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Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature. $$\mathrm{TiO}_{2}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \rightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{TiC}(\mathrm{s})$$ $$\begin{array}{lc} & \text { Free Energies of Formation at } \\ \text { Compound } & 727^{\circ} \mathrm{C}, \mathrm{kJ} / \mathrm{mol} \\ \hline \mathrm{TiO}_{2}(\mathrm{s}) & -757.8 \\ \mathrm{TiC}(\mathrm{s}) & -162.6 \\\ \mathrm{CO}(\mathrm{g}) & -200.2 \end{array}$$ (a) Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) and \(K\) at \(727^{\circ} \mathrm{C}.\) (b) Is the reaction product-favored at equilibrium at this temperature? (c) How can the reactant or product concentrations be adjusted for the reaction to proceed at \(727^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
(a) \(\Delta_{\mathrm{r}} G^{\circ} = 194.8\,\text{kJ/mol}, \;\) \(K \approx 6.56\times10^{-11}\). (b) No, it's not product-favored. (c) Increase C or decrease \(\mathrm{CO}\).

Step by step solution

01

Write the Balanced Reaction Equation

The given reaction is \( \mathrm{TiO}_{2}( ext{s}) + 3 \mathrm{C}( ext{s}) \rightarrow 2 \mathrm{CO}( ext{g}) + \mathrm{TiC}( ext{s}) \). This equation is already balanced.
02

Calculate the Change in Gibbs Free Energy (\(\Delta_{\mathrm{r}} G^{\circ}\))

The change in Gibbs free energy for the reaction is calculated using \(\Delta_{\mathrm{r}} G^{\circ} = \sum \Delta_f G^{\circ}_{\text{products}} - \sum \Delta_f G^{\circ}_{\text{reactants}}\). Here, \(\Delta_f G^{\circ}_{\mathrm{TiC}(\text{s})} = -162.6\,\text{kJ/mol}\), \(\Delta_f G^{\circ}_{2\mathrm{CO}(\text{g})} = 2(-200.2)\,\text{kJ/mol}\), and \(\Delta_f G^{\circ}_{\mathrm{TiO}_2(\text{s})} = -757.8\,\text{kJ/mol}\). Carbon (C) in its elemental form has \(\Delta_f G^{\circ} = 0\).
03

Substitute Values to Compute \(\Delta_{\mathrm{r}} G^{\circ}\)

Substituting the values into the formula gives:\[\Delta_{\mathrm{r}} G^{\circ} = [-162.6 + 2(-200.2)] - [-757.8] = [-162.6 - 400.4] + 757.8 = 194.8 \, \text{kJ/mol}\].
04

Calculate the Equilibrium Constant (\(K\))

The relationship between \(\Delta_{\mathrm{r}} G^{\circ}\) and the equilibrium constant \(K\) is given by the equation \(\Delta_{\mathrm{r}} G^{\circ} = -RT\ln K\). Here \( R = 8.314 \, \text{J/(mol K)}\) and the temperature in Kelvin is \(727 + 273 = 1000\,\text{K}\). Solve for \(K\) using the formula \(K = e^{-\Delta_{\mathrm{r}} G^{\circ}/RT}\).
05

Solve for \(K\)

Convert \(194.8 \, \text{kJ/mol}\) to \(194800 \, \text{J/mol}\) for consistency with \(R\). Calculate \(K\):\[K = e^{-\frac{194800}{(8.314)(1000)}} \] Using a calculator, \(K \approx e^{-23.43} \approx 6.56\times10^{-11}\).
06

Assess Equilibrium Favorability

Since \(K < 1\), the reaction is not product-favored at equilibrium because a small \(K\) indicates reactants are favored.
07

Adjusting Concentrations to Favor Forward Reaction

To proceed forward at \(727^{\circ}\text{C}\), increase the concentration of carbon (C) or reduce the concentration of \(\mathrm{CO}\). Increasing reactants or removing products shifts the equilibrium to the right according to Le Chatelier's Principle.

