Question

Consider the thermodynamics of calcination. (a) The following equation is a generic reaction representing calcination of a metal carbonate: $$\mathrm{MCO}_{3}(s) \longrightarrow \mathrm{MO}(s)+\mathrm{CO}_{2}(g)$$ Would you expect this reaction to become more or less spontaneous as the temperature increases? (b) What is the standard free energy, \(\Delta G^{\circ}\), for the reaction corresponding to the calcination of \(\mathrm{PbCO}_{3}\) (Equation 23.1) at \(25^{\circ} \mathrm{C}\) under standard conditions? Is this reaction spontaneous? If not, at what temperature does this reaction become spontaneous (assuming \(\Delta H^{\circ}\) and \(S^{\circ}\) values do not change with temperature)?

Short answer

The calcination reaction becomes more spontaneous as the temperature increases due to an increase in entropy. The standard free energy, ΔG°, for the calcination of PbCO₃ at 25°C is -751.3 kJ/mol, which indicates that the reaction is spontaneous at this temperature.

Step-by-step solution

Analyze the spontaneity of the reaction as the temperature increases

According to the given reaction, the calcination of a metal carbonate releases carbon dioxide as a gas. The formation of more gas particles leads to an increase in entropy (ΔS°). Since the entropy term in the equation is multiplied by the temperature (T), increasing the temperature will cause a larger effect on the ΔG° value. Recall that a reaction is considered spontaneous when ΔG° < 0. In our equation, ΔG° = ΔH° - TΔS°, if the value of TΔS° increases, then ΔG° will more likely be negative, making the reaction more spontaneous as the temperature increases. Therefore, with an increase in entropy (ΔS° > 0), we can expect this reaction to become more spontaneous as the temperature increases.

Identify the given values for ΔH° and S° for PbCO₃ calcination

For PbCO₃ calcination, the given values in the exercise for the reaction under standard conditions are: ΔH° = -699.5 kJ/mol ΔS° = 174.1 J/mol⋅K

Calculate the standard free energy, ΔG°, for PbCO₃ calcination at 25°C

To calculate ΔG° for the reaction at 25°C, we first need to convert the temperature to Kelvin: T = 25°C + 273.15 = 298.15 K Now, we can use the equation ΔG° = ΔH° - TΔS° to find the standard free energy: ΔG° = (-699.5 kJ/mol) - (298.15 K)(174.1 J/mol⋅K)(1 kJ/1000 J) ΔG° = -699.5 kJ/mol - 51.8 kJ/mol ΔG° = -751.3 kJ/mol

Determine if the reaction is spontaneous at 25°C

Since ΔG° is negative (-751.3 kJ/mol), the reaction is spontaneous at 25°C.

Find the temperature at which the reaction becomes spontaneous (if needed)

As we determined in Step 4, the PbCO₃ calcination reaction is already spontaneous at 25°C. Thus, we do not need to find the temperature at which the reaction becomes spontaneous. In conclusion, the PbCO₃ calcination reaction becomes more spontaneous as the temperature increases, and at 25°C, the reaction is spontaneous with a standard free energy of -751.3 kJ/mol.

Key concepts

Thermodynamics

Thermodynamics is the study of energy, its transformations, and its relation to matter. It helps us understand how and why reactions occur. In the context of chemical processes, thermodynamics focuses on concepts like enthalpy (\( \Delta H \)), entropy (\( \Delta S \)), and Gibbs free energy (\( \Delta G \)).
- **Enthalpy** (\( \Delta H \)): This is the total heat content of a system and indicates whether a reaction absorbs or releases heat. Reactions that release heat are exothermic (negative \( \Delta H \)), while those that absorb heat are endothermic (positive \( \Delta H \)).
- **Entropy** (\( \Delta S \)): Entropy measures the disorder or randomness in a system. Systems tend to favor higher entropy. A positive \( \Delta S \) means the system becomes more disordered.
- **Gibbs Free Energy** (\( \Delta G \)): Gibbs free energy combines enthalpy and entropy to predict reaction spontaneity. The equation is \( \Delta G = \Delta H - T\Delta S \), where \( T \) is the temperature in Kelvin.
A reaction is deemed spontaneous when \( \Delta G \lt 0 \), as it indicates the process is energetically favorable.

Calcination Reaction

Calcination is a thermal decomposition process. It's used extensively to purify ores or drive chemical transformations by heating a solid material. In calcination, thermal energy is provided to cause the desired chemical change.
- A typical calcination reaction involves a metal carbonate decomposing into a metal oxide and carbon dioxide gas, such as \( \text{MCO}_3 (s) \rightarrow \text{MO} (s) + \text{CO}_2 (g) \).
- The process occurs at high temperatures and is often designed to optimize conditions for decomposition. By releasing carbon dioxide, calcination contributes to increasing the entropy of the system, which is favorably impacted by rising temperatures.
Through calcination, the product often becomes more solid and easier to work with for subsequent processing steps. Understanding the thermal and chemical dynamics of calcination is essential in industries like metallurgy and materials science.

Standard Free Energy

Standard free energy change \( \Delta G^\circ \), helps predict the spontaneity of a chemical reaction under standard conditions (1 atm pressure and 25°C). This value reflects the potential work performed by the reaction.
- To determine \( \Delta G^\circ \) for a reaction, we use the formula \( \Delta G^\circ = \Delta H^\circ - T\Delta S^\circ \). The equation factors in enthalpy, entropy, and temperature.
- A negative \( \Delta G^\circ \) indicates that the reaction proceeds spontaneously under standard conditions. Conversely, a positive value suggests the reaction is non-spontaneous unless conditions change, like temperature.
In the example of \( \text{PbCO}_3 \) calcination, \( \Delta G^\circ \) at 25°C is negative, suggesting the reaction naturally occurs at standard temperature. Understanding \( \Delta G^\circ \) is crucial in industrial and laboratory settings for designing efficient chemical processes.

Entropy Changes

Entropy change (\( \Delta S \)) is an important aspect of thermodynamics that refers to changes in the disorder of a system. In chemical reactions, entropy change is a key factor in determining spontaneity.
- A positive \( \Delta S \) indicates that a system becomes more disordered, like when a solid turns into gas. This often favors spontaneity at higher temperatures due to the \( T\Delta S \) term in the Gibbs free energy equation.
- In the calcination reaction, as \( \text{CO}_2 \) gas forms, \( \Delta S \gt 0 \) because gases are highly disordered compared to solids. Therefore, the entropy increase contributes significantly to making the reaction favorable, especially as the temperature rises.
As a system moves towards greater entropy, its free energy decreases, making it more stable and energetically favorable. This understanding is crucial for anticipating the behavior of reactions under various conditions.