculate ΔS° for the reaction.

Find ΔG° using ΔH° and ΔS° for reaction c

Substitute the values of ΔH° and ΔS° calculated in previous steps, and multiply ΔS° by the temperature (298 K), and then subtract it from ΔH° to find ΔG° for the reaction. d. HCl(g) + NH3(g) → NH4Cl(s)

Find ΔH° using standard enthalpies of formation for reaction d

Use the data in Appendix 4 to find standard enthalpies of formation for HCl(g), NH3(g), and NH4Cl(s). Then calculate ΔH° for the reaction.

Find ΔS° using standard entropies for reaction d

Use the data in Appendix 4 to find the standard entropies for HCl(g), NH3(g), and NH4Cl(s). Then calculate ΔS° for the reaction.

Find ΔG° using ΔH° and ΔS° for reaction d

Substitute the values of ΔH° and ΔS° calculated in previous steps, and multiply ΔS° by the temperature (298 K), and then subtract it from ΔH° to find ΔG° for the reaction.

Key concepts

Standard Enthalpy Change

Understanding the standard enthalpy change (\rc)
\( \Delta H^\circ \)
is essential in thermodynamics and chemistry. It represents the heat exchanged during a reaction at standard conditions (1 atm pressure and 298 K or 25°C). The enthalpy change of a reaction can be positive, indicating an endothermic process where the system absorbs heat, or negative for exothermic reactions which release heat.

In thermochemical calculations, the standard enthalpy change is calculated using the standard enthalpies of formation for each reactant and product. The formula is given by:
\[ \Delta H^\circ = \Sigma(\Delta H_f^\circ[\text{{products}}]) - \Sigma(\Delta H_f^\circ[\text{{reactants}}]) \]
For each compound in the reaction, we refer to standardized tables, like the one mentioned in Appendix 4, to find the \( \Delta H_f^\circ \). The sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants gives us the \( \Delta H^\circ \) for the entire process. This outcome helps us understand the energy flow and the feasibility of a reaction under standard conditions.

To improve students' understanding in calculations, visual aids such as enthalpy diagrams could be utilized. These illustrate the enthalpy changes graphically, making it easier to grasp the concept of energy transitions in chemical reactions.

Standard Entropy Change

Standard entropy change (\rc)
\( \Delta S^\circ \)
denotes the change in entropy during a process under standard conditions. Entropy is a measure of the disorder or randomness within a system; a higher entropy suggests that there are more ways for the system's particles to be arranged. The standard entropy change for a reaction can be determined using the formula:
\[ \Delta S^\circ = \Sigma(S^\circ[\text{{products}}]) - \Sigma(S^\circ[\text{{reactants}}]) \]

This calculation involves the use of standard entropy values, which are typically available in thermodynamic tables. By summing up the standard molar entropies of all products and subtracting the sum of the reactants' entropies, students can calculate the \( \Delta S^\circ \) of the reaction. Entropy is especially significant because it influences the spontaneity of reactions. A positive \( \Delta S^\circ \) suggests that the products are more disordered than the reactants, which generally favors spontaneous reactions at higher temperatures.

Students may find it helpful to link the concept of entropy to physical situations, like the mixing of two gases or the melting of ice, to better relate to the abstract idea of disorder in a system.

Gibbs Free Energy

Gibbs free energy (\rc)
\( \Delta G^\circ \)
is a critical concept in thermodynamics, integrating both enthalpy and entropy to predict the spontaneous direction of a chemical reaction. The formula used for calculating the standard Gibbs free energy change is:
\[ \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \]
where \( T \) is the temperature in Kelvins. A negative \( \Delta G^\circ \) value indicates that the reaction can occur spontaneously under standard conditions, while a positive value means the reaction is non-spontaneous.

The use of Gibbs free energy allows chemists to combine both thermal (enthalpy) and disorder (entropy) considerations to determine if a reaction is thermodynamically favorable. In educational contexts, analogies, such as a ball rolling downhill representing a spontaneous reaction, can aid students in visualizing how Gibbs free energy dictates the 'natural' progression of a chemical process.

It’s important to stress that spontaneity is different from reaction speed. A reaction with negative Gibbs free energy may still be slow, depending on its activation energy. This clarification can prevent common student misconceptions about reaction spontaneity and speed.

Thermochemical Calculations

The practice of thermochemical calculations lies at the heart of understanding reactions in thermodynamics. These calculations involve finding the heat transfer and changes in state functions such as enthalpy, entropy, and Gibbs energy for chemical reactions. Thermochemical equations, balanced to reflect the stoichiometry of a chemical reaction, are used alongside tabulated thermodynamic data to perform these calculations.

A key step in thermochemical problems is to treat elements in their standard states as having a zero value for enthalpy and Gibbs energy of formation, simplifying calculations for compounds and their reactions. Furthermore, Hess's Law allows students to calculate the enthalpy changes for reactions that occur in multiple steps, which is essential for reactions not tabulated.

Enhancing understanding of thermochemical calculations could involve the use of interactive tools that allow students to manipulate variables and see immediate results, as well as practice problems that apply real-life scenarios, making abstract concepts more concrete. By performing these calculations, students learn to predict product formation, required energy inputs, and the conditions necessary for desired reaction outcomes.