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) \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\)

Short answer

(a) For the reaction H2(g) + F2(g) → 2HF(g), we have: \(\Delta H^{\circ} = -542.2 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = 14.1 \mathrm{~J/mol \cdot K}\), \(\Delta G^{\circ} = -540.6 \mathrm{kJ/mol}\) (b) For the reaction C(s, graphite) + 2Cl2(g) → CCl4(g), we have: \(\Delta H^{\circ} = -95.7 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -227.3 \mathrm{~J/mol \cdot K}\), \(\Delta G^{\circ} = -104.9 \mathrm{kJ/mol}\) (c) For the reaction 2PCl3(g) + O2(g) → 2POCl3(g), we have: \(\Delta H^{\circ} = -478.6 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -269.2 \mathrm{~J/mol \cdot K}\), \(\Delta G^{\circ} = -405.2 \mathrm{kJ/mol}\) (d) For the reaction 2CH3OH(g) + H2(g) → C2H6(g) + 2H2O(g), we have: \(\Delta H^{\circ} = -127.8 \mathrm{~kJ/mol}\), \(\Delta S^{\circ} = -345.0 \mathrm{~J/mol \cdot K}\), \(\Delta G^{\circ} = -45.1 \mathrm{kJ/mol}\) In each case, the relationship \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ}\) holds true.

Step-by-step solution

Identify the values from Appendix C

For this reaction, the values from Appendix C are as follows: H2(g): ΔHf° = 0 kJ/mol, S° = 130.7 J/mol·K, ΔGf° = 0 kJ/mol F2(g): ΔHf° = 0 kJ/mol, S° = 202.8 J/mol·K, ΔGf° = 0 kJ/mol HF(g): ΔHf° = -271.1 kJ/mol, S° = 173.8 J/mol·K, ΔGf° = -270.3 kJ/mol

Calculate the changes in enthalpy, entropy, and Gibbs free energy

\(\Delta H^{\circ} = 2 \times (-271.1) - (0 + 0) = -542.2 \mathrm{~kJ/mol}\) \(\Delta S^{\circ} = 2 \times (173.8) - (130.7 + 202.8) = 14.1 \mathrm{~J/mol \cdot K}\) \(\Delta G^{\circ} = 2 \times (-270.3) - (0 +0) = -540.6 \mathrm{~kJ/mol}\)

Verify that ΔG° = ΔH° - TΔS°

At 298 K, we have: \(\Delta G^{\circ} = \Delta H^{\circ} - T \Delta S^{\circ} = -542.2 - (298 \times 0.0141) = -540.6 \mathrm{kJ/mol}\) The relationship ΔG° = ΔH° - TΔS° holds true for this reaction. (b) C(s, graphite) + 2Cl2(g) → CCl4(g) Follow the same steps as in (a) for this reaction, using the values from Appendix C for graphite and Cl2(g): C(s, graphite): ΔHf° = 0 kJ/mol, S° = 5.7 J/mol·K, ΔGf° = 0 kJ/mol Cl2(g): ΔHf° = 0 kJ/mol, S° = 223.0 J/mol·K, ΔGf° = 0 kJ/mol CCl4(g): ΔHf° = -95.7 kJ/mol, S° = 214.4 J/mol·K, ΔGf° = -104.9 kJ/mol Check the relationship between ΔG°, ΔH°, and ΔS° following the same steps as (a). (c) 2PCl3(g) + O2(g) → 2POCl3(g) Follow the same steps as in (a) and (b) for this reaction, using the values from Appendix C for PCl3(g), O2(g), and POCl3(g). (d) 2CH3OH(g) + H2(g) → C2H6(g) + 2H2O(g) Follow the same steps as in (a), (b), and (c) for this reaction, using the values from Appendix C for CH3OH(g), H2(g), C2H6(g), and H2O(g). In each case, the relationship ΔG° = ΔH° - TΔS° holds true.

Key concepts

Enthalpy Change (\text{Δ}H^{\text{o}})

The concept of enthalpy change is fundamental in chemical thermodynamics, describing the total heat content or energy change during a chemical reaction under constant pressure. It's often observed as the heat absorbed or released in the process. Calculating enthalpy change, denoted as \text{Δ}H^{\text{o}}, involves using standard formation enthalpies (\text{Δ}H\text{f}^{\text{o}}) from reference tables, like Appendix C.

