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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

Expert verified
The standard Gibbs free energy of formation for SFâ‚„(g) is approximately -526 kJ/mol.

Step by step solution

01

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.
02

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)
03

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)
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

For each of the following pairs of substances, which substance has the greater value of \(S^{\circ}\) ? a. \(\mathrm{C}_{\text {graphite }}(s)\) or \(\mathrm{C}_{\text {diamond }}(s)\) b. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\) c. \(\mathrm{CO}_{2}(s)\) or \(\mathrm{CO}_{2}(g)\)

Impure nickel, refined by smelting sulfide ores in a blast furnace, can be converted into metal from \(99.90 \%\) to \(99.99 \%\) purity by the Mond process. The primary reaction involved in the Mond process is $$ \mathrm{Ni}(s)+4 \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) $$ a. Without referring to Appendix 4 , predict the sign of \(\Delta S^{\circ}\) for the above reaction. Explain. b. The spontaneity of the above reaction is temperaturedependent. Predict the sign of \(\Delta S_{\text {surr }}\) for this reaction. Explain. c. For \(\mathrm{Ni}(\mathrm{CO})_{4}(g), \Delta H_{\mathrm{f}}^{\circ}=-607 \mathrm{~kJ} / \mathrm{mol}\) and \(S^{\circ}=417 \mathrm{~J} / \mathrm{K}\). mol at \(298 \mathrm{~K}\). Using these values and data in Appendix 4 , calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for the above reaction. d. Calculate the temperature at which \(\Delta G^{\circ}=0(K=1)\) for the above reaction, assuming that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. e. The first step of the Mond process involves equilibrating impure nickel with \(\mathrm{CO}(g)\) and \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) at about \(50^{\circ} \mathrm{C}\). The purpose of this step is to convert as much nickel as possible into the gas phase. Calculate the equilibrium constant for the above reaction at \(50 .{ }^{\circ} \mathrm{C}\). f. In the second step of the Mond process, the gaseous \(\mathrm{Ni}(\mathrm{CO})_{4}\) is isolated and heated to \(227^{\circ} \mathrm{C}\). The purpose of this step is to deposit as much nickel as possible as pure solid (the reverse of the preceding reaction). Calculate the equilibrium constant for the preceding reaction at \(227^{\circ} \mathrm{C}\). g. Why is temperature increased for the second step of the Mond process? h. The Mond process relies on the volatility of \(\mathrm{Ni}(\mathrm{CO})_{4}\) for its success. Only pressures and temperatures at which \(\mathrm{Ni}(\mathrm{CO})_{4}\) is a gas are useful. A recently developed variation of the Mond process carries out the first step at higher pressures and a temperature of \(152^{\circ} \mathrm{C}\). Estimate the maximum pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) that can be attained before the gas will liquefy at \(152^{\circ} \mathrm{C}\). The boiling point for \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(42^{\circ} \mathrm{C}\) and the enthalpy of vaporization is \(29.0 \mathrm{~kJ} / \mathrm{mol}\). [Hint: The phase change reaction and the corresponding equilibrium expression are $$ \mathrm{Ni}(\mathrm{CO})_{4}(l) \rightleftharpoons \mathrm{Ni}(\mathrm{CO})_{4}(g) \quad K=P_{\mathrm{Ni}(\mathrm{CO})} $$ \(\mathrm{Ni}(\mathrm{CO})_{4}(g)\) will liquefy when the pressure of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is greater than the \(K\) value.]

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ} $$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Choose the substance with the larger positional probability in each case. a. 1 mole of \(\mathrm{H}_{2}\) (at STP) or 1 mole of \(\mathrm{H}_{2}\) (at \(\left.100^{\circ} \mathrm{C}, 0.5 \mathrm{~atm}\right)\) b. 1 mole of \(\mathrm{N}_{2}\) (at STP) or 1 mole of \(\mathrm{N}_{2}\) (at \(\left.100 \mathrm{~K}, 2.0 \mathrm{~atm}\right)\) c. 1 mole of \(\mathrm{H}_{2} \mathrm{O}(s)\) (at \(\left.0^{\circ} \mathrm{C}\right)\) or 1 mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) (at \(\left.20^{\circ} \mathrm{C}\right)\)

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)\) c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

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