Determine whether titanium dioxide can be reduced by carbon at \(1000 . \mathrm{K}\) in each of the following reactions: (a) \(\mathrm{TiO}_{2}\) (s) \(+2 \mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+2 \mathrm{CO}(\mathrm{g})\) (b) \(\mathrm{TiO}_{2}\) (s) \(+\mathrm{C}(\mathrm{s}) \rightarrow \mathrm{Ti}(\mathrm{s})+\mathrm{CO}_{2}\) (g) given that, at \(1000 . \mathrm{K}, \Delta G_{f}{ }^{\circ}\left(\mathrm{CO}\right.\), g) \(=-200 . \mathrm{kJ} \cdot \mathrm{mol}^{-1}\), \(\Delta G_{f}^{\circ}\left(\mathrm{CO}_{2}, \mathrm{~g}\right)=-396 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} ;\) and \(\Delta G_{f}{ }^{\circ}\left(\mathrm{TiO}_{2}, \mathrm{~s}\right)=\) \(-762 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

Short Answer

Expert verified
Neither reaction (a) nor reaction (b) is spontaneous at 1000K as both have positive Gibbs free energy changes of 362 kJ/mol and 366 kJ/mol, respectively.

Step by step solution

01

Write down the standard Gibbs free energy of formation for each species

Given are the standard Gibbs free energies of formation for CO, CO2, and TiO2 at 1000K. They are: \[\Delta G_f{ }^{\circ}(\mathrm{CO}, \mathrm{g}) = -200 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}, \]\[\Delta G_f{ }^{\circ}(\mathrm{CO}_2, \mathrm{g}) = -396 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}, \]\[\Delta G_f{ }^{\circ}(\mathrm{TiO}_2, \mathrm{s}) = -762 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}.\]Note: Standard Gibbs free energy of formation for elements in their standard states, like Ti(s) and C(s), is zero.
02

Calculate the Gibbs free energy change \(\Delta G\) for reaction (a)

Use the formula \[\Delta G = \Sigma \Delta G_f{ }^{\circ}(products) - \Sigma \Delta G_f{ }^{\circ}(reactants).\]For reaction (a), \[\Delta G = [2 \Delta G_f{ }^{\circ}(\mathrm{CO})] - [\Delta G_f{ }^{\circ}(\mathrm{TiO}_2) + 2 \Delta G_f{ }^{\circ}(\mathrm{C})].\]Plug in the values to get: \[\Delta G = [2(-200 \mathrm{~kJ/mol})] - [-762 \mathrm{~kJ/mol} + 2(0)].\]\[\Delta G = -400 \mathrm{~kJ/mol} + 762 \mathrm{~kJ/mol} = 362 \mathrm{~kJ/mol}.\]Since the \(\Delta G\) is positive, the reaction is not spontaneous at 1000K.
03

Calculate the Gibbs free energy change \(\Delta G\) for reaction (b)

For reaction (b), apply the same formula as in Step 2: \[\Delta G = [\Delta G_f{ }^{\circ}(\mathrm{CO}_2)] - [\Delta G_f{ }^{\circ}(\mathrm{TiO}_2) + \Delta G_f{ }^{\circ}(\mathrm{C})].\]Plugging in the values: \[\Delta G = [-396 \mathrm{~kJ/mol}] - [-762 \mathrm{~kJ/mol} + 0].\]\[\Delta G = -396 \mathrm{~kJ/mol} + 762 \mathrm{~kJ/mol} = 366 \mathrm{~kJ/mol}.\]Since the \(\Delta G\) for reaction (b) is also positive, this reaction is not spontaneous at 1000K either.

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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 the branch of physical chemistry concerned with understanding the energy changes that accompany chemical reactions and the establishment of the conditions under which reactions can occur. A key concept within this field is Gibbs free energy, denoted as \(G\), which provides vital information about the spontaneity of a chemical reaction.

At its core, the Gibbs free energy is a measure of the maximum reversible work that a thermodynamic system can perform at constant temperature and pressure. It is an extensive property, meaning it depends on the quantity of matter in the system. The change in Gibbs free energy (\(\Delta G\)) for a process is a crucial indicator of whether or not a process will occur spontaneously.

If \(\Delta G\) is negative, the process can happen on its own without any energy input from the surroundings. If \(\Delta G\) is positive, the process is non-spontaneous and requires the input of energy to proceed. Lastly, if \(\Delta G\) is zero, the system is at equilibrium, indicating that the forward and reverse processes are occurring at the same rate. This concept is foundational when examining chemical reactions and predicting their behavior under different conditions.
Spontaneous Reactions
Spontaneous reactions are processes that occur naturally without the need for continuous energy input from their surroundings. These reactions exhibit a tendency to proceed in a particular direction until they reach equilibrium. A spontaneous reaction can do work on the surroundings and is associated with a decrease in the Gibbs free energy (\(\Delta G < 0\)).

