Problem 63 - Chemistry and Chemical Reactivity

Question

Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature. $$\mathrm{TiO}_{2}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \rightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{TiC}(\mathrm{s})$$ $$\begin{array}{lc} & \text { Free Energies of Formation } \\ \text { Compound } & \text { at } 727^{\circ} \mathrm{C}, \mathrm{kJ} / \mathrm{mol} \\ \hline \mathrm{TiO}_{2}(\mathrm{s}) & -757.8 \\ \mathrm{TiC}(\mathrm{s}) & -162.6 \\ \mathrm{CO}(\mathrm{g}) & -200.2 \\ \hline \end{array}$$ (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 be spontaneous at \(727^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) \( \Delta_r G^{\circ} = 194.8 \, \text{kJ/mol}, K \approx 7.25 \times 10^{-11} \). (b) Not product-favored at equilibrium. (c) Increase reactants or decrease products.

Step 1: Write the Reaction Equation

The chemical reaction given is \( \mathrm{TiO}_{2}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \rightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{TiC}(\mathrm{s}) \). Write this down to reference the reactants and products for calculations.

Step 2: Determine the Gibbs Free Energy Change Formula

The formula for the Gibbs free energy change of a reaction is \( \Delta_{r} G^{\circ} = \sum \Delta_{f} G^{\circ}_{\text{products}} - \sum \Delta_{f} G^{\circ}_{\text{reactants}} \). Use this to compute \( \Delta_{r} G^{\circ} \).

Step 3: Substitute Values for Free Energies

Substitute the given Gibbs free energies of formation: \( \Delta_{f} G^{\circ}_{\mathrm{TiO}_{2}} = -757.8 \, \text{kJ/mol} \), \( \Delta_{f} G^{\circ}_{\mathrm{TiC}} = -162.6 \, \text{kJ/mol} \), \( \Delta_{f} G^{\circ}_{\mathrm{CO}} = -200.2 \, \text{kJ/mol} \).

Step 4: Calculate the \( \Delta_{r} G^{\circ} \)

Calculate the \( \Delta_{r} G^{\circ} \) of the reaction:\[ \Delta_{r} G^{\circ} = \left((2 \times -200.2) + (-162.6) \right) -(-757.8) \]\[ \Delta_{r} G^{\circ} = (-400.4 - 162.6) + 757.8 = 194.8 \, \text{kJ/mol} \]

Step 5: Relate \( \Delta_{r} G^{\circ} \) to Equilibrium Constant \( K \)

Use the relation \( \Delta_{r} G^{\circ} = -RT \ln K \). Rearrange to express \( K \):\[ \ln K = -\frac{\Delta_{r} G^{\circ}}{RT} \]Convert \(727^{\circ} C\) to Kelvin which is \(1000 K\). Use \(R = 8.314 \times 10^{-3} \, \text{kJ/mol·K}\).

Step 6: Calculate \( K \)

Substitute the known values into the equation:\[ \ln K = -\frac{194.8}{(8.314 \times 10^{-3} \times 1000)} \]\[ \ln K \approx -23.42 \]\[ K \approx e^{-23.42} \approx 7.25 \times 10^{-11} \]

Step 7: Analyze Reaction Favorability

Since \( K \) is very small (\(7.25 \times 10^{-11}\)), the reaction is reactant-favored at equilibrium at \( 727^{\circ} C \).

Step 8: Determine Condition for Spontaneity

For the reaction to be spontaneous, \( \Delta G \) should be negative. This can occur by either increasing the reactant concentration or decreasing the product concentration to drive the reaction forward.

Key Concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a crucial concept in understanding how a reaction behaves under equilibrium conditions. It gives us insights into the extent of a reaction at a given temperature.
When \( K \) is calculated, it tells us the ratio of the concentrations of the products to the reactants when the reaction is at equilibrium. In our example reaction, the equilibrium constant is determined by the formula:

• \( \Delta_{r} G^{\circ} = -RT \ln K \)

Rearranging gives:

• \( \ln K = -\frac{\Delta_{r} G^{\circ}}{RT} \)

where \( R \) is the universal gas constant and \( T \) is the temperature in Kelvin. After performing the calculation in the solution, we found \( K \) to be \( 7.25 \times 10^{-11} \), which indicates a strong tendency towards the reactants under the given conditions.
A small \( K \) value like this suggests that only a tiny amount of product forms; hence, the reaction is heavily reactant-favored. This has important implications for how the reaction is managed in practical settings.

Enthalpy and Entropy

Understanding enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) is key to comprehending the Gibbs Free Energy and, hence, the spontaneity of reactions.
Enthalpy refers to the heat content in a system. During a chemical reaction like the conversion from titanium(IV) oxide to titanium carbide, changes in total energy occur—a gain or loss that can drive reactions forward.
Entropy, on the other hand, represents the disorder or randomness in a system. Reactions often increase entropy, moving towards more disordered states. This alignment with the natural tendency towards disorder influences whether a reaction spontaneously proceeds.
The combination of these factors is captured in Gibbs Free Energy:

• \( \Delta G = \Delta H - T \Delta S \)

where \( T \) is the temperature in Kelvin. Each term plays a role:

• \( \Delta H \) represents energy exchange.
• \( T \Delta S \) reflects changes in disorder.

Together, they determine \( \Delta G \), giving insight into the overall favorability of a reaction.

Reaction Spontaneity

Reaction spontaneity is determined by the sign of the Gibbs Free Energy change (\( \Delta G \)). A negative \( \Delta G \) indicates that a reaction proceeds spontaneously, aligning with nature's tendency towards lesser free energy states.
In the context of our chemical reaction from the exercise, initially we found \( \Delta_{r} G^{\circ} = 194.8 \, \text{kJ/mol} \), which is positive, indicating non-spontaneity under the stated conditions. This means the reaction doesn't naturally create products without further adjustments.
This is where manipulating concentrations can play a role. By increasing the amounts of reactants, or decreasing the concentration of products, we can shift the position of equilibrium to make the reaction more favorable.
This adjustment affects \( \Delta G \) by making it more negative, thus achieving a spontaneous reaction. In practice, understanding these relationships allows chemists to control reactions effectively, ensuring desired outcomes in industrial or laboratory environments.