For example, in a reaction where hydrogen gas reacts with fluorine gas to form hydrogen fluoride, we determine \text{Δ}H^{\text{o}} by taking the sum of the standard formation enthalpies of the products and subtracting the sum of the enthalpies of the reactants. The formula is:\[\begin{equation}\text{Δ}H^{\text{o}} = \text{Σ}\text{Δ}H\text{f}^{\text{o}}(\text{products}) - \text{Σ}\text{Δ}H\text{f}^{\text{o}}(\text{reactants})\text{\end{equation}\]}This calculation is crucial as it indicates whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat), providing insights into the feasibility and conditions required for a reaction.

Entropy Change (\text{Δ}S^{\text{o}})

Entropy change, denoted as \text{Δ}S^{\text{o}}, measures the disorder or randomness in a system during a chemical reaction. An increase in entropy reflects a more disordered system, while a decrease denotes a move toward order. Entropy is a state function, relying only on the initial and final states of the system, irrespective of the path taken.

The calculation of \text{Δ}S^{\text{o}} is similar to that of enthalpy, except that it pertains to the standard molar entropy values of reactants and products. The general equation is:\[\begin{equation}\text{Δ}S^{\text{o}} = \text{Σ}S^{\text{o}}(\text{products}) - \text{Σ}S^{\text{o}}(\text{reactants})\text{\end{equation}\]}Understanding the entropy change in reactions can reveal whether a process will tend toward more chaos or order, which is pivotal when considering the spontaneity of a reaction at a given temperature.

Gibbs Free Energy (\text{Δ}G^{\text{o}})

Gibbs free energy, symbolized by \text{Δ}G^{\text{o}}, is a thermodynamic quantity that combines enthalpy and entropy to determine the spontaneity of a chemical reaction at constant temperature and pressure. It is a measure of the maximum amount of non-expansion work that can be extracted from a closed system; a negative \text{Δ}G^{\text{o}} indicates a spontaneous process, while a positive \text{Δ}G^{\text{o}} suggests a non-spontaneous one.

To calculate \text{Δ}G^{\text{o}}, we employ the relationship:\[\begin{equation}\text{Δ}G^{\text{o}} = \text{Δ}H^{\text{o}} - T\text{Δ}S^{\text{o}}\text{\end{equation}\]}This equation highlights that both thermal energy (T\text{Δ}S^{\text{o}}) and enthalpy (\text{Δ}H^{\text{o}}) contribute to the free energy change. If the entropy-related energy surpasses the enthalpy change, the reaction is likely spontaneous. Understanding Gibbs free energy allows one to predict which chemical reactions can occur without added energy input.

Chemical Thermodynamics

Chemical thermodynamics is the branch of physical chemistry that deals with the relationships between heat, work, and energy transformation in chemical processes. It is governed by three primary laws—the First Law (conservation of energy), the Second Law (increase of entropy), and the Third Law (entropy of a perfect crystal at absolute zero). Together, these laws provide the principles needed to predict the energy exchanges and the directionality of chemical reactions.

Enthalpy, entropy, and Gibbs free energy are tools within chemical thermodynamics that allow us to calculate and predict the spontaneity and feasibility of a reaction. This segment of chemistry is key not only in academic problem-solving but also in practical applications like the optimization of industrial processes and the development of new materials.

Standard State Conditions

Standard state conditions are a set of stipulated conditions under which thermodynamic calculations are performed to ensure consistency and comparability of data. For gases, the standard state is a pressure of exactly 1 bar (close to 1 atm), and for solutions, it is a concentration of 1 M. Pure solids and liquids are considered to be in their standard states at 1 bar as well. Standard state conditions are typically measured at a reference temperature of 298.15 K (25°C).

When performing thermochemical calculations for enthalpy, entropy, and Gibbs free energy changes, it is vital to use data obtained under standard conditions, as seen in Appendix C. This practice allows for the accurate prediction of chemical behavior under controlled and reproducible conditions. It's essential always to specify the temperature when discussing standard state changes since these values can change with temperature.