To determine the spontaneity of a reaction, we consider not only the enthalpy change (\(\Delta H\)) but also the change in entropy (\(\Delta S\)) and the absolute temperature (\(T\)) of the system. The Gibbs free energy equation \(\Delta G = \Delta H - T\Delta S\) allows us to predict if a reaction is spontaneous by considering all these factors. High entropy and low enthalpy (heat released) favor spontaneity, reflecting nature's tendency towards disorder and energy dispersal.

When a reaction has a positive \(\Delta G\), it means that the reaction needs energy to proceed and is considered non-spontaneous under current conditions. However, modifying conditions such as temperature may change the spontaneity. For students, understanding this principle is essential when dealing with thermochemical calculations to predict the feasibility of reactions.
Thermochemical Calculations
Thermochemical calculations involve quantitative assessment of the heat and work associated with chemical processes. These calculations often use the concept of enthalpy (\(\Delta H\)), entropy (\(\Delta S\)), and Gibbs free energy (\(\Delta G\)) in order to predict the direction and extend of chemical reactions.

In the context of the exercise provided, the Gibbs free energy change for reactions (a) and (b) were calculated using the standard free energies of formation of the products and the reactants, using the formula \(\Delta G = \Sigma \Delta G_f{ }^{\circ}(products) - \Sigma \Delta G_f{ }^{\circ}(reactants)\). By substituting the given values for each species into the formula, students were able to determine whether the reactions would occur spontaneously at 1000 K.

The ability to perform these thermochemical calculations is not only a valuable skill for achieving academic success but also for practical applications in various scientific and engineering fields. It involves a clear understanding of tabulated thermodynamic data, such as standard enthalpies of formation, and the application of this information to predict the feasibility of industrial processes.

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Most popular questions from this chapter

Calculate the change in entropy when the pressure of \(3.44 \mathrm{~g}\) of helium gas is decreased from \(125.0 \mathrm{kPa}\) to \(65.0 \mathrm{kPa}\) while the temperature decreases from \(335 \mathrm{~K}\) to \(273 \mathrm{~K}\). Assume ideal behavior.

Without performing any calculations, predict whether there is an increase or a decrease in entropy for each of the following processes: (a) \(\mathrm{Cl}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{HCl}(\mathrm{aq})+\mathrm{HClO}(\mathrm{aq})\); (b) \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}(\mathrm{~s}) \longrightarrow 3 \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{PO}_{4}^{3-}\) (aq); (c) \(\mathrm{SO}_{2}(\mathrm{~g})+\) \(\mathrm{Br}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})+2 \operatorname{HBr}(\mathrm{aq})\)

List the following substances in order of increasing molar entropy at \(298 \mathrm{~K}: \mathrm{CO}_{2}(\mathrm{~g}), \mathrm{Ar}(\mathrm{g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{l}), \mathrm{Ne}(\mathrm{g})\). Explain your reasoning.

The standard entropy of vaporization of acetone is approximately \(85 \mathrm{~J} \cdot \mathrm{K}^{-1} \cdot \mathrm{mol}^{-1}\) at its boiling point. (a) Estimate the standard enthalpy of vaporization of acetone at its normal boiling point of \(56.2^{\circ} \mathrm{C}\). (b) What is the entropy change of the surroundings when 10. \(\mathrm{g}\) of acetone, \(\mathrm{CH}_{3} \mathrm{COCH}_{3}\), condenses at its normal boiling point?

Use the standard Gibbs free energies of formation in Appendix \(2 \mathrm{~A}\) to calculate \(\Delta G^{\circ}\) for each of the following reactions at \(25^{\circ} \mathrm{C}\). Comment on the spontaneity of each reaction under standard conditions at \(25^{\circ} \mathrm{C}\). (a) \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \rightarrow \mathrm{NH}_{3}\) (g) \(+\mathrm{HCl}(\mathrm{g})\) (b) \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{D}_{2} \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{D}_{2}\) (g) \(+\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) (c) \(\mathrm{N}_{2}(\mathrm{~g})+\mathrm{NO}_{2}\) (g) \(\rightarrow \mathrm{NO}\) (g) \(+\mathrm{N}_{2} \mathrm{O}\) (g) (d) \(2 \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})+3 \mathrm{O}_{2}\) (g) \(\longrightarrow 2 \mathrm{CO}_{2}\) (g) \(+4 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\)

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