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Key Concepts

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

Gibbs Free Energy
Gibbs Free Energy, often symbolized as \( G \), is a key concept in thermodynamics that helps determine the spontaneity of a chemical reaction. The change in Gibbs Free Energy, \( \Delta G \), indicates if a reaction will proceed under constant temperature and pressure. If \( \Delta G \) is negative, the reaction occurs spontaneously, meaning it can proceed without any additional energy input.In our exercise, calculating \( \Delta_{\mathrm{r}} G^{\circ} \) involves using the equation:
  • \( \Delta_{\mathrm{r}} G^{\circ} = \sum \Delta_f G^{\circ}_{\text{products}} - \sum \Delta_f G^{\circ}_{\text{reactants}} \)
For the reaction \( \mathrm{TiO}_{2} + 3\mathrm{C} \rightarrow 2\mathrm{CO} + \mathrm{TiC} \), plugging in the given formation energies results in a positive \( \Delta_{\mathrm{r}} G^{\circ} \) of 194.8 kJ/mol.Since \( \Delta_{\mathrm{r}} G^{\circ} \) is positive, this reaction is not spontaneous under standard conditions, implying it is not product-favored.
Equilibrium Constant
The equilibrium constant, \( K \), provides a snapshot of the ratio between products and reactants in a chemical reaction at equilibrium. It gives insight into the position of equilibrium and the extent of the reaction. The relationship between \( \Delta_{\mathrm{r}} G^{\circ} \) and \( K \) is defined by the formula:
  • \( \Delta_{\mathrm{r}} G^{\circ} = -RT\ln K \)
Where \( R \) is the ideal gas constant and \( T \) is temperature in Kelvin. A calculated \( K \) value of \( 6.56 \times 10^{-11} \) in the exercise indicates a reaction highly favoring reactants.When \( K \, << \, 1 \), it means that at equilibrium, the concentration of reactants is much larger than that of products, showing limited reaction progress toward product formation.
Le Chatelier's Principle
Le Chatelier's Principle provides insight on how systems at equilibrium respond to changes, such as in concentration, temperature, or pressure. It predicts that an equilibrium will adjust to counteract changes imposed on it, thereby minimizing the effect of the disturbance.In the given reaction at \( 727^{\circ} \text{C} \), to make the reaction favor the forward direction, we can apply Le Chatelier’s Principle by:
  • Increasing the concentration of reactants, like adding more carbon
  • Decreasing the concentration of products, like removing some \( \mathrm{CO} \)
These adjustments shift the reaction toward forming more products, altering the balance and favoring the forward reaction under constant temperature conditions.
Thermodynamics
Thermodynamics is the study of energy changes accompanying chemical reactions. Within this field, studying reactions like \( \mathrm{TiO}_{2} + 3\mathrm{C} \rightarrow 2\mathrm{CO} + \mathrm{TiC} \) helps us comprehend how energy is transformed and conserved.Key terms often examined include:
  • System: the part of the universe we're focusing on, in this case, our chemical reaction
  • Surroundings: everything outside the system
  • Entropy \( S \): measures randomness or disorder
  • Enthalpy \( H \): represents heat content
  • Gibbs Free Energy \( G \): what we use to predict reaction favorability
Understanding thermodynamics allows us to predict how this reaction behaves at different temperatures, pressures, and concentrations, providing crucial insight into both natural processes and industrial applications.

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Most popular questions from this chapter

Determine if each of the following statements is true or false (a) For a reaction that can yield more than one product, the most stable product is always formed in the greatest amount. (b) For a reaction that can yield more than one product, the pathway that has the lowest activation energy always occurs more quickly. (c) Reactions that are thermodynamically favored are always fast. (d) Diamond is always more stable than graphite.

Predict whether each of the following processes results in an increase in entropy in the system. (Define reactants and products as the system.) (a) Water vapor condenses to liquid water at \(90^{\circ} \mathrm{C}\) and 1 atm pressure. (b) The exothermic reaction of \(\mathrm{Na}(\mathrm{s})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\) forms \(\mathrm{NaCl}(\mathrm{s})\) (c) The endothermic reaction of \(\mathrm{H}_{2}\) and \(\mathrm{I}_{2}\) produces an equilibrium mixture of \(\mathrm{H}_{2}(\mathrm{g})\) \(\mathrm{I}_{2}(\mathrm{g}),\) and \(\mathrm{HI}(\mathrm{g})\) (d) Solid NaCl dissolves in water forming a saturated solution.

Indicate which of the following processes are reversible. (a) Nitrogen and oxygen gases diffuse to give a homogeneous mixture. (b) Ice sublimes at \(-5^{\circ} \mathrm{C}\) and 1.0 atm. (c) Energy as heat is transferred to the surroundings from a mixture of ice and water at \(0^{\circ} \mathrm{C}\) causing more ice to form. (d) Bromine evaporates and the gaseous molecules diffuse into the atmosphere.

Which substance has the higher entropy? (a) dry ice (solid \(\mathrm{CO}_{2}\) ) at \(-78^{\circ} \mathrm{C}\) or \(\mathrm{CO}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C}\) (b) liquid water at \(25^{\circ} \mathrm{C}\) or liquid water at \(50^{\circ} \mathrm{C}\) (c) pure alumina, \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}),\) or ruby (ruby is \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in which some \(\mathrm{Al}^{3+}\) ions in the crystalline lattice are replaced with \(\mathrm{Cr}^{3+}\) ions) (d) one mole of \(\mathrm{N}_{2}(\mathrm{g})\) at 1 bar pressure or one mole of \(\mathrm{N}_{2}(\mathrm{g})\) at 10 bar pressure (both at \(298 \mathrm{K})\)

The ionization constant, \(K_{\mathrm{a}},\) for acetic acid is \(1.8 \times\) \(10^{-5}\) at \(25^{\circ} \mathrm{C} .\) What is the value of \(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}\) for this reaction? Is this reaction product- or reactantfavored at equilibrium?

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