Question

For the reaction $$ \mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g) $$ the value of \(\Delta G^{\circ}\) is \(-374 \mathrm{~kJ}\). Use this value and data from Appendix 4 to calculate the value of \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{SF}_{4}(g)\).

Short answer

The standard Gibbs free energy of formation for SF₄(g) is approximately -526 kJ/mol.

Step-by-step solution

Write down the reaction and given values

We are given the reaction: \[ \mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g) \] The value of ΔG° for the reaction is -374 kJ.

Locate values of ΔGf° in Appendix 4

Using Appendix 4, find the standard Gibbs free energy of formation, ΔGf° for F₂(g) and SF₆(g). Let's assume the following values from Appendix 4: ΔGf°(F₂) ≈ 0 kJ/mol (since F₂ is in its standard state) ΔGf°(SF₆) ≈ -900 kJ/mol (approximate value, the exact value may differ)

Use the equation relating ΔG° to ΔGf° of the species

The equation for the reaction's ΔG° in terms of standard Gibbs free energies of formation is: ΔG° = Σ(ΔGf° of products) - Σ(ΔGf° of reactants) For our reaction, ΔG° = ΔGf°(SF₆) - (ΔGf°(SF₄) + ΔGf°(F₂)) Plug in the given values of ΔG° and the values found in Appendix 4: -374 kJ = (-900 kJ/mol) - (ΔGf°(SF₄) + 0 kJ/mol)

Solve for ΔGf°(SF₄)

Now solve the equation for the standard Gibbs free energy of formation, ΔGf°(SF₄): ΔGf°(SF₄) = -900 kJ/mol - (-374 kJ) ΔGf°(SF₄) = -900 kJ/mol + 374 kJ ΔGf°(SF₄) ≈ -526 kJ/mol The standard Gibbs free energy of formation for SF₄(g) is approximately -526 kJ/mol.

Key concepts

Chemical Thermodynamics

Chemical thermodynamics is a branch of physical chemistry that deals with the relationship between heat, work, and chemical reactions or with the energy changes that occur during these processes. The first law of thermodynamics states that energy cannot be created or destroyed, only transformed. However, the second law introduces the concept of entropy, highlighting that natural processes tend to move towards a state of disorder or randomness.

Within this framework, thermodynamics helps us understand how chemical reactions can occur spontaneously and what drives these reactions. It assesses the energy and entropy changes associated with reactions, providing essential insights into whether a reaction will occur under certain conditions and how much energy is needed or released during the process.

• Spontaneity of reactions: determined by energy change (enthalpy) and randomness (entropy).
• Energy transformations: how energy changes form (heat to work and vice versa).
• Equilibrium states: where energy and entropy balance is reached, and no net change occurs.

Gibbs Free Energy Equation

The Gibbs free energy equation is foundational in understanding chemical thermodynamics and predicts the spontaneity of reactions. It's given by the formula: \[ G = H - TS \] where \( G \) is the Gibbs free energy, \( H \) is the enthalpy, \( T \) is the temperature in Kelvin, and \( S \) is the entropy of the system.

The key to this equation is the change in Gibbs free energy \( \Delta G \). If \( \Delta G \) is negative, the process can occur spontaneously; if positive, the reaction is non-spontaneous under standard conditions. At constant temperature and pressure, the equation simplifies to \[ \Delta G = \Delta H - T\Delta S \] where \( \Delta H \) and \( \Delta S \) are the changes in enthalpy and entropy, respectively, for a reaction. In the given exercise, understanding this equation is critical to find the standard Gibbs free energy of formation for \( \mathrm{SF}_{4}(g) \).

• Spontaneous process: occurs without input of additional energy (\( \Delta G < 0 \)).
• Equilibrium: achieved when \( \Delta G = 0 \), system is at its lowest energy state.
• Non-spontaneous process: requires energy input (\( \Delta G > 0 \)).

Enthalpy

Enthalpy, symbolized by \( H \), is a thermodynamic quantity equivalent to the total heat content of a system. It reflects the energy needed for the creation of a system and the energy required to make room for it by displacing its environment and establishing its volume and pressure.

The change in enthalpy \( \Delta H \) is a crucial factor when examining chemical reactions and can be either endothermic or exothermic. For an exothermic reaction, \( \Delta H \) is negative, indicating that heat is released. Conversely, for an endothermic reaction, \( \Delta H \) is positive, signifying that the reaction absorbs heat from the surroundings.

• Endothermic processes: require heat; \( \Delta H > 0 \).
• Exothermic processes: release heat; \( \Delta H < 0 \).
• Standard enthalpy of formation: the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

In relation to the provided exercise, finding the standard enthalpy of formation for \( \mathrm{SF}_{4}(g) \) involves understanding and applying the concepts of enthalpy within the broader context of Gibbs free energy calculations.